CBSE Sample Papers – Page 48

CBSE Sample Papers for Class 10 Maths Basic Set 5 with Solutions

October 27, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 5 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A, B, C, D, and E.
Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
Section C has 6 Short Answer-ll (SA-II) type questions carrying 3 marks each.
Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section
Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section – A (20 marks)
(Section – A consists of 20 questions of 1 mark each.)

Question 1.
The H.C.F, of the smallest prime number and the smallest composite number is: [1]
(a) 1
(b) 0
(c) 4
(d) 2
Answer:
(d) 2

Explanation:
Smallest prime number = 2;
smallest composite number = 4
∴ HCF (2. 4) = 2.

Question 2.
A quadratic equation, sum of whose roots is – 3√2 and their product is 4 is: [1]
(a) x² + 4x = 0
(b) x² + 4√2 x + 3 = 0
(c) x² + 3√2x + 4 = 0
(d) x² + 3√2x – 4 = 0
Answer:
(c) x² + 3√2x + 4 = 0

Explanation:
A quadratic equation with sum and product of roots as S and P. respectively, is given as x² – Sx + P = 0 So, a quadratic equation in x whose sum of roots is – 3√2 and product of roots is 4,
i.e. x² + 3√2 + 4 = 0

Question 3.
An umbrella has eight evenly spaced ribs (see figure). The space between the umbrella’s two consecutive ribs using the assumption that the umbrella is a flat circle with a radius of 45 cm is: [1]

(a) $\frac{22275}{28}$ cm²
(b) $\frac{23456}{28}$ cm²
(c) 5923 cm²
(d) 2986 cm²
Answer:
(a) $\frac{22275}{28}$

Explanation:
Here, r = 45 cm
and θ = $\frac{360^{\circ}}{8}$ = 45°
Area between two consecutive ribs of the umbrella
= $\frac{\theta}{360^{\circ}}$ × πr²
= $\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7}$ × 45 × 45 = $\frac{22275}{28}$ cm²

Question 4.
From the adjoining figure of a rectangle, the values of x and y is: [1]

(a) 22, 8
(b) 30, 8
(c) 20,6
(d) 25,4
Answer:
(a) 22, 8

Explanation:
Since, the figure given is a rectangle,
∴ x + y = 30; x – y = 14
Solving the two equations, we get
x = 22; y = 8.

Question 5.
A point which divides the join of A (-3, 4) and B (9, 6) internally in the ratio 3 : 2 is: [1]
(a) $\frac{15}{2}, \frac{-16}{3}$
(b) $\frac{21}{5}, \frac{26}{5}$
(c) 0, 0
(d) $\frac{-20}{3}, \frac{25}{3}$
Answer:
(b) $\frac{21}{5}, \frac{26}{5}$

Explanation:
Let P(x, y) be the required point.

Question 6.
If the difference of the roots of the quadratic equation x² – 7x + 2k = 0 is 1, then the value of k is: [1]
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(c) 6

Explanation:
Let α and β be the roots of the given quadratic equation.
Then, α + β = – $\frac{(-7)}{1}$ = 7 ………… (i)
and αβ = $\frac{2 k}{1}$ = 2k …(ii)
It is given that,
α – β = 1
⇒ (α – β)² = 1
⇒ (α – β)² – 4αβ = 1
⇒ (7)² – 4 × 2k = 1 [Using (i) and (ii)]
⇒ 49 – 8k = 1
⇒ – 8k = – 48
⇒ k = 6

Question 7.
If the distance between the points P(2, – 3) and Q(10, y) is 10 units, then the value of ‘y’ is: [1]
(a) -3, 6
(b) 21, 26
(c) 4, 3
(d) – 9, 3
Answer:
(d) – 9, 3

Explanation:
Here,
Distance = $\sqrt{(10-2)^2+(y+3)^2}$ = 10
Squaring both sides, we get
64 + y² + 9 + 6y = 100
⇒ y² + 6y – 27 = 0
⇒ y² + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9)(y – 3) = 0
⇒ y + 9 = 0, y-3 = 0
⇒ y = – 9, 3.

Question 8.
The value of ‘k’ for which the linear equations 3x – y + 8 = 0 and 6x – ky + 16 = 0 represent coincident lines is: [1]
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Explanation:
The given equations represent coincident lines, when
$\frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}$
i.e. when k = 2.

Question 9.
A lighthouse projects a red light over an area of angle 80 at a distance of 16.5 kilometres to alert sailors to the presence of submerged rocks. The area of the sea over which the ships are warned is: [1]
(a) 128.87 km²
(b) 156.87 km²
(c) 5923 km
(d) 189.97 km²
Answer:
(d) 189.97 km²

Explanation:
Here, r = 16.5 km and q = 80°
The area of sea over which the ships are warned
= $\frac{\theta}{360^{\circ}}$ × πr²
= $\frac{80^{\circ}}{360^{\circ}}$ × 3.14 × 16.5 × 16.5
= 189.97 km²

Question 10.
If the curved surface area of a sphere is 4π sq m., then the diameter of the sphere is: [1]
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm
Answer:
(a) 2 cm

Explanation:
Let the radius of the sphere be ‘r ’ cm. Then,
CSA = 4πr²
⇒ 4πr² = 4π
⇒ r² = 1
⇒ r = ± 1
⇒ r = 1 [∵ r cannot be negative]
⇒ Diameter of the sphere is 2 cm.

Question 11.
A ∆ABC is right angled at C, then the value of cos (A + B) is: [1]
(a) 1
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 0
Answer:
(d) 0

Explanation:
In ∆A8C,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B = 180° – ∠C
= 180° – 90°
= 90°
So, cos (A + B) = cos 90° = 0

Question 12.
What is surface area of the resultant cuboid, obtained on joining 2 identical cubes each of edge 2 cm? [1]
(a) 42 sq. cm
(b) 40 sq. cm
(c) 30 sq. cm
(d) 35 sq. cm
Answer:
(b) 40 sq. cm

Explanation:
The dimensions of the resulting cuboid are 4 cm × 2 cm × 2 cm.

So, its total surface area
= 2 (lb + bh + lh)
= 2(8 + 4 + 8)
= 40 sq. cm.

Question 13.
A letter is drawn at random from the letters of the word ERROR. What is probability that the drawn letter is R ? [1]
(a) $\frac{1}{5}$
(b) $\frac{2}{5}$
(c) $\frac{3}{5}$
(d) $\frac{4}{5}$
Answer:
(c) $\frac{3}{5}$

Explanation:
In the given word ‘ERROR’, R occurs three times.
So, P(R) = $\frac{3}{5}$

Question 14.
If cos (A + B) = 0 and sin (A – B) = $\frac{\sqrt{3}}{2}$, then the value of A is: [1]
(a) 30°
(b) 60°
(c) 75°
(d) 80°
Answer:
(c) 75°

Explanation:
cos(A + B) = 0 ⇒ A + B = 90°
sin(A – B) = $\frac{\sqrt{3}}{2}$ ⇒ A – B = 60°
Solving the above equation we get
A = 75°

Question 15.
In ∆ ABC, right angled at B, if tan A = $\frac{1}{\sqrt{3}}$ then the value of (sin A cos C + cos A sin C) is: [1]
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Explanation:
Given:
tan A = $\frac{1}{\sqrt{3}}$
⇒ A = 30°
⇒ C = 60° [∵ ∠B = 90°]
(sin A cos C + cos A sin C)
= sin 30° cos 60° + cos 30° sin 60°

Question 16.
If a fair dice is thrown once, the probability of getting a number which is even as well as prime is: [1]
(a) $\frac{2}{5}$
(b) $\frac{3}{4}$
(c) $\frac{1}{6}$
(d) $\frac{2}{5}$
Answer:
(c) $\frac{1}{6}$

Explanation:
Since 2 is the only number which is even as well as prime,
∴ Required probability is $\frac{1}{6}$

Question 17.
In the given figure PS is the bisector of ∠QPR of ∆PQR. If PQ = 15, PR = 7, QS = 3 + x and SR = x – 3, the value of x is: [1]

(a) $\frac{3}{2}$
(b) 4
(c) $\frac{5}{2}$
(d) $\frac{33}{4}$
Answer:
(d) $\frac{33}{4}$

Explanation:
Since, PS is the bisector of ∠P
∴ By angle-bisector theorem.
⇒ $\frac{P Q}{O S}=\frac{P R}{R S}$
⇒ $\frac{15}{3+x}=\frac{7}{x-3}$
⇒ 15(x – 3) = 7(3 + x)
⇒ 15x – 45 = 21 + 7x
⇒ 8x = 66
⇒ x = $\frac{66}{8}$ = $\frac{33}{4}$

Question 18.
If 6^th term and 8^th term of an A.P. are 12 and 22 respectively, then its 2^nd term is: [1]
(a) – 8
(b) + 8
(c) 0
(d) 1
Answer:
(a) – 8

Explanation:
Given, as = 12 and a8 = 22
Let, the first term of A.P. be ‘a’ and common difference be ‘d
Then,
a + 5d = 12 …(i)
a + 7d = 22 …(ii)
On solving equations (i) & (ii), we get
d = 5 and a = – 13
Then, a₂ = a + d
= – 13 + 5
= – 8

Direction for questions 19 and 20: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Assertion (A) : The values of k for which the quadratic equation 2x² + kx + 2 = 0 has equal roots, is ± 2.
Reason (R) : If roots are equal, the discriminant is 0. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
The equation 2x² + kx + 2 = 0 has equal roots,
When, D = (k)² – 4 (2) (2) = 0
k² = 16
k = ± 4

Question 20.
Assertion (A) : In the given figure, if ∠AOB = 125°, then ∠COD is equal to 55°

Reason (R): Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angle of the centre of circle. [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
In the given figure, ABCD is a quadrilateral circumscribing a circle.
We know that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∴ ∠AOB + ∠COD = 180°
125° + ∠COD = 180°
∠COD = 180° – 125°
= 55°

Section – B (10 marks)
(Section – B consists of 5 questions of 2 marks each.)

Question 21.
Using prime factorisation, find the LCM of 90 and 120. [2]
OR
Explain why 3 × 5 × 7 × 11 + 11 is a composite number.
Answer:
Here, the prime factorisations of 90 and 120 are:
90 = 2 × 3 × 3 × 5,
or = 2¹ × 3² × 5¹
and 120 = 2 × 2 × 2 × 3 × 5,
or = 2³ × 3¹ × 5¹
So, LCM (90, 120) = 23 × 32 × 51, i.e. 360.

(3 × 5 × 7 × 11 + 11) = 11(3 × 5 × 7 × 1 + 1)
= 11 (105 + 1)
= 11 (106)
= 11 × 2 × 53
Since (3 × 5 × 7 × 11 + 11) has more than one factor, so the given number is composite.

Question 22.
Using the quadratic formula, find the roots of the quadratic equation: x² + x – 12 = 0. [2]
Answer:
Using the quadratic formula to the equation:
x² + x – 12 = 0,

Thus, x = – 4 and x = 3 are the two roots of x² + x – 12 = 0.

Question 23.
if tan A = $\frac{7}{24}$, find the value of sin A cos A.
OR
Prove that: sin² A + sin² A tan² A = tan² A. [2]
Answer:
Given,
tan A = $\frac{7}{24}$,
Let AB = 24 K and BC = 7 K
Using Pythagoras theorem is AABC, we get
⇒ AC = 25 K
Thus, sin A = $\frac{B C}{A C}=\frac{7}{25}$

and cos A = $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25}$
Thus, sin A cos A = $\frac{7}{25} \times \frac{24}{25}=\frac{168}{625}$

LH.S. = sin² A + sin² A tan² A
= sin² A (1 + tan² A)
= sin² A sec² A
[∵ sec² A – tan² A = 1]
= sin² A × $\frac{1}{\cos ^2 A}$
= tan² A = R.H.S.

Question 24.
In the given figure, ABP and ACQ are two tangents to a circle with centre O. If ∠TBP = 50° and ∠TCQ = 60°, then find the measure of ∠BTC. [2]

Answer:

Join OB, OT and OC.
We know, tangent is perpendicular to radius at the point of contact.
∴ OB ⊥ AP and OC⊥ AQ
∴ ∠O BP = 90°
⇒ ∠OBT + ∠TBP = 90°
⇒ ∠OBT + 50° = 90°
⇒ ∠OBT = 90° – 50° = 40°
Similarly, ∠OCQ = 90° and ∠TCQ = 60°
∴ ∠OCT = 30°
Now, in ∆OBT
OB = OT [Radii]
∴ ∠OTB = ∠OBT
[Equal angles opposite to equal sides]
⇒ ∠OTB =40°
Similarly, in ∆OTC
OT = OC
⇒ ∠OCT = ∠OTC = 30°
So, ∠BTC = ∠OTB + ∠OTC
= 40° + 30° = 70°

Question 25.
The perimeter of a sheet of paper which is in the shape of a quadrant of a circle, is 75 cm. Find its area. [2]

Answer:
Here, let OB = OA = r cm. Then,
Perimeter of quadrant = OA + $\widehat{A B}$ + BO
= r + $\frac{90^{\circ}}{360^{\circ}}$ × 2πr + r
= 2r + $\frac{\pi r}{2}$ = r$\left(2+\frac{\pi}{2}\right)$
Equating it to 75 cm, we have
r$\left(2+\frac{\pi}{2}\right)$ = 75
⇒ r = $\frac{2 \times 75}{4+\pi}$
= $\frac{2 \times 75 \times 7}{28+22}$ = $\frac{2 \times 75 \times 7}{50}$
= 21 cm.
Thus, area of the quadrant = $\frac{\pi}{4}$ (r)²
= $\frac{\pi}{4}$ (21)² = 346.5 sq cm.

Section – C (18 marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
If sin θ + cos θ = √3, prove that tan θ + cot θ = 1. [3]
OR
Evaluate: $\frac{\sin 30^{\circ}+\tan 45^{\circ}-{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
Answer:
Given: sin θ + cos θ = √3
squaring both sides, we get
(sin θ + cos θ)² = (√3)²
⇒ sin² θ + cos² θ + 2 sin θ cos θ = 3
⇒ 1 + 2 sin θ cos θ = 3 [∵ sin² θ + cos² θ = 1]
⇒ sin θ cos θ = 1 ….(i)
Now, tan θ + cot θ = $\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$
= $\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}$
= $\frac{1}{1}$
= 1

Question 27.
A computer animation below shows a cat moving in a straight line.

Its height, h metres, above the ground, is given by 3s – 3h = – 9, where s is the time in seconds after it starts moving. In the same animation, a mouse starts to move at the same time as the cat and its movement is given by – 3s + h = 1.
(A) Draw the graph of the two equations on the same sheet of graph paper; [1]
(B) Will the mouse be able to catch the cat? [1]
(C) If yes, after how much time and at what height? [1]
Answer:
(A)

(B) Yes, if find the several values of the variables s and h for cat as well as mouse, then the same values of s and h show their intersection point. It mean that the cat will definitely catch the mouse.

(C) As mentioned in above statement, the intersection point defines their time and height. Hence, after 1 second at a height of 4 m, the cat will catch the mouse.

Question 28.
If the points A (1, – 2), B (2,3), C (a, 2) and D (-4, -3) from a parallelogram, find the value of a. [3]
Answer:
We know that the two diagonals AC and BD of a parallelogram ABCD bisect each other.
So, mid-point of AC = mid-point of BD = P.

⇒ a = – 3.

Question 29.
In the figure, all three sides of a triangle ABC touch the circle at points P, Q and R. Find the value of x. [3]

Answer:
From the figure,
AR = AQ, BQ = BP, CP = CR
⇒ BQ = 10 cm and hence
AQ = (18 – 10) cm, i.e. 8 cm.
Also, CR = 6 cm
Thus, AC = x = AR + CR = AQ + CR
= (8 cm + 6 cm) = 14 cm.

Question 30.
Find the value of x for which DE || AB in the figure given below: [3]

Answer:
Since, DE is parallel to AB.
∴ By BPT, we have,
$\frac{\mathrm{CD}}{\mathrm{DA}}=\frac{\mathrm{CE}}{\mathrm{EB}}$
⇒ $\frac{x+3}{3 x+19}=\frac{x}{3 x+4}$
⇒ (x + 3)(3x + 4) = x(3x + 19)
⇒ 3x² + 13x + 12 = 3x² + 19x
⇒ 6x = 12
or x = 2

Question 31.
The product of the LCM and HCF of two numbers is 24. If the difference of the two number is 2, find the numbers.
OR
How many spherical lead shots each 4.2 cm in diameter can be obtained from a rectangular solid of lead having dimensions 66 cm × 42 cm × 21 cm. [3]
Answer:
Let the two numbers be a and b. Then,
a × b = HCF (a, b) × LCM (a, b) = 24 …….. (i)
Also, a – b = 2 …….. (ii)
From equation (i) and (ii), we have
a – $\frac{24}{a}$ = 2
⇒ a² – 24 = 2a
⇒ a² – 2a – 24 = 0
⇒ a² – (6 – 4) a – 24 = 0
⇒ a – 6a + 4a – 24 = 0
⇒ a (a – 6) + 4 (a – 6) = 0
⇒ (a – 6) (a + 4) = 0
⇒ a = 6, – 4
But a cannot be negative
a = 6
So, b = $\frac{24}{a}$ = $\frac{24}{6}$ = 4

Volume of lead, obtained on melting the rectangular solid = (66 × 42 × 21) cu. cm
Volume of one spherical lead shot
= $\frac{4}{3}$π(2.1)³ cu.m
So, Number of spherical lead shots that can be obtained
= $\frac{\text { Volume of the rectangular solid }}{\text { Volume of } 1 \text { spherical lead shot }}$
= $\frac{66 \times 42 \times 21}{\frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10}}$ = 1500

Section – D (20 marks)
(Section – D consists of 4 questions of 5 marks each.)

Question 32.
The first term of an A.P. is 5, the last term is 45 and the sum of its all terms is 400. Find the number of terms of the A.P. and also the common difference. [5]
Answer:
Let ‘a’ be the first term and ‘d be the common difference of AP.
Then, a = 5
Here, last term (l) = 45
and Sum of all terms = 400
Let the A.P. contains ‘n’ terms. Then
S_n = $\frac{n}{2}$(a + l)
⇒ $\frac{n}{2}$(5 + 45) = 400
⇒ n = 16
As last term is the n^th term, we have
a + (n – 1)d = 1
⇒ 5 + (16 – 1)d = 45
⇒ d = $\frac{40}{15}$ or $\frac{8}{3}$
Thus, n = 16 and d = $\frac{8}{3}$

Question 33.
A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 30 metres from its root. Find the whole height of the tree [Use √3 = 1.732]. [5]
OR
In the figure, RQ ⊥ PQ, PQ ⊥ PT and ST ⊥ PR. Prove that: ST × QR = PS × PQ.

Answer:
Let the tree AB is broken by the wind at P.
Then, PX = PB
Let ∠PXA = θ = 30°
Let PA = x and PB = h
From the figure, in ∆AXP,

⇒ h = $\frac{60}{\sqrt{3}}$ = 20√3 m
Thus, the totat height of the tree = x + h
= (1o√3 + 20√3) m
= (30√3) m
= 51.96 m.

Consider ∆s PST and RQP, we have
∠PST = ∠RQP (90° each)
∠TPS = ∠PRQ (alternate angles)
⇒ By AA similarity criterion,
∆PST ~ ∆RQP
Thus, $\frac{\mathrm{PS}}{\mathrm{ST}}=\frac{\mathrm{QR}}{\mathrm{PQ}}$
or ST × QR = PS × PQ

Question 34.
Sumit arranges to pay off a debt of ₹ 3,60,000 by 40 installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one – third of the debt unpaid. Determine the first installment’s value.
OR
Calculate the median for the following data: [5]

Class	Frequency
More than or equal to 150	0
More than or equal to 140	12
More than or equal to 130	27
More than or equal to 120	60
More than or equal to 110	105
More than or equal to 100	124
More than or equal to 90	141
More than or equal to 80	150

Answer:
Let, the values of first installment be ₹ a.
The monthly installments form an AP, so let us suppose the man increases the value of each installment by ₹ d every month.
∴ The common difference of arithmetic series
= ₹ (- d)
Amount paid in 30 installments
= ₹ 36,000 – $\frac{1}{3}$ × 36,000
= ₹ 24,000
Let S_n denotes the total amount of money paid in the n installments.
Then, S₃₀ = ₹ 24,000
⇒ $\frac{30}{2}$ [2a + (30 – 1 )d] = 24,000
⇒ 15[2a + 29d] = 24,000
⇒ 2a + 29 d = 1600 ………… (i)
Also, S₄₀ – ₹ 36,000
$\frac{40}{2}$ [2a + (40 – 1 )d] = 36,000
⇒ 20[2a + 39 d] = 36,000
⇒ 2a + 39d = 1800 ………… (ii)
Applying (ii) – (i), we get
(2a + 39d) – (2a + 29d)= 1800 – 1600
⇒ 10d = 200
⇒ d = 20
Put d = 20, in (i), we get
2a + 29 × 20 = 1600
⇒ a + 580 = 1600
⇒ 2a = 1020
a = 510
Hence, the value of first installment is ₹ 510.

Converting the given distribution into continuous distribution

Here, N = 150, $\frac{\mathrm{N}}{2}$ = 75
∴ Median class is 110 – 120
Here, l = 110; cf = 45, f = 45, h = 10
Median = l + $\frac{\left(\frac{N}{2}-c f\right)}{f}$ × h
= 110 + $\frac{(75-45)}{45}$ × 10
= 110 + $\frac{300}{45}$
= 110 + 6.67
= 116.67 (approx)

Question 35.
Compute the mean and mode of the following data: [5]

Answer:
Calculation of Mode:
Here, the modal class is 45 – 55
For this class,
l = 45, h = 10, f = 33, f₀ = 31, f₂ = 17, h = 10

Calculation of Mean:

Section – E (12 marks)
(Case Study Based Questions)
(Section – E consists of 3 questions. All are compulsory.)

Question 36.
Uttar Bantra Sarbojanin Durgotsav Committee had started planning for their Durga puja a year in advance with a mega budget in mind.

Bholeram Tents is given a contract by the municipal corporation of Budaun (Uttar Pradesh), India to setup a mega function pandal (tent). The architect has designed a tent of height 7.7 m in the form of a right circular cylinder of diameter 36 m ana height 4.4 m surmounted by a right circular cone. This tent is setup in a rectangular park of dimensions 70 m × 60 rn as shown below.
The tent is made of canvas. (Take π = 3.14)

On the basis of the above information, answer the following questions:
(A) For the workers to finalise the purchase of material find the height of the conical, part. [1]
(B) Find the slant height of the conical part. [1]
(C) To purchase the canvas, what is the area of the canvas to be used approx in making the tent.
OR
Find the cost of canvas at ₹ 4.50 sq m and the area of the rectangular park outside the tent? [2]
Answer:
(A) Height of conical part,
h = 7.7 – 4.4
= 3.3 m

(B) Slant height of conical part,

= 18.3 m

(C) Area of canvas used in making tent
= πrl + 2πrH
= πr(l + 2H)
= 3.14 × 18 (18.3 + 2 × 4.4)
= 3.14 × 18 (18.3 + 8.8)
= 1531.692
≈ 1533 sq m

Cost of canvas = ₹ 1533 × 4.50 = ₹ 6898.5
Area of rectangular park
= Area of park – Area of circular base
= 60 × 70 – 3.14 × 182
= 3182.64 sq m

Question 37.
Ramesh places a mirror on level ground to determine the height of a pole (with traffic light fired on it) (see the figure). He stands at a certain distance so that he can see the top of the pole reflected from the mirror. Ramesh’s eye level is 1.8 m above the ground. The distance of Ramesh and the pole from the mirror are 1.5 rn and 2.5 m respectively.

On the basis of the above information, answer the following questions:
(A) Which criterion of similarity is applicable to similar triangles? [1]
(B) Find the height of the pole. [1]
(C) If Ramesh’s eye level is 1.2 m above the ground, then find the height of the pole.
OR
If the distance of Ramesh and the pole from the mirror are 2.5 m and 1.5 m respectively, then find the height of the pole. [2]
Answer:
(A) Since, angle of incidence and angle of reflection are the same, ∠AMB = ∠CMD
Also, ∠ABM = ∠CDM = 90
So, by AA similarity criterion
∆ABM ~ ∆CDM

(B) As ∆ABM ~ ∆CDM,
$\frac{A B}{C D}=\frac{B M}{{D M}}$
$\frac{A B}{1.8}=\frac{2.5}{1.5}$
⇒ AB = $\frac{5}{3}$ × 1.8
⇒ AB = 3
Thus, the height of the pole is 3 metres.

(C) As ∆ABM ~ ∆CDM,
$\frac{A B}{C D}=\frac{B M}{{D M}}$
$\frac{A B}{1.2}=\frac{2.5}{1.5}$
⇒ AB = $\frac{5}{3}$ × 1.8
= 2

As ∆ABM ~ ∆CDM,
$\frac{A B}{C D}=\frac{B M}{{D M}}$
$\frac{A B}{1.8}=\frac{1.5}{2.5}$
⇒ AB = $\frac{1.5 \times 1.8}{2.5}$
= $\frac{5.4}{5}$ = 1.08

Question 38.
4 boys are having a night in and one of the boy’s mother decides to play a game. 17 cards numbered 1, 2, 3 … 17 are put in a box and mixed thoroughly.
The mother asks each boy to draw a card and after each draw, the card is replaced back in the box. She shows some magic tricks and at the end, decides to test their mathematical skills.

On the basis of the above information, answer the following questions:
(A) Find the probability of drawing an odd number card in the first draw by the first boy. [1]
(B) Find the probability of drawing a prime number card in the second draw by the second boy. [1]
(C) If the card is not replaced after the second draw, find the probability of drawing a card bearing a multiple of 3 greater than 4 in the third draw by the third boy.
OR
If the card is replaced after the third draw, find the probability of drawing a card bearing a number greater than 17 in the fourth draw by the fourth boy and if the card is replaced after the fourth draw, find the probability of drawing a card bearing a multiple of 3 or 7 in the fifth draw by the fourth boy. [2]
Answer:
(A) Number of possible outcomes = 17
Number of favourable outcomes = 9{1, 3, 5, 7, 9,11,13,15,17}
∴ P(getting on odd number on card) = $\frac{9}{17}$

(B) Number of possible outcomes = 17
Number of favourable outcomes = 7{2, 3, 5, 7, 11, 13,17}
∴ P(getting a prime number) = $\frac{7}{17}$

(C) If the card drawn is not replaced, then total number of cards remaining is 16.
Now, total number of outcomes = 16
Favourable outcomes = 4{6, 9,12, 15}
∴ P{getting a multiple of 3} = $\frac{4}{16}$ = $\frac{1}{4}$

Since, there is no card with a number greater than 17.
Total number of outcomes = 17
Favourable outcomes = 7{3, 6, 7, 9, 12, 14,15}
∴ P{getting a multiple of 3 or 7} = $\frac{7}{17}$

CBSE Sample Papers for Class 10 Maths Basic Set 11 with Solutions

October 27, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 11 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A, B, C, D, and E.
Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
Section C has 6 Short Answer-ll (SA-II) type questions carrying 3 marks each.
Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section
Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
In the following frequency distribution, what is the upper limit of the median class? [1]

(a) 18.5
(b) 20.5
(c) 25.5
(d) 17.5
Answer:
(d) 17.5

Explanation:

Class	Frequency	Cumulative
– 0.5 – 5.5	13	13
5.5 – 11.5	10	23
11.5 – 17.5	15	38
17.5 – 23.5	8	46
23.5 – 29.5	11	57

Here, N = 57 So, $\frac{N}{2}$ = 28.5
The cumulative frequency, just greater than 28.5, is 38 which belongs to class 11.5 – 17.5.
So, the median class is 11.5 – 17.5 Its upper limit is 17.5

Question 2.
If the perimeter of o circle is equal to that of a square, then the ratio of the area of circle to the area of the square [1]
(a) 14: 11
(b) 12: 13
(c) 11:14
(d) 13:12
Answer:
(a) 14: 11

Explanation:
Here, it is given that
4s = 2πr

So, the ratio of the area of the circle to the area of square is 14 : 11.

Question 3.
A, B, and C alt begin running in the same direction on a circular track at the same moment. A fin ¡shes a round in 252 seconds, B in 308 seconds, and C In 198 seconds. When will they meet at the beginning point? [1]
(a) 46 min 12 sec
(b) 42 min 6 sec
(c) 52 min 12 sec
(d) 56 min 10 sec
Answer:
(a) 46 min 12 sec

Explanation:
∴ 252 = 22 × 32 × 7
308 = 22 × 7 × 11 198 = 22 × 32 × 11
⇒ Required time = LCM (252, 308, 198)
= 22 × 32 × 7 × 11
= 2772 s
Now, 1 min = 60 s
1 s = $\frac{1}{60}$min
∴ 2772 s = $\frac{2772}{60}$min = 46 min 12 s

Question 4.
In the figure, ABCÐ is a square of side, 14 cm. If APD and BPC are semi-circles, then the area of the shaded region is [1]

(a) 40 sq. cm
(b) 46 sq. cm
(c) 42 sq. cm
(d) 50 sq. cm
Answer:
(c) 42 sq. cm

Explanation:
Area of the shaded region = Area of the square – 2 × Area of semi-circular regions n
= 14 × 14 – 2 × $\frac{π}{2}$(7)²
= 196 – 154
= 42 sq. cm

Question 5.
The coordinates of the points which trisect the line segment joining (1, – 2) and (- 3, 4) is: [1]
(a) (0, 0) (-1, 0)
(b) (-2, 3) (0, -2)
(c) $\left(-\frac{1}{4}, 0\right)\left(-\frac{4}{3}, 2\right)$
(d) $\left(-\frac{1}{3}, 0\right)\left(-\frac{5}{3}, 2\right)$
Answer:
(d) $\left(-\frac{1}{3}, 0\right)\left(-\frac{5}{3}, 2\right)$

Explanation:
Let A (1, – 2) and B (- 3, 4) be the given two points.
Let P and Q be the points of trisection of $\overline{A B}$. P trisects $\overline{A B}$ in the ratio 1 : 2; and Q trisects $\overline{A B}$ in the ratio 2 : 1.

Question 6.
In a ΔABC if BD ⊥ CA and CE ⊥ BA. Then ΔABD congruent to: [1]
(a) ΔACB
(b) ΔACE
(c) ΔAED
(d) ΔABD
Answer:
(b) ΔACE

Explanation:
In Δs ABD and ACE, we have

∠D = ∠E (each is of 90°)
∠A = ∠A (common)
So, by AA similarity criteria,
ΔABD ~ ΔACE
Hence Proved.

Question 7.
The ratio in which x-axis divides the join of (2, -3) and (5, 6) is: [1]
(a) 1: 2
(b) 3 : 4
(c) 1: 3
(d) 1: 5
Answer:
(a) 1 : 2

Explanation:
Let P(x, 0) be the point on x-axis which divides the join of (2, -3) and (5, 6) in the ratio m : n.
∴ By section formula,
P(x, 0) = $\left(\frac{5 m+2 n}{m+n}, \frac{6 m-3 n}{m+n}\right)$
$\frac{6 m-3 n}{m+n}$ = 0
6m = 3n
⇒ $\frac{m}{n}=\frac{1}{2}$ is the required ratio.

Question 8.
If cosec A = $\frac{13}{12}$, then the value of $\frac{2 \sin A-3 \cos A}{4 \sin A-9 \cos A}$ [1]
(a) 4
(b) 5
(c) 6
(d) 3
Answer:
(d) 3

Explanation:
Given cosec A = $\frac{13}{12}$

Question 9.
If the angle of elevation of the top of a tower from a point of observation at a distance of 100 m from its base is 45°, then the height of the tower is: [1]
(a) 160 m
(b) 100 m
(c) 200 m
(d) 150 m
Answer:
(b) 100 m

Explanation:
Here, PQ is the tower and A is a point of observation at a distance of 100 m from PQ.
From right ΔAPQ,

Thus, the height of tower is 100 metre.

Question 10.
If the 6th term and 11th term of an A.P. are 12 and 22 respectively, then its 2nd term is: [1]
(a) 8
(b) -5
(c) -8
(d) 16
Answer:
(c) -8

Explanation:
Here, 6th term = a₆ = a + 5d = 12 . . .(i)
and 8th term = a₈ = a + 7d = 22 . . .(ii)
Subtracting equation (i) from equation (ii), we get
2d = 10, i.e. d = 5
a = – 13
So, 2nd term = a₂ = a + d
= (- 13) + 5
= – 8

Question 11.
The value of x and y is: x – 2y = 4 and 6x – 4y = 8 [1]
(a) x = 2, y = 0
(b) x = 1, y = 3
(c) x = 0, y = -2
(d) x = -1, y = -31
Answer:
(c) x = 0, y = -2

Explanation:
Given equations are
x – 2y = 4 …(i)
And 6x – 4y = 8 …(ii)
Multiply the eq. (i) x 2 then subtract eq. (ii) from eq. (i), we get
x = 0, y = -2

Question 12.
How many tangents can be drawn to a circle from a point lying inside the circle? [1]
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(a) 0

Explanation:
Zero, as tangent can be drawn to a circle from a point Lying inside a circle.

Question 13.
The degree of the polynomial : (x + 1) (x – x + x – 1) is: [1]
(a) 4
(b) 3
(c) 5
(d) 2
Answer:
(c) 5

Explanation:
The given ponynomial is (x + 1) (x² – x + x⁴ – 1)
⇒ x³ + x² – x² – x + x⁵ + x⁴ – x – 1
So, the degree of this polynomial is 5.

Question 14.
If r = 3 is a root of quadratic equation kr2 -kr- 3 = 0, then the value of k is: [1]
(a) $\frac{1}{2}$
(b) 3
(c) $\frac{1}{3}$
(d) $\frac{1}{4}$
Answer:
(a) $\frac{1}{2}$

Explanation:
Given, equation is kr² – kr – 3 = 0
If, r = 3
Then, k(3)² – k(3) – 3 = 0
⇒ 9k – 3k – 3 = 0
⇒ 6k =3
⇒ k = $\frac{1}{2}$

Question 15.
Two different dice are thrown together. The probability of getting the sum of the two numbers less than 7 is: [1]
(a) $\frac{5}{12}$
(b) $\frac{7}{12}$
(c) $\frac{12}{5}$
(d) $\frac{3}{11}$
Answer:
(a) $\frac{5}{12}$

Explanation:
Total outcomes = 36
Outcomes in which sum of two numbers is less than 7 = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 2) (2, 3) (2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1) i.e., Total number of possible outcomes = 15
∴ Required probability = $\frac{15}{36}$
= $\frac{5}{12}$

Question 16.
If x = 1 is a root of quadratic equation x² – ax – 1 = 0, then the value of ‘o’ is: [1]
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

Explanation:
Put x = 1, in the given equation x² – ax – 1 = 0.
We have
(1)² – a × 1 – 1 = 0
⇒ 1 – a – 1 = 0
⇒ a = 0

Question 17.
Convert the following statement into a pair of linear equations in x and y (x > y)
“The sum of 2 numbers is 58. The greater number exceeds twice the smaller number by 1.” [1]
(a) x + 2y = 1, x + y
(b) x – 2y = 3, x + y = 58
(c) x = 3y + 2, x – y = 58
(d) x = 2y + 1, x + y = 58
Answer:
(d) x = 2y + 1, x + y = 58

Explanation:
x + y = 58
x = 2y + 1

Question 18.
A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. The probability of getting a red or white ball is: [1]
(a) $\frac{2}{5}$
(b) $\frac{3}{5}$
(c) $\frac{4}{5}$
(d) $\frac{1}{5}$
Answer:
(b) $\frac{3}{5}$

Explanation:
Total balls = 5 + 8 + 7 = 20
Number of Red or White balls = 5 + 7 = 12
∴ Required probability = $\frac{12}{20}$
= $\frac{3}{5}$

Question 19.
Assertion (A): If a and (3 are the zeroes of 2x² + 5x + 9, then the 9 value of αβ is $\frac{9}{2}$
Reason (R): If a and p are the zeroes of p(x) = ax² + bx + c, then αβ = $\frac{c}{a}$. [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Here, quadratic equations 2x² + 5x + 9. Then α.β = $\frac{c}{a}=\frac{9}{2}$.

Question 20.
Assertion (A) :The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10.
Reason (R) : Distance between two points is given by $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$ [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
For y = 3
Distance PQ = $\sqrt{(10-2)^2+(y+3)^2}$
= $\sqrt{8^2+6^2}$
= $\sqrt{64+36}$
= $\sqrt{100}$
= 10 units

SECTION – B (10 Marks)
(Section – B consists of 5 questions of 2 mark each.)

Question 21.
Prove that the number 4ⁿ, n being a natural number, can never end with the digit 0.
OR
Draw a factor tree for the number 546. [2]
Answer:
If 4ⁿ ends with 0, then it must have 2, 5 as its factors.
But, (4)ⁿ = (2²)ⁿ = 2²ⁿ, i.e, the only prime factor of 4ⁿ is 2.

Also, we know from the Fundamental Theorem of Arithmetic that the prime factorisation of each number is unique.
∴ 4ⁿ can never end with digit 0.

Question 22.
The angles of a triangle are in A.P., the least being half the greatest. Find the angles. [2]
Answer:
Let, the angles be a, a + d, a + 2d.
∴ a = $\frac{1}{2}$(a + 2d)
⇒ 2a = a + 2d
⇒ a = 2d …(i)
Now, a + a + d + a + 2d = 180° (Angles sum property of a triangle)
⇒ 3a + 3d = 180°
⇒ 3 (2d) + 3d = 180° (Using (i))
⇒ 9d = 180°
⇒ d = 20°
⇒ a = 40°
Thus, the measure of the angles of the triangle are 40°, 60° and 80°.

Question 23.
What number should be added to the polynomial x² – 5x + 4 so that 3 is the zero of the polynomial? [2]
Answer:
Let k be the number to be added to the given polynomial.
Then the polynomial becomes x² – 5x + (4 + k)
As 3 is the zero of this polynomial, we get:
(3)² – 5(3) + (4 + k) = 0
⇒ (4 + k) = 15 – 9
⇒ 4 + k = 6
⇒ k = 2
Thus, 2 is to be added to the given polynomial.

Question 24.
Evaluate: 3 cos² 60° sec² 30° – 2 sin² 30° tan² 60°. [2]
OR
A coin is tossed twice. Find the probability of getting at-most 2 heads.
Answer:
3 cos²60° sec²30° – 2 sin²30° tan²60°
= 3$\left(\frac{1}{2}\right)^2\left(\frac{2}{\sqrt{3}}\right)^2$ – 2$\left(\frac{1}{2}\right)^2(\sqrt{3})^2$
= 1 – $\frac{3}{2}=-\frac{1}{2}$
OR
Sample space = (H, H) (H, T) (T, H) (T, T)
∴ Possible outcomes = 4
Favourable outcomes = (H, H) (H, T) (T, H) (T,T)
All are favourable cases;
Required probability = $\frac{4}{4}$ = 1

Question 25.
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. How much canvas cloth is required to just cover the heap. [2]
Answer:
Slant height of cone

∴ Cloth required
= πrl = $\frac{22}{7}$ × 12 × 12.5 = 471.42 sq.m

SECTION – C (18 marks)
(Section – C consists of 6 questions of 3 mark each)

Question 26.
Knowing that √5 is an irrational number, show that 3 + 2√5 is an irrational number. [3]
Answer:
Let, 3 + 2√5 be a rational number. Then, it can be written in the form , where p, q are co-primes and q ≠ 0.
Now, 3 + 2√5 = $\frac{p}{q}$,
2√5 = $\frac{p}{q}$ – 3
√5 = $\frac{p-3 q}{2 q}$
Since, $\frac{p-3 q}{2 q}$ is a rational number, √5 is an irrational number (as, L.H.S. = R.H.S.) which is a contraction to the given fact 45 is an irrational number.
This concludes that 3 + 2√5 is an irrational number.

Question 27.
Taking A as the origin, find the coordinates of the vertices P, Q and R of the triangle PQR. The class X students of a secondary school have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 metre from each other. There is a triangular grassy lawn in the plot as shown in the figure. [3]

The students have to sow seeds of flowering plants on the remaining area of the plot.
(A) Taking A as the origin, find the coordinates of the vertices P, Q and R of the triangle PQR;
(B) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Answer:
(A) From the figure, the coordinates of R Q, R are:
P = (1, 4)
Q = (4, 3)
R = (3, 1)

(B) Taking C as origin, CB as X-axis and CD as Y-axis, the coordinates of the vertices P, Q and R are (8, 4) (5, 5) and (6, 7) respectively.

Question 28.
A 90 cm tall female is standing next to a lamppost. She now moves 1.2 metres per second away from a lamp post’s base. What is the length of her shadow after 4 seconds, if the lamp is 3.6 m above the ground? [3]
OR
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Answer:
Speed of girl = 1.2 m/s
∴ In 4 seconds, travels

distance = 1.2 x 4 = 4.8 m
∴ After 4 seconds, she reaches at D.
∴ BD = 4.8 m
Let CD be the length of her shadow.
Now, ∠ABD = ∠EDC = 90°
∴ AB ∥ ED
Hence, by BPT
$\frac{A B}{E D}=\frac{B C}{D C}$
$\frac{A B}{E D}=\frac{B C}{D C}$
⇒ 4x = 4.8 + x
⇒ x = 1.6 m
OR
We draw OX ⊥ AB.
Since, AOB is an isosceles triangle, X bisects AB, i.e. AX = XB
Also, ∠AOX = ∠BOX = 60°

From right-angled triangle AXO, we have:
$\frac{\mathrm{AX}}{\mathrm{AO}}$ = sin 60 and $\frac{O X}{A O}$ = cos 60
AX = AO × $\frac{\sqrt{3}}{2}$
and OX = AO × $\frac{1}{2}$
⇒ AX = 6√3 cm and OX = 6 cm
AB = 12√3 cm
Now,
Area of the shaded segment
= Area of sector AOB – Area of AAOB
= $\frac{120^{\circ}}{360^{\circ}}$π(12)² – $\frac{1}{2}$ × AB × OX
= (48π – 36√3) sq. cm
= [48 (3.14) – 36(1.73)] sq. cm
= (150.72 – 62.28) sq. cm
= 88.44 sq. cm

Question 29.
As seen in the figure, a medicine capsule is shaped like a cylinder with two hemispherical ends. Find the capsule’s capacity if its length is 14 mm and its diameter is 6 mm. [3]

Answer:
Diameter of capsule = 6 mm
⇒ Radius of capsule (r) = = 3 mm
So, Length of cylindrical part (h)
= Length of capsule – 2 × Radius of hemispherical ends
= 14 – 2 × 3
= 14 – 6 = 8 mm
So, capacity of the capsule
= 2 × Volume of hemispherical ends + Volume of cylinderical part
= 2 × $\frac{2}{3}$πr³ + πr²h
= $\frac{4}{3}$πr³ + πr²h
= $\frac{4}{3}$ × π × (3)³ + π × (3)² × 8
= 36π + 72π
= 108π mm³

Question 30.
The distribution below gives the makes of 100 students of a class, if the median makes are 24, find the frequencies f₁ and f₂

All the black face cards are removed from a pack of 52 playing cards. The reaming cards are well shuffled and then a card is drawn at random. Find the probability of getting o: [3]
(A) face card
(B) red card
(C) black card.
Answer:

Class	Frequency	Cumulative
0 – 5	4	4
5 – 10	6	10
10 – 15	10	20
15 – 20	f₁	20 + f₁
20 – 25	25	45 + f₁
25 – 30	f₂	45 + f₁ + f₂
30 – 35	18	63 + f₁ + f₂
35 – 40	5	68 + f₁ + f₂

Now, Median = 24 (Given)
So,median class = 20 – 25
For this class,
I = 20. h = 5, $\frac{N}{2}$ = 50, cf = 20 + f₁, f= 25
We know, Median = l + $\frac{\frac{N}{2}-c f}{f}$ × h
⇒ 24 = 20 + $\frac{50-\left(20+f_1\right)}{25}$ × 5
⇒ 4 = $\frac{\frac{N}{2}-c f}{f}$
⇒ 20 = 30 – f₁

Also, sum of frequencies = 100
⇒ 68 + f₁ + f₂ = 100
⇒ f₁ + f₂ = 32
⇒ 10 + f₂ = 32
⇒ f₂ = 22
∴ f₁ = 10, f₂ = 22.
OR
When all the three black face cards are removed,
Remaining number of cards = 52 – 3 = 49
(A) Number of face cards in the remaining deck = 9
∴ P(getting a face card) = $\frac{9}{49}$

(B) Number of red cardsin the remaining deck = 26
∴ P (getting a red card) = $\frac{26}{49}$

Question 31.
In the figure, two chords AB and CD intersect each other at the point P. [3]
Prove that
(A) ΔAPC ~ ΔDPB
(B) AP. PB = CP. DP

Answer:
(A) Consider Δs APC and DPB
∠APC = ∠DPB
(Vertically opopsite angles)
Also,
∠CAP = ∠PDB

(Angles made by the same arc CB)
So, by AA similarity criteria,
ΔAPC ~ ΔDPB.
Hence Proved.

(B) This further concludes that
$\frac{\mathrm{AP}}{\mathrm{PC}}=\frac{\mathrm{DP}}{\mathrm{PB}}$
or AP.PB = CP.DP
Hence Proved.

SECTION – D
(Section – D consists of 4 questions of 5 mark each)

Question 32.
A man sold a chair and a table for f 1520, thereby making a profit of 25% on the chair and 10% on the table. By selling together for ₹ 1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each. [4]
OR
If Zeba was younger by 5 years than what she really is, then the square of her are (in years) would have been 11 more than five times her actual age. What is her age now?
Answer:
Let the cost of 1 chair be ₹ p and of cost 1 table is ₹ q.
Then,
(p + 25% p) + (q + 10% q) = 1520;
(p + 10% p) + (q + 25% q) = 1535
⇒ $\frac{5 p}{4}+\frac{11}{10}$q = 1520
$\frac{11 p}{10}+\frac{5}{4}$q = 1535
⇒ 25p + 22q = 30400; …(i)
22p + 25q = 30700 …(ii)

Adding the two equations, we get 47 p + 47 q = 61100
⇒ 47p + 47q = 61100
⇒ p + q = 1300 …(iii)
From Eq. (i) and Eq. (ii), we get
25p + 22 (1300 – p) = 30400
⇒ 3p = 1800
or p = 600
From equation (iii), we have q = 700.
Thus, the cost of 1 chair is ₹ 600 and of 1 table is ₹ 700.
OR
Let Zeba’s present age (in years) be x.
Then (x – 5)² = 11 + 5x
⇒ x² – 10x + 25 = 11 + 5x
⇒ x² – 15x + 14 = 0
⇒ (x – 14) (x – 1) = 0
⇒ x – 14 = 0 [∵ (x – 1) ≠ 0)]
⇒ x = 14
Thus, her present age is 14 years.

Question 33.
In the figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90° If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle. [4]

Answer:
Join OP and OQ.
We know, Lengths of tangents drawn from an external point to a circle are equal.
So,
Here, AR = AQ, BP = BQ,
CP = CS, DR = DS
Here, ∠B = 90° and OP = OQ = r (say)
OPBQ is a square and QB = r.

Now, AD = AR + RD
⇒ AQ + SD
⇒ 23 = AQ + 5
⇒ AQ = 18 cm
Also 29 = AB = AQ + BQ
⇒ 18 + r
⇒ r = 11
Thus, radius of the circle is 11 cm.

Question 34.
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1. [4]
Answer:
Given, sin θ + cos θ = √3
(sin θ + cos θ)² = (√3)²
⇒ sin²θ + cos²θ + 2 sin θ cos θ = 3
⇒ 1 + 2 sin θ cos θ = 3 (∵ sin²θ + cos²θ = 1) …(i)

Question 35.
Three uniformly flowing pipes fill a paddling pool. While the third pipe is the only one filling the paddling pool, the first two pipes are simultaneously filling the paddling pool. Five hours faster than the first pipe and four hours later than the third pipe, the second pipe fills the paddling pooL Calculate how long it takes for each pipe to fill the paddling pool independently. [4]
OR
A cylindrical hole with a diameter of 7 cm is drilled out of a cuboid solid metallic block with dimensions of 15 cm 10 cm 5 cm. Find the block’s remaining surface area.
Answer:
Let V be the volume of the pool and x the number of hours required by the second pipe alone to fill the pool. Then, the first pipe takes (x + 5) hours, while the third pipe takes (x – 4) hours to fill the pool. So, the parts of the pool filled by the first, second and third pipes in one hour are respectively.
$\frac{V}{x+5}, \frac{V}{x}, \frac{V}{x-4}$
Let the time taken by the first and second pipes to fill the pool simultaneously be t hours.
Then, the third pipe also takes the same time to fill the pool

⇒ (2x + 5) (x – 4) = x2 + 5x
⇒ x² – 8x – 20 = 0
⇒ x² – 10x + 2x – 20 = 0
⇒ (x – 10) (x + 2) = 0
⇒ x = 10 or x = -2
But x cannot be negative. so x = 10
Hence the timings required by first, second and third pipes to fill the pool individually are 15 hours, 10 hours and 6 hours respectively.
OR
Length of solid mettalic block l = 15 cm
Breadth of solid mettalic block, b = 10 cm
Height of solid mettalic block, h = 5 cm
Diameter of cylinderical block (d) = 7 cm

Radius (r) = $\frac{d}{2}=\frac{7}{2}$ = 3.5 cm
Height of cyLindericaL block, h = 5 cm
Surface area of cuboid = 2(lb + bh + hl)
= 2(15 × 10 + 10 × 5 + 5 × 15)
= 2(150 + 50 + 75)
= 2 × 375 = 550 cm2
Curved surface area of cylinder k = 2prh
= 2 × $\frac{22}{7}×\frac{7}{2}$ × 5 = 11 × 7
= 77 cm²
Surface area of the reamining block
= Surface area of cuboidal block + CSA of cylinder – Area of 2 cylinderal holes
= 550 + 110 – 77
= 660 – 77
= 583 cm²
Hence the surface area of remaining block = 583 cm².

SECTION – E (12 marks)
(Case Study Based Questions)
(Section – E consists of 3 questions. All are compulsory)

Question 36.
The diagram shows a grid of squares. A button is placed on one of the squares. A fair dice is thrown. If 1,2,3 or 4 is thrown, the button is moved one square to the left. If 5 or 6 is thrown, the button is moved one square to the right.

On the basis of the above information, answer the following questions:
(A) The button is placed on square X.
the dice is thrown once. Then find the probability that the button is moved to the right. [1]
(B) On the other occasion, the button is placed on square Y. The dice is thrown once and the button is moved. The die is thrown a second time and the button is moved again. Find the probability that the button is moved at P. [1]
(C) Find the probability that the button is moved at Q and also find the probability that the botton is moved at Y.
OR
Find the probability that the button is moved at P, Q or Y. [2]
Answer:
(A) P(right) = $\frac{2}{6}=\frac{1}{3}$

(B) P(gatting 1, 2, 3 or 4 twice) = $\frac{2}{6}=\frac{1}{3}$
= $\frac{1}{3} \times \frac{1}{3}$
= $\frac{1}{9}$

(C) As per the situation given in part (B), if the button is at Y and it is moved twice, then there is no possibility that it reach to Q.
As per the situation given in part (B), if the button is at Y and it is moved twice, then it will reach to Y again if it is first move to left then to right or vice-versa.
P(Y) = $\frac{2}{6} \times \frac{4}{6}+\frac{4}{6} \times \frac{2}{6}=\frac{16}{36}$
= $\frac{4}{9}$
OR
The probability that the button is moved at P = $\frac{1}{9}$
The probability that the button is moved at Q = 0
The probability that the button is moved at y = $\frac{4}{9}$

Question 37.
Earth is excavated to make a railway tunnel. The tunnel is a cylinder of radius 5 m and length 450 m. A level surface is laid inside the tunnel to carry the railway lines. The Diagram 1 shows the circular cross – section of the tunnel. The level surface is represented by AB, the centre of the circle is 0 and ∠AOB = 90°. The space below AB is filled with rubble (debris from the demolition buildings).

Steel girders are erected above the tracks to strengthen the tunnel. Some of these are shown in Diagram 2. The girders are erected at 6 m intervals along the length of the tunnel, with one at each end.
On the basis of the above information, answer the following questions:
(A) Find the volume of earth removed to make the tunnel. [1]
(B) Find the area of ΔAOB shown in Diagram 1. [1]
(C) Find the length of each girder.
OR
How many girder are erreced and find total length of steel required in the 450 m length of tunnel. [2]
Answer:
(A) Volume of tunnel = πr²h
= π × 5 × 5 × 450
= 11250π cu cm

(B) Area of ΔAOB = $\frac{1}{2}$ × AO × OB
= $\frac{1}{2}$ × 5 × 5
= 12.5 sq.m

(C) As steel girders are in the shape of a circle or a sector of the circle with centra/angle 270°.
So, circumference of the circle with central angle 270°
= $\frac{270^{\circ}}{360^{\circ}}$ × 2πR
= $\frac{3}{4}$ × 2πR
= 1.5π × 5
= 7.5π cm
OR
No. of girders = $\frac{450}{6}$
= 75
∴ Toatl grider = 75 + 1 (one at least)
= 76
Toatl length = 7.5π × 76
= 570 π m

Question 38.
A radio mast PQ , of height ‘h’ metres, is standing vertically on the horizontal ground. From A, the angle of elevation of the top of the mast is found to be 45°. On moving 50 m up a slope of 15°, the angle of elevation ofP is found to be 75° from B, The horizontal through B is BC.

On the basis of the above information, answer the following questions:
(A) What is the measure of ∠APC? [1]
(B) Find the measure of ∠BPC. [1]
(C) Find the length BP.
OR
Find the length AP and the approaximat value of h. [2]
Answer:
(A) In ΔAPQ
∠A + ∠P + ∠Q = 180°
⇒ 45° + ∠P + 90° = 180°
∠P = 180° – 90° – 45°
= 180° – 135°
= 45°

(B) In ΔBPC
∠B + ∠BPC + ∠C = 180°
⇒ 75° + ∠BPC + 90° = 180°
⇒ ∠BPC = 180° – 165°
= 15°

In ΔABQ’,
sin 15° = $\frac{\mathrm{BQ’}}{\mathrm{AB}}$
BQ’ = 50 sin 15°
= 13 m
(Taking sin 15° = 0.26)
Then, BQ’ = CQ = 13 m
PC = h – 13

In ΔABQ’
cos 15° = $\frac{AQ’}{AB}$
AQ’ = 50 cos 15°
= 48.25 ≈ 49
(Taking cos 15° = 0.965)
QQ’ = AQ – AQ’
= h – 49
BC = QQ’ = h – 49

Now, in ΔPBC
tan 75° = $\frac{P C}{B C}$
3.73 = $\frac{h-13}{h-49}$
3.73h – 182.77 = h – 13
2.73h = 169.77
h = 62.186
sin 75° = $\frac{\mathrm{PC}}{\mathrm{PB}}$
1 = $\frac{62.2-13}{P B}$
PB = 49.2 = 50 m

In ΔAPQ
sin45° = $\frac{h}{\mathrm{AP}}$
AP = $\frac{h}{\frac{1}{\sqrt{2}}}$ = √2h

In ΔPBC
tan 75 = $\frac{\mathrm{PC}}{\mathrm{BC}}$
3.73 = $\frac{\mathrm{PC}}{\mathrm{BC}}$
3.73h – 182.77 = h – 13
h = 62.186
h ~ 60 m

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

October 27, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 6 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A, B, C, D, and E.
Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
Section C has 6 Short Answer-ll (SA-II) type questions carrying 3 marks each.
Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section
Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
The distance of the point (7, -8) from the origin is: [1]
(a) $\sqrt{112}$
(b) $\sqrt{115}$
(c) $\sqrt{113}$
(d) 96
Answer:
(c) $\sqrt{113}$

Explanation:
Distance of (7, – 8) from origin
= $\sqrt{(7-0)^2+(-8-0)^2}$
= $\sqrt{49+64}$
= $\sqrt{113}$ units

Question 2.
The ratio in which the line segment joining the points ( -1, 7) and ( 4, -3) is divided by the point (1, 3) is: [1]
(a) 2 : 3
(b) 3 : 2
(c) 1: 4
(d) 2 : 5
Answer:
(a) 2:3

Explanation:
Let P(1, 3) divide AB in the ratio k : 1

Then, P(1, 3) = P$\left(\frac{4 k-1}{k+1}, \frac{-3 k+7}{k+1}\right)$
⇒ $\frac{4 k-1}{k+1}$ = 1 and $\frac{-3 k+7}{k+1}$ = 3
3k = 2 and -6k = -4
⇒ k = $\frac{2}{3}$
Thus,the ratio is 2: 3

Question 3.
For the following distribution, [1]

The sum of the lower limits of the median class and the modal class is:
(a) 20
(b) 35
(c) 30
(d) 25
Answer:
(d) 25

Explanation:

	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Here, the modal class is 15 – 20 and the median class is 10 – 15.
So, the sum of the two lower limits = 25

Question 4.
An equilateral triangle ABC is inscribed in a circle with centre O. The measure of ∠BOC is: [1]
(a) 120°
(b) 130°
(c) 60°
(d) 45°
Answer:
(a) 120°

Explanation: As ΔABC is equilateral

∴ ∠A = 60°
⇒ ∠BOC = 2 × ∠A = 120°

Question 5.
The value of 4 tan²A – 4 sec²A is: [1]
(a) -3
(b) 2
(c) -4
(d) 4
Answer:
(c) -4

Explanation:
4 tan²A – 4 sec²A
= 4 tan²A – 4 (1 + tan²A)
= – 4.

Question 6.
What is the perimeter of triangle with vertices (0, 0) (1, 0) and (0,1)? [1]
(a) (1 + √2) units
(b) (3 + √2) units
(c) √2 units
(d) (2 + √2) units
Answer:
(d) (2 + √2) units

Explanation:
Perimeter of Δ AOB

= $\overline{O A}+\overline{A B}+\overline{O B}$
= 1 + √2 + 1
= (2 + √2) units.

Question 7.
If 2 cos 3θ = √3 (0° ≤ θ ≤ 90°), then the value of θ is: [1]
(a) 10°
(b) 40°
(c) 20°
(d) 60°
Answer:
(a) 10°

Explanation: Given, 2 cos 3θ = √3
⇒ cos 3θ = $\frac{\sqrt{3}}{2}$ = cos 30°
⇒ 3θ = 30°
⇒ θ = 10°

Question 8.
If the area of a sector of a circle of radius 2 cm is 7t sq m, then what is the central angle of the sector ? [1]
(a) 90°
(b) 45°
(c) 30°
(d) 60°
Answer:
(a) 90°

Explanation:
Let the central angle be θ. Then, area of the sector
= [$\frac{\theta}{360}$ × π(2)²] sq.cm

Equating it to π sq cm, we have,
$\frac{\theta}{360}$ × π × 4 = π
⇒ θ = 90°

Question 9.
If 18, a, b, – 3 are in A.P., then the value of a + b is: [1]
(a) 15
(b) 20
(c) 25
(d) 30
Answer:
(a) 15

Explanation:
Since 18, a, b, – 3 are is A.P, so their common difference will be same.
a – 18 = b – a = – 3 – b
So, a – 18 = – 3 – b
⇒ a + b = 18 – 3
= 15

Question 10.
In ΔABC, D and E are prints on the sides AB and AC respectively, such that DE ∥ BC
If AD = 2.5 cm, BD = 3 cm and AE = 3.75 cm, then the value of AC is: [1]
(a) 8 cm
(b) 9.1 cm
(c) 8.25 cm
(d) 9 cm
Answer:
(c) 8.25 cm

Explanation:
Since DE ∥ BC
So, by B.P.T, we have

Question 11.
In the given figure, if ∠MPQ = 40°, then ∠OPM is: [1]

(a) 30°
(b) 20°
(c) 50°
(d) 60°
Answer:
(c) 50°

Explanation:
In the figure,
∠OPQ = 90° [∵ Tangent ⊥ Radius]
⇒ ∠OPM = 90° – ∠MPQ
= 90° – 40°
= 50°

Question 12.
A box contains 7 red balls and 6 blue balls. A ball is drawn at random from the box. The probability that this drawn is a red or blue ball is: [1]
(a) 2
(b) 4
(c) 1
(d) 3
Answer:
(c) 1

Explanation:
P (a red or blue ball)
= $\frac{7+6}{13}=\frac{13}{13}$ = 1

Question 13.
The empirical relationship between mean, median and mode is: [1]
(a) Mode + Mean = Median
(b) 3 Median = Mode – Median
(c) Mode = Median + 2 Mean
(d) 3 Median = Mode + 2 Mean
Answer:
(d) 3 Median = Mode + 2 Mean

Explanation:
The emperical relation between the measures of central tendency which is given by
Mode = 3 Median – 2 Mean

Question 14.
If the product of the zeros of the polynomial ax² – 6x – 12 is 4, then the value of ‘a’ is: [1]
(a) 3
(b) 2
(c) 4
(d) -3
Answer:
(d)-3

Explanation:
Given equation: ax² – 6x – 12
Product of zeros = $\frac{\text { Constant term }}{\text { Coefficient of } x^2}$
⇒ $\frac{(-12)}{a}$ = 4
a = – 3

Question 15.
The value of x, in the adjoining figure, if DE ∥ BC, is: [1]

(a) 8
(b) 9
(c) 10
(d) 11
Answer:
(a) 8

Explanation:
In ΔABC, DE ∥ BC

(By Thales theorem)

(2x – 1) (x – 1) = (2x + 5) (x – 3)
⇒ 2x² – 2x – x + 1 = 2x² + 5x – 6x – 15
⇒ 2x = 16
⇒ x = 8

Question 16.
What is value of x + y, if ΔABC and ΔPQR are similar ? [1]

(a) 12.8 cm
(b) 14.3 cm
(c) 12.5 cm
(d) 14 cm
Answer:
(b) 14.3 cm

Explanation:
As, ΔABC and ΔPQR are similar

⇒ y = 12.8
∴ x + y = 1.5 + 12.8
= 14.3 cm

Question 17.
A quadratic polynomial whose zeros are – 7 and 5 is: [1]
(a) x² + 2x – 35
(b) x² – 2x + 35
(c) x² + 3x – 25
(d) x² – 3x – 35
Answer:
(a) x² + 2x – 35

Explanation:
Sum of zeroes = – 7 + 5 = -2
Product of zeroes =-7 x 5 = -35
A quadratic polynomial with sum and product of zeroes is given as,
x² – (sum of zeroes) x + Product of zeroes.
⇒ x² – (- 2) x – 35
⇒ x² + 2x – 35

Question 18.
The value for x and y:
x + y = 2 and 2x – y = 1 is: [1]
(a) 2
(b) 1
(c) 4
(d) 3
Answer:
(b) 1

Explanation:
On adding both the equations, we get

⇒ x = 1
Then, y = 2 – x = 2 – 1 = 1
⇒ x = y = 1 is the required solution.

Question 19.
Assertion (A) : In rhomus the diagonals are 20 cm and 10√6 cm in length; then side length of rhombus is 15.8 cm.
Reason (R) : The sum of the sides of a rhomus is equal to the sum of the squares of its diagonals. [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
The diagonals of a rhombus bisect each other at right angles.
∴ In ΔAOD, using pythagoras theorem,

AD² = OA² + OD²
= 10² + (5√6)²
= 100 + 150
= 250
= 15.8 cm

Question 20.
Assertion (A) : sec²θ = 1 + tan²θ is a trigonometric identity.
Reason (R) : An equation involving trigonometric ratios of an angle is called trigonometric identity, which is true for all values of the angles involved. [1]
Answer:
(a) Both assertion (A) and reason (R) are correct and reason (R) is correct explanation of assertion (A).

Explanation:
Here, sec²θ = 1 + tan²θ
Put, θ = 45°
sec²45° = (√2)² = 2
1 + tan²θ = 1 + (tan 45°)
= 1 + 1 = 2
Thus, sec θ = 1 + tan²θ

SECTION – B (10 Marks)
(Section – B consists of 5 questions of 2 marks each)

Question 21.
Write any two irrational numbers whose product is a rational number. [2]
Answer:
Consider two irrationals as, 5 – 2√2 and 5 + 2√2
Here,
(5 – 2√2)(5 + 2√2) = 5² – (2√2)²
= 25 – 8 = 17 (a rational number)

Question 22.
If the zeros of the polynomial x³ – 3x² + x + 1 are a – b, a and a + b, then find the values of a and b. [2]
Answer:
As (a – b), a and (a + b) are zeroes of x³ – 3x² + x + 1, we have:
a – b + a + a + b = 3 [∵ Sum of zeroes = $-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}$]
⇒ 3a = 3, or a = 1 …(i)
Also, a (a – b) + a (a + b) + (a – b) {a + b) = 1 [∵ Sum of product of zeroes = $-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}$
⇒ 3a² – b² = 1 …….(ii)
(a – b) a (a + b) = -1 [∵ Product of Zeroes = $-\frac{\text { Constant term }}{\text { Coefficient of } x^3}$]
⇒ a(a² – b²) = -1 …(iii)
From (i) and (ii), we have b = ± √2
Thus, a = 1, b = ± √2

Question 23.
The product of A’s age 5 years ago with his age 9 years later is 15. Find A’s present age. [2]
OR
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in the same two variables such that the geometrical representation of the pair of equations so formed is:
(A) parallel lines
(B) coincident lines
Answer:
Let A’s present age (in years) be x. Then,
(x – 5) (x + 9) = 15 •
⇒ x² + 4x – 45 = 15
⇒ x² + 4x – 60 = 0
⇒ x² + 10x – 6x – 60 = 0
⇒ x (x + 10) – 6 (x + 10) = 0
⇒ (x + 10) (x – 6) = 0
⇒ x – 6 = 0 (∵ x+ 10 ≠ 0)
⇒ x = 6
Thus, A’s present age is 6 years.

OR
(A) For parallel lines, we must have equation ax + by + c = 0
which must satisfy $\frac{2}{a}=\frac{3}{b} \neq \frac{-8}{c}$
So, we can write the required equation as
2x + 3y – 2 = 0

(B) For coincident lines, we must have equation
ax + by + c = 0
which must satisfy $\frac{2}{a}=\frac{3}{b}=\frac{-8}{c}$
So, we can write the required equation as
4x + 6y – 16 = 0

Question 24.
A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 × 16 m. Find the area of the field, that a cow can graze. [2]
OR
Prove that: (tan θ + 2) (2 tan θ + 1) = 5 tan θ + 2 sec²θ
Answer:
Shaded area can be grazed by the cow tied at the corner B. .

= 154 sq cm.
OR
(tan 0 + 2) (2 tan 0 + 1)
= 2 tan2 0 + 4 tan 0 + tan 0 + 2 = 2 tan2 0 + 5 tan 0 + 2 = 2 (tan2 0 + 1) + 5 tan 0 = 2 sec2 0 + 5 tan 0

Question 25.
Determine the mean of the following data : [2]

Answer:

X	f	fx
2	5	10
4	6	24
3	8	24
7	12	84
9	10	90
5	7	35

Here, Σf = 48 and Σfx = 267
We know,
mean = $\frac{\Sigma f x}{\Sigma f}=\frac{267}{48}$ = 5.5625

SECTION – C (20 Marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
Prove that √3 is an irrational number. [3]
OR
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Let us suppose that √3 is a rational number.
Then √3 can be written in the form $\frac{p}{q}$ where p, q are co-prime i.e., they do not have common factor other than 1.
Now, √3 = $\frac{p}{q}$
⇒ 3 = $\frac{p^2}{q^2}$[squaring both sides]
⇒ p² = 3q²
⇒ 3 divides p²
⇒ 3 divides p
⇒ 3 is factor of p.

∴ Let p = 3m.
⇒ P² = 3 q²
⇒ (3m)² = 3q²
⇒ 3m² = q²
⇒ 3 divides q²
⇒ 3 divides q
It means, 3 is a factor of both p and p and q cannot have any common other than 1.
It means, our assumption is wrong.
Hence, √3 is an irrational number.
OR
Number are of two types – prime and composite
Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78 = 13 × 13 × 6
The given expression has 6 and 13 as its factors.
Therefore, it s a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 cannot be factorized further
Therefore, the given expression has 5 and 1009 as its factors.
Hence, it is a composite number.

Question 27.
Two friends Reena and Sunita applied for the post of Computer Engineer in two different companies and got selected. Reena has been offered a job with a starting monthly salary of ₹ 48000, with an annual increment of ₹1400in her salary. Sunita has been offered a job with a starting monthly salary of ₹ 40000, with an annual increment of ₹ 1800 in her salary.
(A) Determine their monthly salaries for the 13th year. [1]
(B) Find each of them’s total salary of 13 years. [1]
(C) Who will get more salary in 13 years? And how much more? [1]
Answer:
Reena’s yearly amount (in ₹) of monthly salary.
48000, 49400, 50800, ………………..
It is A.P. with a = 48000 and d = 1400.
Sunita’s yearly amount (in ₹) of monthly salary.
40000, 41800, 43600, ………………..
It is A.P. with a’ = 40000 and d’ = 1800.

(A) So, Reena’s 13th year monthly salary = a + 12d
= 48000 + 12 x 1400
= ₹ 64800
Sunita’s 13th year monthly salary = a + 12d
= 40000 + 12 x 1800
= ₹ 61600

(B) Reena’s total salary of 13 years
= $\frac{13}{2}$[a₁ + a₁₃] x 12
= 78 [48000 + 64800]
= ₹ 8,7,98,400
Similarly, Sunita’s total salary of 13 years
= $\frac{13}{2}$ [40000 + 61600] x 12
= ₹ 79,24,800

Question 28.
Find the roots of the equation
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$ (x ≠ -4, 7)
OR
Find a fraction which becomes $\frac{1}{2}$ when the denominator is increased by 4, and $\frac{1}{8}$ when the numerator is decreased by 5. [3]
Answer:
The given equation is:
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
⇒ $\frac{x-7-x-4}{(x+4)(x-7)}=\frac{11}{30}$
⇒ $\frac{x-7-x-4}{(x+4)(x-7)}=\frac{11}{30}$
⇒ (x + 4)(x – 7) = 30
⇒ x² + 4x – 7x – 28 = – 30
or x² – 3x + 2 = 0
or (x – 2) (x – 1) = 0
⇒ x – 2 = 0, x – 1 = 0
⇒ x = 2,1
So, the two roots are x = 2 and x = 1.
OR
Let the required fraction be $\frac{p}{q}$.
Then,
$\frac{p}{q+4}=\frac{1}{2}$
⇒ 2p – q – 4 = 0 …(i)
$\frac{p-5}{q}=\frac{1}{8}$
8p – q – 40 = 0 …(ii)
Subtracting the first equation from the second equation, we get
6p -36 = 0
⇒ p = 6.
Substituting this value p = 6 in either equations, we get
q = 8
Thus, the required fraction is $\frac{6}{8}$.

Question 29.
In the figure, a square OABC is inscribed in a quadrant OPBQ.
If OA = 20 cm, find the area of the shaded region. (Use π = 3.14) [3]

Answer:
As, OA = 20 cm
∴ Diagonal

So, radius of the quadrant = 20√2 cm
Area of the quadrant
= $\frac{\pi}{4}$(20√2)² sq cm
= 200π sq cm Area of the square
= (20)² sq cm, i.e. 400 sq cm

Area of the shaded region
= (200π – 400) sq cm
= (628 – 400) sq cm
= 228 sq cm

Question 30.
Given that ΔPQR is similar to ΔBAR. Find: [3]

(A) the value of x;
(B) the value of y;
Answer:
(A) Here,
∠PRQ = 54° (vertically opposite angles)
Now, In APQR,
∠PQR = 180° – (68° + 54°) = 58°
∵ ΔPQR – ΔBAR
∠Q = ∠A
⇒ x = 58°

(B) Again, ΔPQR – ΔBAR,
∴ $\frac{P Q}{B A}=\frac{Q R}{A R}$
⇒ $\frac{2 y}{y+2}=\frac{y+3}{y}$
⇒ 2 y² = y² + 5y + 6
⇒ y² – 5y – 6 = 0
⇒ (y – 6) (y + 1) = 0
y = 6 (∵ y ≠ -1)

Question 31.
The annual rainfall record of a city for 66 days is given below in the table: [3]

Calculate the median rainfall, using the formula.
Answer:
The cumulative frequency table for the given data is:

Ranifall (in cm)	Frequency	Cumulative frequency
0-10	22	22
10-20	10	32
20-30	8	40
30-40	15	55
40-50	5	60
50-60	6	66

Here, N = 66
So, $\frac{N}{2}$ = 33
A cumulative frequency just greater than 33 is 40, which belongs to class 20 – 30.
So, the median class is 20 – 30.
For this class,
l = 20, cf = 32, f = 8, $\frac{N}{2}$ = 33 and h = 10
So, Median = l + $\frac{\frac{\mathrm{N}}{2}-c f}{f}$ × h
= 20 + $\frac{33-32}{8}$ × 10
= 20 + $\frac{1}{8}$ × 10
= 21.25
Thus the median rain fall (in cm) is 21.25

SECTION – D
(Section – D consists of 4 questions of 5 marks each)

Question 32.
Draw the graphs of the equations:
4x – y = 4 and 4x + y = 12
Hence, determine the vertices of the triangle formed by the lines representing these equations and the x – axis. Shade the triangular region so formed.
OR
Solve 2x + 3y = 11 and 2x-4y = -24 and hence find the value of‘m’ for which y = mx + 3. [5]
Answer:
Table of values of
4x – y = 4

Table of Values of
4x + y = 12

From the graph, we find the vertices A, B, C of Δ ABC as A(2, 4), B(l, 0), C(3, 0).
Also, triangular region ABC is shaded.
OR
2x + 3y = 11 …(i)
2x – 4y = -24 …(ii)
Using equation (ii), we can say that
2x = – 24 + 4y
⇒ x = -12 + 2y
Putting this in equation (i), we get
2(-12 + 2y) + 3y = 11
⇒ -24 + 4y + 3y = 11
⇒ 7y = 35
⇒ y = 5

Putting value of y in equation (i), we get
2x + 3(5) = 11
⇒ 2x + 15 = 11
⇒ 2x = 11 – 15 = -4
⇒ x = -2
Therefore, x = -2 and y = 5
Putting values of x and y in y = mx + 3, we get
5 = m(-2) + 3
⇒ 5 = -2m + 3
⇒ -2m = 2
⇒ m = -1

Question 33.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Using the above result, prove the following:
If a circle touches all the four sides of a quadrilateral ABCD, prove that:
AB + CD = BC + DA. [5]
Answer:
I-Part:
We are given a circle with centre 0, a point A lying outside the circle and two tangents AX and AY on the circle from the point A.
We need to prove that
AX = AY
Join OX, OY and AO.
We know, tangent is perpendicular to radius, at the point of contact.
∠AXO = ∠AYO = 90°

Now, in right triangles AXO and AYO we have
AO = AO (common)
XO = YO (radii of the same cirde)
∠AXO ≅ ∠AYO (each 90°)
Therefore, by R.H.S. congruence criterion,
ΔAXO ≅ ΔAYO
⇒ AX = BY

IInd-Part:
Here, a circle touches all the four sides of quadrilateral ABCD.
From the figure, using the above result we have,
AS = AP, BP = BQ, CQ = CR
and DS = DR

Now, AB + CD = (AP + BP) + (CR + DR)
= (AP + DR) + (BP + CR)
= (AS + DS) + (BQ + CQ)
= AD + BC

Question 34.
Prove that:
$\frac{\cot \theta+{cosec} \theta-1}{\cot \theta-{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}$
OR
Prove that $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ = 1 + tan A + cot A. [5]
Answer:

OR

Question 35.
Two stations are located at a distance of ‘a’ and ‘b’ from the foot of a leaning tower that leans in the direction of the north. If a and p be the elevations of the top of the tower from these stations, show that the inclination 6 to the horizontal is given by cot θ = $\frac{b \cot \alpha-a \cot \beta}{b-a}$. [5]
Answer:
Let, the height of the tower DE = h
Distance between first station to foot of tower AD = a + x

Distance between second station to foot of tower BD = b + x
Distance between C and D = x
Given, a and p are angles of elevation of two station to the top of the tower.
i.e., ΔDAE = α, ΔDBE = β, αDCE = 0
In ΔCDE,
cot θ = $\frac{x}{h}$ …(i)

In ΔBDE,
cot b = $\frac{b+x}{h}$
⇒ b + x = h cot β

In ΔADE
cot α = $\frac{a+x}{h}$
⇒ a + x = h cot α

Multiply ‘b’ on both sides
ba + bx = bh cot α …(iii)

Subtract (iii) from (ii)
(b – a)x = h (b cot β – a cot β)
$\frac{x}{h}=\frac{b \cot \alpha-a \cot \beta}{b-a}$
cot θ = $\frac{b \cot \alpha-a \cot \beta}{b-a}$ [from (i)]
Hence, proved

SECTION – E (12 marks)
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory)

Question 36.
To make the learning process more interesting, creative and innovative, Amayra’s class teacher brings clay in the classroom, to teach the topic. Surface Areas and Volumes. With clay, she forms a cylinder of radius 6 cm and height 8 cm. Then she moulds the cylinder into a sphere and asks some questions to students.
(A) Find the radius of the sphere so formed.
OR
Find the total surface area of the cylinder. [2]
(B) What is the volume of the sphere so formed? [1]
(C) Find the ratio of the volume of sphere to the volume of cylinder. [1]
Answer:
(A) Since, volume of sphere = volume of cylinder
⇒ $\frac{4}{3}$R³ = πr²h, where R, r are the radii of sphere and cylinder respectively.
⇒ R³ = $\frac{6 \times 6 \times 8 \times 3}{4}$ = (6)³
R = 6 cm
∴ Radius of sphere = 6 cm
Total surface area of the cylinder = 2πr (r + h)
= 2 × $\frac{22}{7}$ × 6(6 + 8)
= 2 × 2 × $\frac{22}{7}$ × 6 × 14 = 528 cm²

(B) Volume of sphere = $\frac{4}{3}$πR³
= $\frac{4}{3} \times \frac{22}{7}$ × 6 × 6 × 6 = 905.14 cm3

Question 37.
A linguist is performing a statistical analysis of word frequency distributions as part of her quantitative stylistics to understand the measurable aspects of lexical structure. She picks a random newspaper sentence (structure of which is shown below) that has 20 words in it.

The number of letters in each word is counted and the table below shows the frequency distribution:

On the basis of the above information, answer the following questions:
(A) A word is chosen a random from the whole sentence. What is the probability that it has 4 letters? [1]
(B) A word is chosen at random from those with an odd number of letters. What is the probability that it has 7 letters? [1]
(C) First person chooses a word at random from the whole sentence, Another person chooses a word at random from the whole sentence. What is the probability that one person chooses a 2-letter word and the other chooses a 6-letter word?
OR
Find the mean number of letters in the whole sentence. [2]
Answer:
(A) No. of four letter words = 5
Total number of words = 20
∴ P(4 letter words) = $\frac{5}{20}=\frac{1}{4}$

(B) No. of words with 7 letters =2
Total no. of words with odd letters = 9
∴P(word has 7 letter) = $\frac{2}{9}$

(C) If first person chooses a 2-letter word then second person chooses a 6-letter word or vice-versa.
∴ Required probability

Question 38.
Resident Welfare Association (RWA) of a M2K Society in Azadpur have put up three electric poles A, B and C in a society’s common park near Tower A. Despite these three poles, some parts of the park are still in dark.
So, RWA decides to have one more electric pole D in the park.

On the basis of the above information, answer the following questions:
(A) What is the distance of the pole B form the corner O of the park? [1]
(B) Find the position of the fourth pole D so that four points A, B, C and D form a parallelogram.
OR
Find the distance between poles A and C. [2]
(C) Find the distance between poles B and D. [1]
Answer:
Coordinates of B are (6. 6) Distance from origin
= $\sqrt{(6-0)^2+(6-0)^2}$
= $\sqrt{(6-0)^2+(6-0)^2}$
= $\sqrt{72}$ units

(B) If ABCD forms a parallelogram, then the diagonals bisects each other.
Mid point of AC = $\left(\frac{2+5}{2}, \frac{7+4}{2}\right)$ = (3.5, 5.5)
Now, mid-point of diagonal, BD will be same. Let, the coordinates of D be (x, y)
Then,
$\frac{6+x}{2}$ = 3.5 and $\frac{6+y}{2}$ = 5.5
Coordinates of A are (2, 7)
Coordinates of C are (5, 4)
Distance of AC
= $\sqrt{(5-2)^2+(4-7)^2}$
= $\sqrt{9+9}$
= $\sqrt{18}$ units

(C) Coordinates of B(6, 6)
Coordinates of D(1, 5)
Distance between BD
= $\sqrt{(6-1)^2+(6-5)^2}$
= $\sqrt{5^2+1^2}$
= $\sqrt{25+1}$
= $\sqrt{26}$ units

CBSE Sample Papers for Class 10 Maths Standard Set 7 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 7 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A-E.
Section A has 20 MCQs carrying 1 mark each
Section B has 5 questions carrying 02 marks each.
Section C has 6 questions carrying 03 marks each.
Section D has 4 questions carrying 05 marks each.
Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E
Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

Section – A
(Section A consists of 20 questions of 1 mark each.)

Question 1.
The prime factorisation of 96 is: (1)
(a) 2⁵ × 3
(b) 2⁶
(c) 2⁴ × 3
(d) 2⁴ × 32
Answer:
(a) 2² × 3

Explanation: The prime factorisation of 96 is:
96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

Question 2.
The zeroes of the poLynomial p(x) = 4x² – 12x +9 is: (1)
(a) $\frac{1}{2}$, 0
(b) 0,$\frac{1}{2}$
(c) $\frac{5}{6}$,$\frac{1}{2}$
(d) $\frac{3}{2}$,$\frac{3}{2}$
Answer:
(d) $\frac{3}{2}$,$\frac{3}{2}$

Explanation: p(x) = 4x² – 12x + 9 = (2x – 3)² = 0
Thus, x= $\frac{3}{2}$ and $\frac{3}{2}$ are the two zeroes of p(x).

Question 3.
If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4. then the values of a and b is: (1)
(a) a = 3, b = 1
(b) a = 5, b = 2
(c) a = 1, b = 3
(d) a =4, b = 2
Answer:
(a) a = 3, b = 1

Explanation: As x = o, y = b is the solution of: x – y = 2 and x + y = 4
we have,
a – b = 2 and a + b = 4
⇒ a = 3 and b = 1

Question 4.
Discriminant of the quadratic equation 2x² + 4x -7=0 is: (1)
(a) 28
(b) 72
(c) 36
(d) 48
Answer:
(b) 72

Explanation:
The discriminant of 2x² + 4x- 7 = 0 is [(4)² – 4(2)(-7)] = 16 + 56 = 72

Question 5.
The roots of quadratic equation x² – 4x + 2 = 0 (1)
(a) 2 ± $\sqrt{2}$
(b) $\sqrt{2}$,2
(c) 3,2
(d) 3 ± $2 \sqrt{2}$
Answer:
(a) 2 ± $\sqrt{2}$

Explanation:
Here, quadratic equation is: p(x) = x² – 4x + 2 On comparing it with ax² + fax + c = 0
Then, a = 1, b = – 4, c = 2
D = b² – 4ac = (- 4)² – 4 × 1 × 2
= 16 – 8 = 8
Then, roots, x = $\frac{-b+\sqrt{D}}{2 a}$
= $\frac{-(-4)±\sqrt{8}}{2}$
= $\frac{4±2\sqrt{2}}{2}$
= 2 ± $\sqrt{2}$
Hence, roots of the equation are 2 + $\sqrt{2}$ and 2 – $\sqrt{2}$.

Question 6.
If S_n = 5n² + 3n, then its nth term is: (1)
(a) 5n – 1
(b) 10n²
(c) 10n – 2
(d) 8n – 3
Answer:
(c) 10n – 2

Explanation:
a_n = S_n – S_n-1
= 5n² + 3n – [ 5 (n – 1)² + 3(n – 1)]
= 5n² + 3n – [ 5n2 + 5 – 10n + 3n – 3]
= 10n – 2

Question 7.
If the common difference of an A.P. is 5, then find a₁₈ – a₁₃. (1)
(a) 38
(b) 40
(c) 18
(d) 25
Answer:
(d) 25

Explanation:
a₁₈ – a₁₃ = a + 17d – a – 12d
= 5d = 5 × 5
= 25

Question 8.
In an A.P., a = -6 and d = 2. The sum of its first 20 terms is: (1)
(a) 270
(b) 200
(c) 195
(d) 260
Answer:
(d) 260

Explanation: S_n = $\frac{n}{2}$ [2a + (n – 1)d]
So, S₂₀ = $\frac{20}{2}$ [2(-6)+(20 – l)(2)]
= 10[-12 + 38]
= 10 × 26 = 260

Question 9.
Write the relationship between the coefficients, if the following pair of equations is inconsistent. ax + by + c = 0, a’x+b’y + d = 0. (1)
(a) $\frac{a}{a^{\prime}} \neq \frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}$
(b) $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}} \neq \frac{c}{c^{\prime}}$
(c) $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}$
(d) 0
Answer:
(b) $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}} \neq \frac{c}{c^{\prime}}$

Explanation:
the required relationship is:
$\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}} \neq \frac{c}{c^{\prime}}$

Question 10.
In a ∆ABC, right-angled at B, if AB : AC = 1 : 2, then the value of $\frac{2 tan A}{1+tan^{2} A}$ is: (1)
(a) $\frac{1}{2}$
(b) $\frac{5}{2}$
(c) $\frac{\sqrt{3}}{2}$
(d) $\frac{2}{\sqrt{3}}$
Answer:
(c) $\frac{\sqrt{3}}{2}$

Explanation:

Question 11.
If a tower 6 metres high casts a shadow on the ground that is 2-^3 metres long, then the elevation of the sun is: (1)
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer:
(a) 60°

Explanation:
As per the given question:

Question 12.
The area of a circle, whose circumference is 22 cm, is: (1)
(a) 38 cm
(b) 36 cm²
(c) 38.5 cm²
(d) 40 cm²
Answer:
(c) 38.5 cm²

Explanation:
Let r be the radius of the circle.
Then,
2πr = 22 cm
⇒ r = $\frac{22}{2 \times \frac{22}{7}}$
⇒ r = $\frac{7}{2}$ cm or 3.5 cm
Thus,
Area = πr² = $\frac{22}{7}$ × 3.5 × 3.5 cm² = 38.5 cm²

Question 13.
If the ratio between the volumes of two spheres is 8 : 27, then the ratio between their surface areas is: (1)
(a) 2 : 3
(b) 1 : 2
(c) 25 : 16
(d) 4 : 9
Answer:
(d) 4 : 9

Explanation:
Let, the radius of 2 sphere be r and R

Question 14.
The class-mark of the class interval 10-25 is: (1)
(a) 17.5
(b) 16
(c) 14
(d) 18
Answer:
(a) 17.5

Explanation: The class mark of 10 – 25 is $\frac{10 + 25}{2}$ = 17.5

Question 15.
One card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is either red or a queen. (1)
(a) $\frac{5}{14}$
(b) $\frac{1}{14}$
(c) $\frac{5}{13}$
(d) $\frac{7}{13}$
Answer:
(d) $\frac{7}{13}$

Explanation:
In a pack of 52 cards, there are 26 red cards and 2 black queens.
So, total possible outcomes = 26 + 2 = 28
P(a red or a queen) = $\frac{28}{52}$ = $\frac{7}{13}$

Question 16.
How many face cards are there in a pack of 52 cards? (1)
(a) 12
(b) 10
(c) 14
(d) 16
Answer:
(a) 12

Explanation: Face cards are jacks, queen and kings
∴ Total face cards = 4 + 4 + 4=12

Question 17.
Determine the upper limit of the modal class of the following frequency distribution:

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

(a) 16
(b) 19.5
(c) 18
(d) 17.5
Answer:
(d) 17.5

Explanation:
The given frequency distribution in the exclusive form is:

Class	0.5 – 5.5	5.5-11.5	11.5- 17.5	17.5 – 23.5	23.5 – 29.5
Frequency	13	10	15	8	11

Here, the modal class is 11.5 – 17.5
So, the upper limit of the modal class is 17.5.

Question 18.
The empirical relationship among the three measures of central tendency mean, mode and median. (1)
(a) 3 Median = Mode + 2 Mean
(b) Mode = 2 Median – Mean
(c) Mean = 3 Mode + 2 Median
(d) Median = Mode – Mean
Answer:
(a) 3 Median = Mode + 2 Mean

Explanation: The required relationship is: 3Median = Mode + 2Mean

DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (IQ are true and reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Statement A (Assertion): If the circumference of a circle is 176 cm, then its radius is 28 cm.
Statement R (Reason): Circumference = 2π × radius. (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Circumference (C) = 176 cm
2πr= 176
⇒ 2 × $\frac{22}{7}$ × r = 176
⇒ $\frac{44}{7}$ × r = 176
⇒ r = 176 × $\frac{7}{44}$ = 28 cm
∴ The radius of the circle is 28 cm.

Question 20.
Statement A (Assertion): If the value of mode and mean is 60 and 66 respectively, then the value of median is 64.
Statement R (Reason): Median = mode + 2 mean (1)
Answer:
(c) Assertion (A) is true but reason(R) is false.

Explanation: Median = -(Mode + 2 mean) = (60 + 2 × 66) = 64

Section – B
(Section B consists of 5 questions of 2 marks each)

Question 21.
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, find the coordinates of P and Q.
OR
Find the third vertex of a triangle, if two of its vertices are at (-3,1) and (0, -2) and the centroid is at the origin. (2)
Answer:
Letthe coordinates of P and Q be (0, y) and (x,0) respectively.
∴ Mid-point of PQ : ($\frac{0+x}{2}$, $\frac{y+0}{2}$)
i.e ($\frac{x}{2}$,$\frac{y}{2}$)
Equating it with (2, -5), we have,
$\frac{x}{2}$ = 2; $\frac{y}{2}$ = -5
⇒ x = 4, y = -10
Thus, the coordinates of P and Q are (0, -10) and (4, 0) respectively.
OR
Given that two points are (- 3, 1) and (0, -2) and its centroid is (0, 0).
Let the third vertex be (x, y). Then,
($\frac{x-3+0}{3}$,$\frac{y+1-2}{3}$) = (0,0)
⇒ $\frac{x-3}{3}$ = 0 and $\frac{y-1}{3}$ = 0
⇒ x = 3 and y = 1
Thus, the third vertex is (3,1).

Question 22.
Explain why 3 × 5 × 7 × 9 × 11 + 11 is a composite number.
OR
If n = 2³ × 3⁴ × 5⁴ × 7, where n is a natural number, then find the number of consecutive zeros in n. (2)
Answer:
11(3 × 5 × 7 × 9 + 1) = 11 (945 + 1)
= 11 × 946 = 11 × 2 × 11 × 43
Since, it has more than 2 factors.
Therefore, it is a composite number.
OR
According to questions,
n = 2³ × 3⁴ × 5⁴ × 7
= 2³ × 3⁴ × 5³ × 5 × 7
= (2 × 5)³ × 3⁴ × 5 × 7
= (10)³ × 3⁴ × 5 × 7
= 3⁴ × 5 × 7 × 1000
Thus, in the given natural number ‘ri there are 3 zeros.

Question 23.
A tangent from point A, which is placed 5 cm away from the circle’s centre, has a length of 4 cm. Find the circle’s radius. (2)

Answer:
We know that the tangent at any point of a circle is ⊥ to the radius through the point of contact.
∴ ∠OPA = 90°
∴ OA² = OP² = AP²
(By Pythagoras theorem)
⇒ (5)² = (OP)² + (4)²
⇒ 25 = (OP)² + 16
⇒ OP² = 9
⇒ OP = 3 cm
(OP = – 3 cm is moved)

Question 24.
Find the missing frequency for the given data if mean of distribution is 52. (2)

Wages in Rs	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of Workers	5	3	4	f	2	6	13

Answer:
Given data is
Find the missing frequency for the given data is mean of distribution is 52.

Wages in Rs	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of Workers	5	3	4	f	2	6	13

Given data is

Wages	f_i	x_i	f_ix_i
10-20	5	15	75
20-30	3	25	75
30-40	4	35	140
40-50	f	45	45 f
50-60	2	55	110
60-70	6	65	390
70-80	13	75	975
Total	33 + f		1765 + 45f

Then, Mean = $\frac{\Sigma f_i x_i}{\Sigma f_i}$
$52=\frac{1765+45 f}{33+f}$
⇒ 7f = 1765 – 1716 = 49
⇒ f = 7
Hence, missing frequency is 7.

Question 25.
If a number x is chosen at random from the numbers -2, -1, 0,1, 2, then what is the probability that x² < 2 ? (2)
Answer:
Total outcomes = 5
Favourable outcomes (i.e., x² < 2) are 4, 1, 0, 1, 4 i.e. 3
∴ P(x² < 2) = $\frac{3}{5}$
Hence, the required probability is $\frac{3}{5}$

Section – C
(Section C consists of 6 questions of 3 marks each.)

Question 26.
If p is a prime number, then prove that $\sqrt{p}$ is irrational. (3)
Answer:
Let us assume, to the contrary that $\sqrt{p}$ is rationaL
So, we canfind co-prime integers a’ and ‘b’ (b* 0),
such that $\sqrt{p}$ = $\frac{a}{b}$
⇒ $\sqrt{p}$ b = a
⇒ pb² = a² …(i)
⇒ a² is divisible by p
⇒ a is divisible by p.
So, we can write a = pc for some integer c.
Therefore, a² = p²c² (squaring both side)
⇒ pb² = p²c² (from (i))
⇒ b² = pc²
⇒ b² is divisible by p
⇒ b is divisible by p
⇒ p divides both a and b.
⇒ ‘a’ and ‘b’ have at least p’ as a common factor
But this contradicts the fact that ‘a and ‘b’ are coprime.
This contradiction arises because we have assumed that $\sqrt{p}$ is rational.
Therefore, $\sqrt{p}$ is irrational.

Question 27.
Find the zeros of the polynomial 2x² – (1 + 2$\sqrt{2}$)x + $\sqrt{2}$. (3)
Answer:
p(x) = 2x² – (1 + 2$\sqrt{2}$)x + $\sqrt{2}$ = 0
= 2x² – x + 2$\sqrt{2}$x + $\sqrt{2}$ = 0
= x(2x – 1) – $\sqrt{2}$ (2x – 1) = 0
= (2x – 1) (x – $\sqrt{2}$) = 0
Either, 2x – 1 = 0 or x – $\sqrt{2}$ = 0
⇒ x = $\frac{1}{2}$ or x = $\sqrt{2}$
Thus, zeros of the given polynomial are $\frac{1}{2}$ and $\sqrt{2}$.

Question 28.
Solve for x:$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \cdot x \neq 0,1,2$ (3)
Answer:
Here,
$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \cdot x \neq 0,1,2$
$\frac{(x-1)+2(x-2)}{(x-2)(x-1)}$ = $\frac{6}{x}$
$\frac{x-1+2x-4}{(x^{2}-2x-x+2)}$ = $\frac{6}{x}$
⇒ (3x – 5)x = 6(x² – 3x + 2)
⇒ 3x² – 5x = 6x² – 18x + 12
⇒ 3x² – 13x + 12 = 0
⇒ 3x² – 9x – 4x + 12 = 0
⇒ 3x(x – 3) – 4(x – 3) = 0
⇒ (3x – 4) (x – 3) = 0
⇒ x = $\frac{4}{3}$,3
Hence, the values of x are $\frac{4}{3}$ and 3.

Question 29.
In the figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with the chord PQ, then determine ∠POQ.

OR
O is point inside a triangle ABC The bisectors of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA at point D, E and F respectively, show that AD x BE x CF = DB x EC x FA. (3)
Answer:
Here, ∠QPR = 50°
and ∠OPR = 90°
[∵ Radius is perpendicular to tangent]
∴ ∠OPQ + ∠QPR = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
Now, OP = OQ [Radii of same circle]

∴ ∠OPQ = ∠OQP = 40°
In ∆OPQ,
∠POQ + ∠OPQ + ∠OQP = 180°
⇒ ∠POQ + 40° + 40° = 180°
⇒ ∠POQ = 180° – 80° = 100°
OR
Given : A AABC, in which O is a point inside it. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CD at points D, E and F respectively.

=> DB × EC × FA = AD × BF × CF
or AD × BE × CF = DB × EC × FA
Hence, proved.

Question 30.
The diagram shows a round about at the junction of four roads (of equal width).
The central park is in the form of a circle with centre O and radius 14 m.
The curbs BC, DE, FG and HA are in the form of arcs that lie on a circle with centre O and radius 21 m. The angles subtended by these curbs at O are 60°, 45°, 45°, 90°.

(A) Find the total lengths of the curbs; 1 1/2
(B) Find the area of the circular road surrounded the central park. 1 1/2
Answer:

(B) Area of the circular road surrounding the central park
= [π(21)² – π(14)²] m²
= $\frac{22}{7}$× (21 + 14) (21 – 14) m²
= 770 m$\frac{22}{7}$.

Question 31.
Rnd the median of the following data:

Marks (out of 90)	No. of Students
0-10	2
10-20	2
20-30	4
30-40	6
40-50	6
50-60	5
60-70	2
70-80	4
80-90	4
Total	35

OR
Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. Find the probability that both will visit the shop:
(A) the same day
(B) different days
(C) consecutive days. (3 )
Answer:

Then, $\frac{N}{2}$ = $\frac{35}{2}$ = 17.5
Then, median class is 40 – 50
l = 40 f = 6, c.f. = 14, h = 10

Then Me = l + $\frac{N/2 – cf}{f}$ × h
= 40 + $\frac{17.5 – 14}{6}$ × 10
= 40 + $\frac{3.5}{6}$ × 10 = 40 + 5.83
= 45.83
Thus, median of the given data is 45.83
OR
Two customer can visit the shop in 6 × 6 ways = 36 ways.
Total no. of events = 36

(A) Two customer can visit the shop on the same day in one of the following ways, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
∴ P(of visiting on same day) = $\frac{6}{36}$ = $\frac{1}{6}$

(B) ∴ P(of visiting on different day) = 1 – $\frac{1}{6}$ = $\frac{5}{6}$

(C) Two customers can visit the shop on 2 consecutive days in the following ways : (Mon, Tue), (Tue, Wed), (Wed, Thurs), (Thurs, Fri), (Fri, Sat) = 5 ways ,
∴ P (of visiting on consecutive days) = $\frac{5}{36}$

Section D
(Section D consists of 4 questions of 5 marks each.)

Question 32.
State and prove basic proportionality theorem.
OR
D and E are points on the sides CA and CB respectively of a AABC, right-angled at C. Prove that: AE² + BD² = AB² + DE² (5)

Answer:
Statement “If a side is parallel to one side of a triangle and it intersects the other two sides in 2 distinct points, then it divides the other 2 sides in same proportion.”
Proof:

Given: A ∆ABC, a line DE parallel to BC intersect AB and D and AC at E.
To prove: $\frac{AD}{DB}$ = $\frac{AE}{EC}$
Construction : Draw EF ⊥ AB and DG ⊥ AC Join BE and CE
Proof: Since, EF ⊥ AB
EF is the height of triangle ADE and DBE.
Area of ∆ADE = $\frac{1}{2}$ × b × h = $\frac{1}{2}$ × AD × EF
Area of ∆DBE = $\frac{1}{2}$ × DB × EF

But ∆DBE and ∆DCE are on the same base DE and between the same parallel straight lines BC and DE.
Area of ∆DBE = Area of ∆DCE …(iii)
From, (i), (ii) and (iii), we have
$\frac{AD}{DB}$ = $\frac{AE}{EC}$
Hence proved.
OR
In the figure, ACB is a right angled triangle, right-angled at C. D and E are points on sides CA and BC respectively.
We Join DE, BD and AE.
In right triangle ACE, we have:
AE² = AC² + EC² …(i)
In right triangle BCD, we have:
BD² = BC² + DC² …(ii)
Adding (i) and (ii), we get:
AE² + BD² = (AC² + EC²) + (BC² + DC²)
= (BC² + CA²) + (CE² + CD²)
= AB² + DE².
Hence proved.

Question 33.
Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot A at the rate of 2 for 3 bananas and the second lot B at the rate of ₹ 1 per banana, and got a total collection of f 400. If he had sold the first lot A at the rate of f 1 per banana, and the second lot B at the rate of 4 for 5 bananas, his total collection would have been ₹ 460.
Determine the total number of bananas he had. (5)
Answer:
Let lot A contains ‘x’ bananas; and lot B contains ‘y’ bananas. Then, according to the question,
$\frac{2}{3}$x + y = 400 ; x + $\frac{4}{5}$ y = 460
⇒ 2x + 3y = 1200 ; 5x + 4y = 2300
⇒ 10x + 15y = 6000 …(i)
and 10x + 8y = 4600 …(ii)
Subtracting (ii) from (i), we get
⇒ 7 y = 1400
⇒ y = 200.
From (i), we get
10x + 15 × 300 = 6000
⇒ 10x = 1500
⇒ x = 300
Thus, lot A contains 300 bananas and lot B contains 200 bananas.
Vijay had 500 bananas in all.

Question 34.
Two tangents are drawn from a point P to a circle with a centre of O. Prove that ∆APB is equilateral, if OP = diameter of the circle. (5)

Answer:
Join OP.
Suppose OP meets the circle at Q. Join AQ.
We have
i.e., OP = diameter
∴ OQ + PQ = diameter
∴ OQ + PQ = diameter
PQ = Diameter – radius [••• OQ = r]
∴ PQ = radius
Thus, OQ = PQ = radius
Thus, OP is the hypotenuse of right triangle
OAP and Q is the mid-point of OP
∴ OA = AQ = OQ
[v mid-point of hypotenuse of a right triangle is equidistant from the vertices]
⇒ AOAQ is equilateral
⇒ ∠AOQ = 60°
So, ∠APO = 30°
∴ ∠APB = 2∠APO = 60°
Also PA = PB
⇒ ∠PAB = ∠PBA
But ∠APB = 60°
∴ ∠PAB = ∠PBA = 60°
Hence, ∆APB is equilateral triangle.

Question 35.
Trigonometric ratios sin A, sec A, and tan A should be expressed in the form of cotA.
OR
The angles of depression of the top and bottom of building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distances between the building and the tower. (5)
Answer:
For sin A,
Bu using identity, cosec² A – cot² A = 1 .

OR
Let PQ be the tower of height h m and AB be the building of height 50 m.

From right triangle BDQ,
$\frac{DQ}{BD}$ = tan 30° = $\frac{1}{\sqrt{3}}$
⇒ BD = $\sqrt{3}$ DQ …..(i)
Also, from right triangle APQ,
$\frac{PQ}{AP}$ = tan 60° = $\sqrt{3}$
or $\frac{PQ}{BD}$ = $\sqrt{3}$
or BD = $\sqrt{3}$ …..(ii)
From (i) and (ii), we have
DQ = $\frac{PQ}{3}$
Further,
PQ = PD + DQ

SECTION – E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
Selvi is setting up a water purifier system in her house which includes setting up an overhead tank in the shape of a right circular cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid.

The underground water tank (sump) is a sturdy single moulded piece built to with stand underground pressure and is available in the storage capacity of 2000 L
These, along with hassle-free installation and minimum maintenance needs make it the ideal water storage solution.
Dimensions (sump):
1.57 m × 1.44 m × 95 cm.
Dimensions (overhead tank):
Radius is 60 cm and Height is 95 cm

Water flow conditions at the required overload capacity should be checked for critical pressure drop to ensure that valves are adequately sized.
On the basis of the above information, answer the following questions:
(A) Find the ratio of the capacity of the sump to the capacity of the overhead tank. (1)
(B) If overhead tank need to be painted to save it from corrosion, how much area need to be painted? (1)
(C) If water is filled in the overhead tank at the rate of 20 litre per minute, the tank will be completely filled in how much time?
OR
If the amount of water in the sump, at an instant, is 1500 litres , then find the water level in the sump at that instant? (2)
Answer:

(A)

(B) CSA. of cylindrical tank = 2πrH
= 2 × 3.14 × 60 × 95
= 35,796 cm²
= 3.5796 m²
= 3.6 m²

(C) Volume of water in cylindrical tank = πr²h
= 3.14 × 60 × 60 × 95
= 1073880 cm³
Now, 1l = 1000 cm³
∴ Volume of tank = 1073.88l
20l tank is filled in 1 minute
∴ 1073.88l tank is filled in $\frac{1073.88 l}{20}$
= 53.69
= 54 minutes

Question 37.
Rishu ¡s riding in a hot air bafloon. After reaching a point P, he spots a car parked at B on the ground at an angle of depression of 30°. The balloon rises further by 50 metres and now he spots the same car at an angle of depression of 45° and a lorry parked at B’ at an angle of depression of 30°. (Use $\sqrt{3}$ = 1.73)

The measurement of Rishu facing vertically is the height. Distance is defined as the measurement of car/lorry from a point in a horizontal direction. If an imaginary line is drawn from the observation point to the top edge of the car/lorry, a triangle is formed by the vertical, horizontal and imaginary line.
On the basis of the above information, answer the following questions:
(A) If the height of the balloon at point P is ‘h’m and distance AB is ‘x’ m, then find the relation between ‘x’ and ‘h’. (1)
(B) Find the relation between the height of the balloon at point P’ and distance AB. (1)
(C) Find the height of the balloon at point P, and the distance AB on the ground.
OR
Find the distance B’B on the ground. (2)
Answer:

(C) On solving equation obtained in (A) and (B), we get
$\sqrt{3}$h = h + 50
⇒ h($\sqrt{3}$ – 1) = 50
⇒ h = $\frac{50}{0.732}$ = 68.25

In ∆APB.
tan 30° = $\frac{AP}{AB}$
⇒ AB = $\frac{AP}{tan 30°}$ = $\frac{68.25}{1/\sqrt{3}}$
= 68.25 × 1.732
= 118 m

In ∆AP’B’
tan 30° = $\frac{AP’}{AB’}$
$\frac{1}{\sqrt{3}}$ = $\frac{68.25+50}{AB’}$
⇒ AB’ = 118.25 × 1.732 = 204.809
BB’ = AB’ – AB
= 204.809 – 118
= 86.80 = 87 m

Question 38.
To conduct sport day activities in the rectangular school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each along DB.
100 flower pots have been placed at a distance of 1 m from each other along DA as shown in the figure below.

Radha runs $\frac{1}{4}$th of the distance DA on 2nd line and post a green flag at X . Preeti runs $\frac{1}{5}$th of the distance DA on other line post a red flag at Y.
On the basis of the above information, answer the following questions:
(A) Treating DB as x-axis and DA as y-axis, then find the position of green flag. (1)
(B) Treating DB as x-axis and DA as y-axis, the find the position of red flag. (1)
(C) Find the distance (in complete metres) between the two flags.
OR
Find the perimeter (in complete metres) of the triangular region OXY. (2)
Answer:
(A) Radha’s distance on x-axis is 2 and on y-axis, she is at $\frac{1}{4}$ × 100 = 25°
Green flag coordinates are (2,25)

(B) X-coordinate = 8
Y-coordinate = $\frac{1}{5}$ × 100 = 20
Coordinates of red flag (8,20).

(C) coordinates of green flag is (2,25)
∴ coordinates of red flag is (8,20)
∴ By distance formula
Distance = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
= $\sqrt{\left(8-2\right)^2+\left(20-25\right)^2}$
= $\sqrt{36 + 25}$
= $\sqrt{61}$
= 7.8 cm
= 8 cm (approx)
OR

Perimeter = OX + OY + YY
= 25.07 + 21.54 + 7.81
= 54.42
= 55 m

CBSE Sample Papers for Class 10 Social Science Set 11 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Social Science Set 11 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
Section A – From question 1 to 20 are MCQs of 1 mark each.
Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
What, led to the coIIapse of fixed exchange rates system?
(a) GoId no Ionger commanded the respect of the markets.
(b) DoIIar had Iost its respect in the market as the principIe currency.
(c) Other currencies had over taken DoIIar in the market.
(d) SiIver was now used to fix the price of a singIe doIIar instead of goId. [1]
Answer:
(b) Dollar had lost its respect in the market as the principle currency.
Explanation: From the 1960s, the rising costs of the US’s overseas involvements weakened its finances and competitive strength. The US dollar now no Longer commanded confidence as the world’s principal currency. It could not maintain its value in relation to gold.

Related Theory
It eventually led to the collapse of the system of fixed exchange rates and the introduction of a system of floating exchange rates.

Question 2.
Which of the foIIowing subjects is incIuded in the state list as mentioned in the Indian Constitution?
(a) Marriages
(b) Adoption
(c) PoIice
(d) Banking [1]
Answer:
(c) Police
Explanation: The seventh schedule under Article 246 of the constitution deals with the division of powers between the union and the states. It contains three lists.

The union list has subjects on which Parliament may make laws. The state list includes subjects under the purview of state legislatures like policing, liquor, land, and local government. The concurrent list has subjects in which both Parliament and state legislatures have jurisdiction.

Related Theory:
The Constitution accords supremacy to the Parliament on concurrent list items in case of a conflict between the central and state laws.

Question 3.
Study the picture beIow and identify the sector that the depicted activity beIong to:

(a) Primary sector
(b) Secondary sector
(c) Tertiary sector
(d) Quaternary Sector [1]
Answer:
(d) Secondary Sector
Explanation: A woman can be seen making clothes out of the raw material along with other workers surrounding her. This shows she is manufacturing a good which classifies her into the category of secondary sector.

Related Theory:
A sector which provides services and support to the Primary and Secondary Sector is called the Tertiary Sector.

Caution:
Picture related questions are not asked directly. They focus on the concept and ask questions based on its application. Observing and deciphering the message of the picture is extremely important.

Question 4.
Arrange the foIIowing institutions of power in the increasing order of the size of their jurisdiction:
(I) Panchayat Samiti or MandaI
(II) State Government
(III) Gram Panchayat
(IV) ZiIIa Parishad Options:
(a) (I)-(II)-(III)-(IV)
(b) (IV)-(I)-(II)-(III)
(c) (III)-(I)-(IV)-(II)
(d) (II)-(I)-(III)-(IV) [1]
Answer:
(c) (III)-(I)-(IV)-(II)
Explanation: Gram Panchayat heads a village or groups of villages.
Samitis head groups of gram panchayats.
Zilla Parishad heads all panchayat samitis of a district.
State governments head zilla parishads.

Question 5.
Identify the name of the country using the given hints:
(1) This country cut off its cotton suppIies to Britain when a civiI war broke out.
(2) The country was discovered by Christopher CoIumbus.
(a) BeIgium
(b) India
(c) USA
(b) ScotIand [1]
Answer:
(c) USA
Explanation: The American Civil War broke out and thus the cotton supplies to Britain were cut off completely. The UK turned to India to fill this deficiency.

Related Theory:
When Britain turned to India, raw cotton exports from India increased and the price of raw cotton shot up.

Caution:
Questions which require identification of countries, cities or any other things need to solved carefully. Each hint should be connected with the other to finally identify the mentioned subject.

Question 6.
What are SEZs?
(a) forums for SociaI and Economic reforms
(b) speciaI Zones with Economic and Tax benefits for investors
(c) Conferences for EnvironmentaI safety and concerns
(d) Seminars for Eco-friendIy and SustainabIe DeveIopment [1]
Answer:
(b) special Zones with Economic and Tax benefits for investors.
Explanation: Special Economic Zones (SEZs), have world class facilities: electricity, water, roads, transport, storage, recreational and educational facilities. Companies who set up production units in the SEZs do not have to pay taxes for an initial period of five years. Government has also allowed flexibility in the labour laws to attract foreign investment.

Question 7.
Match the items in coIumn A with those of coIumn B and find the most appropriate code which refIects the correctIy matched pairs.

	Column A		Column B
(A)	Cotton	(I)	Uttar Pradesh
(B)	Jute	(II)	Maharashtra
(C)	Wheat	(III)	Rajasthan
(D)	Bajra	(IV)	West Bengal

Codes:
(a) A-(I), B-(III), C-(IV), D-(II)
(b) A-(III), B-(IV), C-(II), D-(I)
(c) A-(II), B-(IV), C-(I), D-(III)
(d) A-(IV), B-(III), C-(II), D-(I) [1]
Answer:
(c) A-(II), B-(IV), C-(I), D-(III)
Explanation: Wheat is the main food crop, in north and north-western part of the country. Bajra grows well on sandy soils and shallow black soil. Major Bajra producing states are Rajasthan, Uttar Pradesh, Maharashtra, Gujarat and Haryana.

Question 8.
Which one of these is NOT a function of a poIiticaI party?
(a) to refIect fundamental poIiticaI divisions in a society
(b) to contest eIections
(c) to pIay a decisive roIe in making Iaws
(d) to form and run governments [1]
Answer:
(a) To reflect fundamental political divisions in a society
Explanation: Political Parties fill political offices and exercise political power by contesting elections.
They are created to expand democracy to the grass root Level. They don’t reflect any political divisions in a society.

Question 9.
How did the OraI cuIture enter the worId of print?
(a) PeopIe began debating about what was printed in the newspaper.
(b) PeopIe taIked about important issues and then wrote about their opinions in the pamphIets.
(c) Printers began pubIishing popuIar baIIads which were profuseIy iIIustrated with pictures. These were then sung and recited at gatherings in viIIages and taverns.
(d) PeopIe were writing new prose work which interested a Iot of peopIe and they began to interview peopIe about it. [1]
Answer:
(c) Printers began publishing popular ballads which were profusely illustrated with pictures. These were then sung and recited at gatherings in villages and taverns.
Explanation: Oral culture thus entered print and printed material was orally transmitted. The line that separated the oral and reading cultures became blurred. And the hearing public and reading public became intermingled.

Question 10.
Which of the foIIowing factors determines the use of Iand in India?
(a) PhysicaI factors such as topography
(b) Human factors such as popuIation density
(c) SpirituaI factors Iike Vastu
(d) Both (a) or (b) [1]
Answer:
(d) Both (a) or (b)

Question 11.
Choose the incorrect pair:

	Column A		Column B
(A)	Community Government	(I)	Power is shared among different social groups.
(B)	Coalition Government	(II)	Many political Government
(C)	Civil War	(III)	A war in which two countries participate.
(D)	Checks and Balances	(IV)	Horizontal system of power sharing.

Answer:
(c) Civil War-(III) A war in which two countries participate.
Explanation: A civil war is a violent conflict between opposing groups within a country that becomes so intense that it appears like a war. So, it is fought inside a country.

Question 12.
There are two statements marked as Assertion (A) and a Reason (R). Mark your answer as per the codes provided beIow: Assertion (A): Dams aIso fragment rivers making it difficuIt for aquatic fauna to migrate.
Reason (R): Dams are created to reroute rivers in the first pIace.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct expIanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct expIanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is faIse.
(d) Assertion (A) is faIse but Reason (R) is true. [1]
Answer:
(c) Assertion (A) is true but Reason (R) is false.
Explanation: Regulating and damming of rivers affect their natural flow causing poor sediment flow and excessive sedimentation.
They are not used to reroute rivers but to store water.

Related Theory:
Multi-purpose projects and large dams have also been the cause of many new environmental movements like the ‘Narmada Bachao Andolan’
and the ‘Tehri Dam Andolan’.

Question 13.
Choose the odd one out.
(a) Infant MortaIity Rate, Iiteracy Rates, Mean years of schooIing
(b) Average years of schooIing, MaternaI mortaIity rate
(c) Gross NationaI Income, Mean years of SchooIing at 25 years, Iife expectancy at birth
(d) BMI, Infant mortaIity rates, deveIopmentaI goaIs [1]
Answer:
(c) Gross National Income, Mean Years of Schooling at 25 years, Life expectancy at birth
Explanation: This option includes all the necessary indicators required to measure HDI by the UNDP. Rest of the options show indicators which are either not calculated or are not adequate for calculating the HDI.

Question 14.
Read the given tabIe about TrishaIa’s terms of credit agreement with the Bank of Bhopal.

Principal	1,00,000 rupees payable at the end of the third year directly
Rate of Interest	12% pa simple interest.
Duration of the loan	3 years
Collateral	Documents of the new car Trishala will buy.

CaIcuIate the amount TrishaIa owes to the bank at the end of three years.
(a) Rs 148000
(b) Rs 124000
(c) Rs 136000
(d) Rs 150000 [1]
Answer:
(c) 136000
Explanation: Simple Interest = $\frac{P \times R \times T}{100}$
Amount payable = P + SI
SI = $\frac{100000 \times 12 \times 3}{100}$
= Rs 36000
Amount = 100000 + 36000
Amount = Rs 136000

Question 15.
Rampur is an area where 80% peopIe borrow money from the bank whiIe 10% don’t borrow and the rest 10% take it from their friends, reIatives or IocaI moneylenders. Based on this arrangements where wiII Rampur be situated?
(a) In an Urban Region
(b) Semi-Urban area
(c) In a Rural region
(d) CapitaI of a country [1]
Answer:
(a) In an Urban Region

Question 16.
Which of the foIIowing countries is an exampIe of a country having the muIti-party system?
(a) USA
(b) SwitzerIand
(c) India
(d) United Kingdom [1]
Answer:
(c) India
Explanation: If several parties compete for power, and more than two parties have a reasonable chance of coming to power either on their own strength or in alliance with others, we call it a multi-party system.

Related Theory:
The United States of America and the United Kingdom are examples of two-party system.

Question 17.
Fill in the bIank by choosing the most appropriate option:
The BeIgian government shifted to a federal from a ………………… form of government.
(a) ParIiamentary
(b) Presidential
(c) Aristocratic
(d) Unitary [1]
Answer:
(d) Unitary
Explanation: In a federal form of govt, the state/regional government are not subordinate to the central government.

Related Theory:
In Belgium, in 1993, through constitutional amendment, the regional governments state were made independent of central government
completely.

Question 18.
How did the Bretton Woods system affect the gIobaI market of the 20th century?
(a) It caused extreme economic depression.
(b) It inaugurated an era of unprecedented growth of trade and incomes for the Western industriaI nations and Japan.
(c) It affected the trade fIows and caused extreme unempIoyment.
(d) It discouraged global investments and trade. [1]
Answer:
(b) It inaugurated an era of unprecedented growth of trade and incomes for the Western industrial nations and Japan.
Explanation: The Bretton Woods system inaugurated an era of unprecedented growth of trade and incomes for the Western industrial nations and Japan. World trade grew annually at over 8 per cent between 1950 and 1970 and incomes at nearly 5 per cent. The growth was also mostly stable, without large fluctuations.

Related Theory:
These decades also saw the worldwide spread of technology and enterprise. Developing countries were in a hurry to catch up with the advanced industrial countries.

Question 19.
UNDP can be expanded as:
(a) United Nations DeveIopment Programme
(b) United Nations Diversity Programme
(c) United Nations Democratic Programme
(d) United Nations Diversity Network Programme [1]
Answer:
(a) United Nations Development Programme
Explanation: The United Nations Development Programme is a United Nations agency tasked with helping countries eliminate poverty and achieve sustainable economic growth and human development.

Related Theory:
It is the largest UN development aid agency, with offices in 170 countries.

Question 20.
Read the foIIowing statements carefuIIy and identify the statements which is/are reIated to unorganised sector based on the codes that foIIow:
(I) A teacher taking cIasses in a schooI.
(II) A head Ioad worker carrying a bag of cement on his back in a market.
(III) A farmer irrigating her fieId.
(IV) A doctor in a hospitaI treating a patient.
(V) A daiIy wage Iabourer working under a contractor.
(VI) A factory worker going to work in a big factory.
(VII) A handIoom weaver working in her house.
Codes:
(a) I, III, V, VII
(b) I, II, IV, VI
(c) II, III, IV, VII
(d) II, III, V, VII [1]
Answer:
(d) II, III, V VII
Explanation: The unorganised sector is characterised by small and scattered units which are largely outside the control of the government. Jobs here are low-paid and often not regular. There is no provision for overtime, paid leave, holidays, leave due to sickness etc. employment is not secure.

Organised sector covers those enterprises or places of work where the terms of employment are regular and people have assured work. They get paid leaves, payment during holidays, PF, gratuity, pension after they retire, medical facilities, etc.

Section -B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
How did the IocaI peopIe in the areas conquered by NapoIion react to French ruIe? ExpIain.
OR
Why was the Swaraj Party formed? [2]
Answer:
Local people in the areas ruled by Napoleon reacted in the following ways:

Peasants, artisans, workers and businessmen enjoyed the freedom they had found after abolition of the feudal system.
However, as the initial enthusiasm wore off, people began hating his rule and army. They began to realise that for administrative reforms, their political freedom had been compromised.
Increased taxation, censorship, forced conscription into the French armies required to conquer the rest of Europe made them very hostile towards Napoleon and the French army. (Mention any 2 points)

OR
The Swaraj Party was formed by C.R. Das and Motilal Nehru within the Congress to contest for the council elections.
Leaders were tired of protesting for a long time and wanted to attempt to bring a change from inside the administration by participating in the council.

Question 22.
Mention in brief two characteristics of Joint Forest Management. [2]
Answer:
Two characteristics of the Joint Forest Management are:

It involves local communities in the management and restoration of degraded forests.
Local village people undertake protection activities mostly on degraded forest land managed by the forest department. In return, they are entitled to intermediary benefits like non-timber forest produces.

Question 23.
Deposits with the banks are beneficial to the depositors as weII as to the nation. Do you agree? ExpIain. [2]
Answer:
Banks benefit the depositors and the nation simultaneously:

Banks accept the deposits and pay interest to the depositors, like a bonus on their savings, the money deposited is in turn used to extend loans to the needy.
Banks keep the deposits of people safe and simultaneously they also provide capital for various small and big businesses, encouraging industrialists and native craftsmen, helping the economy of the country to grow.
Banks meditate between those who have surplus funds and those are in need of these funds, thus it helps in promoting equitable distribution. (Mention any 2 points)

Question 24.
How is economic inequaIity reIated to caste inequaIity? [2]
Answer:

Economic Inequality is closely connected to casteism because people belonging to disadvantaged castes are deprived of various economic resources in their everyday lives.
They are deprived of economic growth and development which pushes them into a vicious circle of poverty.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
Who organised the DaIits into the Depressed CIasses Association in 1930? Why did he cIash with Gandhi ji and what was the resuIt?
OR
Describe the events of French RevoIution which had infIuenced the peopIe beIonging to other parts of Europe. [3]
Answer:
Dr. B.R. Ambedkar who had himself been severely looked down upon due to his caste, raised the demand for separate electorates for Dalits which would confirm their representation in the associations.
(1) Dalits were highly discriminated against and Dalit leaders were afraid of being subjugated by majority of higher classes.

(2) They were sure they would be neglected completely and no heed will be paid towards their issues in the legislature unless they received proper representation in the lawmaking body.

(3) Thus, Ambedkar clashed with Mahatma Gandhi at the second Round Table Conference for Separate electorates for Dalits to ensure their true representation and also organised the Dalits into the Depressed classes association in 1930.
OR
Following are the events which affected the perspective of European people :
(1) With the news of the events of French Revolution reaching other cities, students and other members were encouraged to rebel for equality and liberty. The educated middle class began to set up Jacobin clubs in various cities.

(2) They began to organise activities and campaigns which made it easy for French armies to influence people.

(3) They helped the French armies to carry the idea of nationalism and spread it abroad. They encouraged the spread of these ideas through national symbols. A sense of collective identity was thus created.

Question 26.
“IndustriaIisation and urbanisation go hand in hand”. ExpIain and vaIidate the statement. [3]
Answer:
As soon as any industrial activity starts in a town, urbanisation follows.
(1) Industry provides employment to skilled and semi-skilled labourers in large numbers. Population migrates from rural hinterlands to seek jobs in these industries.

(2) Once mass migration occurs, the town develops its Housing and transportation facilities to accommodate these people.

(3) Schools, Colleges, Markets, Hospitals and other infrastructural facilities follow once the living standard of the inhabitants is capable enough to support them. Slowly, an uninhabited town becomes a fully functional city with all kinds of facilities available for its citizens.

Question 27.
ExpIain the three components of a ‘poIitical party’. [3]
Answer:
Three components of a political party are:
(1) The Leaders: Most prominent leaders, usually the founder, and most trustworthy who formulate policies and programmes of the party are chosen for contesting elections. These are the topmost base of every party.

(2) The Active Members: These people are involved in different committees of the party and participate directly in campaigning. They link the leaders to followers.

(3) The Followers: They believe in the party’s ideology and support the party by casting their votes in favour of the party at the time of elections. They are not associated with politics directly.

Question 28.
Why did the Indian government put barriers on foreign trade and foreign investments after independence? ExpIain. [3]
Answer:
Developmental goals can be different for different people in the following ways:
(1) Different people have different priorities for some job, house or food is important, for others respect, security at the workplace or opportunities to study.

(2) For example, a poor man may want a job to support his livelihood and his family while a businessman from a rich urban family would want to secure all possible kinds of luxury for his family and himself

(3) Developmental Goals can thus be different for people of different economic, mental and social standards. Thus what may be the developmental goal for one may not be a developmental goal for the other.

Question 29.
How can different peopIe have different development goaIs? [3]
Answer:
Indian Government put barriers against foreign trade and investments after independence because:

Indian products and producers needed protection against cheaper foreign alternatives.
As Indian industries were just coming up in 1950s and 1960s, the competition from more developed foreign industries at that stage would not have allowed these industries to come up.
To stabilize its economy, India needed to export more and import less. Putting barriers discouraged the people from buying imported goods thus helping India to make it balance of trade less negative.

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
How do democracies accommodate sociaI diversities? ExpIain any two conditions.
OR
Describe any five characteristic features of democracy. [5]
Answer:
Democrarcies accommodate social diversities in various ways. This can be asserted through the following statements:

Democrarcies provide solutions for living a harmonious social life even when the citizens are ethnically diverse. Belgium has done the same for its diverse population.
Democrarcy provides opportunity to negotiate differences and deliberate right choices for everyone.
Democrarcies provide inclusion for every race, ethnicity, caste, gender and creed.
Non Democratic regimes do not pay attention to internal social problems. They suppress them.
In a Democracy, opinions of the minorities are also heard and hence every voice is represented and accounted for.

OR
The five characteristics of democracy are as follows:

Democracy gives power to people to elect their government, control it and remove it from power. It ensures that all its citizens have a say in the functioning of the government.
No person holds all the power in a democracy. Different political parties compete with each other for authority and power. If the government does not fulfil its promises or fails to live up to the expectations of the people, then it is very likely that it will not get re-elected.
Democracy ensures that the rights of its people are protected by the state and the government functions according to the laws.
In a democratic set up, there is no discrimination between people based on race, religion, caste, colour or birth.
Criticism and feedback are two important mechanisms of democracy to control the government.
The government’s policies are scrutinised and evaluated by the people through the media and the opposition parties.

Question 31.
Compare the geographicaI conditions required for the growth of rice and wheat.
OR
“RaiIways are the principaI mode of transportation for freight and passengers in India.” Justify the statement with arguments. [5]
Answer:
(1) Rice is a kharif crop and requires a hot and humid climate for cultivation while Wheat is a rabi crop.

(2) Temperature above 25°C and high humidity with annual rainfall above 100 cm are favourable for growth of rice while wheat requires a cool growing season and a bright sunshine at the time of ripening.

(3) Rich alluvial soils of the floodplains and deltaic areas which are renewed every year are ideal for rice cultivation while wheat is grown in alluvial soil basins.

(4) Rice requires abundant rainfall or good water supply through irrigation and flooded fields during the earlier part of its growing season in June-July while Wheat requires 50 to 75 cm of annual rainfall evenly- distributed during its growing season.

(5) Rice requires plenty of cheap labour as most of the farming activities involves manual labour while wheat is not a labour intensive.
OR
Railways are the principal mode of trans-portation for freight and passengers in India as:
(1) The Indian railways is the largest public sector undertaking in the country.

(2) Trains cover long distances.

(3) Trains transport a large number of passengers as well as goods at the same time.

(4) Superfast passenger trains and goods trains provide comfortable journey.

(5) Freight trains transport heavy and bulky raw material to the manufacturing industries and finished goods to the market.

(6) They also help in strengthening national integration as people of different regions, languages and religion travel together and learn new ideas and ways of living from one another. (Mention any five points)

Question 32.
‘MusIims in India were Iukewarm in their response to Civil Disobedience Movement, Examine the statement.
OR
ExpIain the ways through which British manufacturers attempted to take over the Indian market. [5]
Answer:
Muslims felt alienated from Hindus due to the decline of the Khilafat-Non Cooperation Movement.

Congress became closer to be identified with radical Hindu groups like Hindu Mahasabha, etc.
The British tried to widen the gap between these communities by favouring one and alienating the others and hence, the communities grew further apart.
Religious protests and militant fervour, provoking communal sentiments and riots were organised in various cities.
Muslims were scared of being alienated and subdued by the Hindu Majority.
Thus they gave an extremely lukewarm response to the Non-cooperation movement.

OR
Industrialiste from Manchester, used advertisements to take over Indian market:

They put labels on cloth bundles before sending them to India which made the place of manufacture of the cloth and name of the company familiar to the buyer.
The “Made in Manchester” label was also supposed to be a mark of quality and guarantee.
Labels also carried images of Indian gods and goddesses printed on them to lure Indians.
The images of Krishna or Saraswati were also intended to make the manufacture from a foreign land appear familiar to people.
The manufactures also printed calendars with images of gods/goddesses (because they were used by people of all statuses) to popularize their products.
The figures of important personages also began to be used in advertisements and calendars. The message they conveyed was that when the product was being used by kings, or produced under royal command, its quality could not be questioned. (Mention any 5 points)

Question 33.
Has GIobaIisation equaIIy benefited peopIe from every economic cIass or status? Justify the statement with five arguments.
OR
Why is agricuIture the most Iabour absorbing sector in India? How does disguised unempIoyment make it worse? ExpIain with an exampIe. [5]
Answer:
No, Globalisation has not been able to proportionally benefit all classes and groups.
(1) Globalisation and greater competition among both local and foreign producers has been of great advantage to consumers. They have greater alternatives to choose from and a comparatively lesser price to pay.

(2) Investment by MNCs in certain sectors like automobiles, technology, fast food, banking and other such services have helped them massively.

(3) Globalisation has caused many problems for craftsmen, native artisans and small industrialists because of severe competition of quality and prices. Batteries, capacitors, plastics, toys, tyres, are some examples.

(4) Globalisation has created new opportunities for companies providing services, particularly those involving IT, it has endangered job security of labourers who, faced stiff competition, are willing to work flexibly even at lower costs than before.

(5) Sometimes the workers who work in shifts on a regular basis during the peak season and help MNCs to make large profits are denied their fair share of benefits brought about by globalisation.
OR
(1) Agriculture is the most Labour absorbing sector because there are not many jobs created in the tertiary and the secondary sector.

(2) Farming does not require a lot of initial investment or very hard training.

(3) Families are involved in Agriculture and hence, it becomes an inherited skill.

(4) Agricultural products have huge demand in the market and can be sold without any processing.

(5) Disguised unemployment which involves employment of more than the required people in a job reduces the productivity of the people and their effort gets divided without earning much.

Question 34.
Read the given source and answer the foIIowing questions:
She is foIIowed by the peopIes of Germany, bearing the bIack, red and goId fIag. InterestingIy, at the time when Sorrieu created this image, the German peopIes did not yet exist as a united nation – the fIag they carry is an expression of IiberaI hopes in 1848 to unify the numerous German-speaking principaIities into a nation-state under a democratic constitution.
(A) Which country is referred to as ‘she’ in the first sentence? [1]
(B) Name the image that the author is taIking about in the given source? [1]
(C) What can be clearly inferred about the notions of collective identity and common cuIture prevaIent among Europeans? [2]
Answer:
(A) This source mentions France as ‘she’ in the first line. France was followed by Germany in the picture of Sorrieu. France was just about to become a nation-state when the picture was printed and thus is seen approaching the Statue of Liberty which represents liberty from absolutist notions.

Related Theory:

Switzerland had already become a nation-state and thus has passed the Statue of Liberty as a symbol of gaining Liberty.
England was preceded by Austria, the Kingdom of the Two Sicilies, Lombardy and Poland in this journey.

(B) Frederic Sorrieu created an image titled as Democratic and Social Republics featuring various countries of the world in a march to the Statue of Liberty representing their struggle to convert to Republics from monarchical states. This paragraph talks about that visual series of four prints.

(C)

We can clearly infer that the Nations did not share a common identity or culture and were fragmented amongst them.
Countries in Eastern and Central Europe were ruled by autocratic monarchies with diverse people as citizens who did not follow similar religion or spoke the same language as described in the paragraph.

Related Theory:
These differences did not promote any political unity and they did not share any collective identity. Polish was spoken by the aristocracy of Galicia, German was spoken in Bohemia of the alpine region and French was spoken in France. People of Hungary spoke a variety of dialects in other parts of the country. However Magyar dominated the country.

Question 35.
Read the given source and answer the foIIowing questions:
The totaI voIume of workabIe mineraI deposits is an insignificant fraction i.e. one per cent of the earth’s crust. We are rapidIy consuming mineraI resources that required miIIions of years to be created and concentrated. The geoIogicaI processes of mineraI formation are so sIow that the rates of repIenishment are infiniteIy smaII in comparison to the present rates of consumption. MineraI resources are, therefore, finite and non-renewabIe. Rich mineraI deposits are our country’s extremeIy vaIuabIe but short-Iived possessions.

Continued extraction of ores Ieads to increasing costs as mineraI extraction comes from greater depths aIong with decrease in quaIity.
(A) Give a reason why mineraI resources are finite. [1]
(B) How does the continued extraction of ores affect the quaIity of mineraI resources? [1]
(C) Suggest two measures to conserve mineraI resources. [1]
Answer:
(A) Mineral resources are finite because the rates of their formation are very slow when compared to the rates of their usage. This means they are replenished slowly and should be conserved.

(B) Continued extraction of ores makes the ore/mineral extraction more expensive as they come from greater depths and are deteriorated in quality.

Metals should be recycled and upcycled to reduce dependence upon new extraction.
Improved technologies should be evolved to allow use of low grade ores at low costs.

Question 36.
Read the given source and answer the foIIowing questions:
The idea of power-sharing has emerged in opposition to the notions of undivided poIiticaI power. For a Iong time it was beIieved that aII power of a government must reside in one person or group of persons Iocated at one pIace. It was feIt that if the power to decide is dispersed, it wouId not be possibIe to take quick decisions and to enforce them.

But these notions have changed with the emergence of democracy. One basic principIe of democracy is that peopIe ruIe themseIves through institutions of seIf-governance. In a good democratic government, due respect is given to diverse groups and views that exist in a society. Everyone has a voice in the shaping of pubIic poIicies. Therefore, it foIIows that in a democracy poIiticaI power shouId be distributed among as many citizens as possibIe.
(A) What is the basic principIe of democracy? [1]
(B) Why shouId poIiticaI power be distributed in a democracy? [1]
(C) Why do peopIe ruIe themseIves through institutions of seIf-governance in a democracy? [2]
Answer:
(A) The basic principles of Democracy is that the people are the source of all political power.

(B) Political power should be distributed among various people and institutions in a democracy because:

This prevents the possibility of despotism and tyranny of the majority community.
This makes the Democratic government more representative and inclusive.

(C) People rule themselves through institutions of self-governance because people are the source of all power in a democracy. They self -govern as they impart power and authority to these institutions and the individuals governing them by pooling the will and sovereignty.

In other forms of government, the ruler/ despot or the powerful group rules over the people.

Section – F
Map Based Questions (2 + 3 = 5)

Question 37.
a. On the given poIiticaI map of India, identify the pIaces marked as A and B with the heIp of the foIIowing information and write their correct names on the Iines drawn near them.
(A) The pIace where the session of Indian NationaI Congress was heId in September 1920.
(B) A pIace in Gujarat where peasant Satyagraha was organised. [2]

b. On the same outIine map of India Iocate and IabeI any three of the foIIowing with suitabIe symboIs
(a) KaIpakkam – NucIear Power PIant
(b) New MangaIore – Seaport
(c) Pune – Software TechnoIogy Park
(d) Tehri – Dam [3]

Answer:
a. (A) Nagpur (B) Kheda
b. Located and labelled on the map

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 4 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A, B, C, D, and E.
Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
Section C has 6 Short Answer-ll (SA-II) type questions carrying 3 marks each.
Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section
Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section – A (20 marks)
(Section – A consists of 20 questions of 1 mark each.)

Question 1.
The HCF of 40 and 54 is: [1]
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Explanation:
The prime factorisations of 40 and 54 are:
40 = 2 × 2 × 2 × 5, or 2³ × 5¹
54 = 2 × 3 × 3 × 3, or 2¹ × 3³
So, HCF (40, 54) = 2¹, i.e. 2.

Question 2.
The value of k for which the polynomial 2kx² – 3kx + 7 has real roots is: [1]
(a) 35
(b) 34
(c) 19
(d) None of these
Answer:
(d) None of these

Explanation:
For real roots,
Discriminant > 0
⇒ b² – 4ac ≥ 0
Here, a = 2k, b = – 3k and c = 7.
∴ (- 3k)² – 4 × 2k × 7 ≥ 0
⇒ 9k² – 56 k ≥ 0
⇒ k (9k – 56) ≥ 0
But k ≠ 0
⇒ k > 0 or k ≥$\frac{56}{9}$

Question 3.
If the value of ‘x’ in the equation 2x + 3y = 13 is 2, then the corresponding value of y is: [1]
(a) 1
(b) 2
(c) 4
(d) 3
Answer:
(d) 3

Explanation:
Putting the value of x in the equation 2x + 3y = 13, we have
2(2) + 3y = 13
⇒ 4 + 3y = 13
⇒ 3y = 13 – 4 = 9
⇒ y = 3

Question 4.
The ratio in which x-axis divides the join of points (2, – 3) and (5, 6) internally is: [1]
(a) 2 :1
(b) 1 : 3
(c) 1 : 2
(d) 1 : 4
Answer:
(c) 1 : 2

Explanation:
Let P (x, 0) be a point on x-axis which divides A (2, – 3) and B (5, 6) in the ratio k : 1.
Then, by section formula
⇒ P(x, 0) = $\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)$
⇒ $\frac{6 k-3}{k+1}$ = 0
⇒ 6k – 3 = 0
⇒ k = $\frac{1}{2}$
⇒ Required ratio = k : 1 = $\frac{1}{2}$ : 1 = 1 : 2.

Question 5.
The ratio of the height of a tower and the length of its shadow is Vi: 1. The angle of elevation of the Sun is: [1]
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Answer:
(a) 60°

Explanation:
Let the angle of elevation be θ.
Then, tan θ = $\frac{B C}{A B}$

⇒ θ = 60°
So, the angle of elevation of the sun is 60°.

Question 6.
The value of sec² 60° cos 45° – cosec² 30° tan 45° is: [1]
(a) 2√2 – 4
(b) 3√2 + 2
(c) 2√2
(d) 7√2 – 2
Answer:
(a) 2√2 – 4

Explanation:
sec² 60° cos45° – cosec² 30° tan 45°
= (2)² × $\frac{1}{\sqrt{2}}$ – (2)² × 1
= 2√2 – 4

Question 7.
The 10^th term of the A.P.: 2, 7, 12, …………. is: [1]
(a) 47
(b) 36
(c) 45
(d) 30
Answer:
(a) 47

Explanation:
Here, a = 2, d = 5
So, a₁₀ = a + 9d
= 2 + 45 = 47

Question 8.
The sum of the first 10 multiples of 2 is: [1]
(a) 110
(b) 210
(c) 150
(d) 140
Answer:
(a) 110

Explanation:
First 10 multiple of 2 are 2, 4, 6, 8, ….. 20.
This is an A.P. with a = 2 and d = 2.
So, their sum = $\frac{10}{2}$ [2 × 2 + 9 × 2] = 110

Question 9.
A quadratic polynomial whose zeros are 2 and -5 is: [1]
(a) x² + 10x – 3
(b) x² + 3x – 10
(c) x² – 3x – 10
(d) x² + 5x + 4
Answer:
(b) x² + 3x – 10

Explanation:
A quardratic polynomial with sum and product of zeroes as S and P, respectively is given as,
x² – Sx + P
∴ Required polynomial is x² + 3x – 10.
[∵ sum of zeroes = -3 and product of zeroes = -10]

Question 10.
The sum of the digits of a 2-digit number is 10. A number is selected at random. The probability of the chosen number to be divisible by 3 is: [1]
(a) 1
(b) 3
(c) 4
(d) 0
Answer:
(d) 0

Explanation:
2-digit numbers, where the sum of the digits is 10, are 19, 28, 37, 46, 55, 64, 73, 82 and 91.
Of these nine numbers, no number is divisible by 3.
So, the required probability = $\frac{0}{9}$ i.e. 0.

Question 11.
The median of the given data is: [1]
2, 4, 6, 12, 3, 5, 10, 8, 2, 4, 9, 2, 10
(a) 4
(b) 9
(c) 5
(d) 10
Answer:
(c) 5

Explanation:
Rearranging the given data in ascending order, we have,
2, 2, 2, 3, 4, 4, 5, 6, 8, 9,10, 10, 12
Here, number of terms (n) = 13
∴ Median = $\left(\frac{n+1}{2}\right)^{\text {th }}$ term
= 7^th term = 5.

Question 12.
In a single throw of an unbiased die with 6 faces, what is the probability of getting a prime number ? [1]
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{1}{5}$
Answer:
(a) $\frac{1}{2}$

Explanation:
Out of the six numbers 1, 2, 3, 4, 5, 6, three are prime numbers, namely 2, 3 and 5.
So, P (a prime number) = $\frac{3}{6}$, i.e. $\frac{1}{2}$

Question 13.
If θ is the angle (in degrees) of a sector of a circle of radius ‘r’, then what is the area of the sector ? [1]
(a) $\frac{\theta}{360^{\circ}}$ × πr²
(b) $\frac{\theta}{360^{\circ}}$ × 2πr
(c) $\frac{360^{\circ}}{\theta}$ × πr²
(d) $\frac{360^{\circ}}{\theta}$ × 2πr
Answer:
(a) $\frac{\theta}{360^{\circ}}$ × πr²

Explanation:
Area of sector with angle θ and radius r is $\frac{\theta}{360^{\circ}}$ × πr².

Question 14.
A cylindrical pencil sharpened at one edge is the combination of which two solid figures ? [1]
(a) Sphere and cone
(b) Cone and triangle
(c) Cuboid and cone
(d) Cylinder and cone
Answer:
(d) Cylinder and cone

Explanation:

So, it a combination of cylinder and a cone.

Question 15.
If one zero of P(y) = 4y² – 8ky – 9 is negative of other, then the value of ‘k’ is: [1]
(a) 1
(b) 4
(c) 0
(d) 7
Answer:
(c) 0

Explanation:
Since one zero is the negative of the other zero, so the sum of two zeroes is 0.
⇒ $\frac{8 k}{4}$ = 0 ⇒ k = 0

Question 16.
The value of ‘k’ if the given system of equations 5x + ky = – 7 and x + 2y = 3 is inconsistent, is: [1]
(a) 10
(b) 20
(c) 15
(d) 7
Answer:
(a) 10

Explanation:
A pair of system of equations is inconsistent, it

Question 17.
AT is a tangent to circle with centres such that OT = 4 cm and ∠OTA = 30°. The length of AT is: [1]

(a) 2 cm
(b) √3 cm
(c) 4√3 cm
(d) 2√3 cm
Answer:
(d) 2√3 cm

Explanation:
Join OA,
Then ∠OAT = 90° (Since, OA ⊥ AT)
Now, cos 30° = $\frac{\mathrm{AT}}{\mathrm{OT}}$
⇒ $\frac{\sqrt{3}}{2}=\frac{A T}{4}$
⇒ AT = 2√3

Question 18.
If ∆ABC ~ ∆PQR, perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then the length of AC is: [1]
(a) 3 cm
(b) 6 cm
(c) 5 cm
(d) 4 cm
Answer:
(d) 4 cm

Explanation:
∴ ∆ABC ~ ∆PQR

Direction for questions 19 & 20: In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of Assertion (A)
(b) Both assertion (A) and reason (R) are true But reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Assertion (A): The polynomial 2x² + 14x + 20 have two zeroes.
Reason (R): A quadratic polynomial can have two or more than two zeroes. [1]
Answer:
(c) Assertion (A) is true but reason (R) is false.

Explanation:
Here,
2x² + 14x + 20 = 0
⇒ x² + 7x + 10 = 0
⇒ x² + 5x + 2x – 10 = 0
⇒ (x + 5) (x + 2) = 0
⇒ x = – 5, – 2

Question 20.
Assertion (A): The two tangents are drown to a circle from an external point, than they subtend equal angles at the centre.
Reason (R): A parallelogram circumscribing a circle is a rhombus. [1]
Answer:
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)

Explanation:
From an external point the two tangents drawn subtend equal angles at the centre. So A is true. Also, a parallelogram circumscribing a circle is a rhombus.

Section – B (10 marks)
(Section – B consists of 5 questions of 2 mark each.)

Question 21.
If HCF (150, 210) = 30, then find LCM (150, 210). [2]
Answer:
We know that
LCM (150, 210) = $\frac{150 \times 210}{{HCF}(150,210)}$
= $\frac{150 \times 210}{30}$ = 1050

Question 22.
Find the value of x for which 2x, (x + 10) and (3x + 2) are three consecutive terms of an A.P.
OR
If the first term of on A.P. is P and its common difference is q. then find its 6th term. [2]
Answer:
Since, 2x, (x + 10) and (3x + 2) are three consecutive terms of A.P.
∴ (x + 10) – 2x = (3x + 2) – (x + 10)
⇒ 10 – x = 2x – 8
⇒ 3x = 18
⇒ x = 6
OR
Let a be the first term and d be the common difference of the A.P.
Then, a = p and d = q
∴ 6th term = a + 5d
= p + 5q

Question 23.
Find a relationship between x and y such that the point (x, y) is equidistant from the points (3, 6) and (- 3, 4). [2]
Answer:
Let P (x, y), A (3, 6) and B (-3, 4).
As P is equidistant from A and B,
So PA = PB or PA² = PB²
i.e., (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
⇒ (x² – 6x + 9) + (y² – 12y + 36)
= (x² + 6x + 9) + (y² – 8y + 16)
⇒ 12x + 4y – 20 = 0.
or 3x + y – 5 = 0
which is the required relationship between x and y.

Question 24.
The shadow of a 5 m long stick is 2 m long. At the same time, find the length of the shadow of a 12.5 m high tree. [2]
Answer:
Let the length of a shadow of 12.5 m high tree x m
Now, ratio of lengths of objects = Ratio of lengths of their shadows
$\frac{5}{12.5}$ = $\frac{2}{x}$
x = $\frac{2 \times 12.5}{5}$ = $\frac{25}{5}$ = 5 m

Question 25.
The area of a circle is 154 sq. cm. Find its circumference.
OR
A bag contains 3 red and 5 blue balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(A) red?
(B) yellow?
Answer:
Let ‘r’ cm be the radius of the circle
Then, πr² = 154
⇒ r² = $\frac{154 \times 7}{22}$ = 49 ⇒ r = 7 cm
So, circumference
= 2πr = 2 × $\frac{22}{7}$ × 7 = 44 cm.

Total balls = 3 + 5 = 8
(A) P (red ball) = $\frac{3}{8}$
(B) P (yellow ball) = $\frac{0}{8}$, i.e. 0. (As there is no yellow ball in the bag.)

Section – C (18 marks)
(Section – C consists of 6 questions of 3 mark each.)

Question 26.
Find the greatest 4-digit number which is divisible by 15, 24 and 36. [3]
OR
Solve for x and y:
3x + 2y = 11, 2x + 3y = 4
Answer:
LCM (15, 24, 36) is the smallest number which is divisible by 15, 24, 36.
Also, every multiple of the LCM (15, 24, 36) is also divisible by 15, 24 and 36.
Now,
15 = 3 × 5
24 = 2³ × 3
36 = 2² × 3²
∴ LCM (15, 24, 36) = 2³ × 3² × 5 i.e. 360
The greatest 4-digit number which is the multiple of 360, is 9720 (27 × 360), which is the required number.

3x + 2y = 11 …….(i)
2x + 3y = 4 …… (ii)
Adding (i) and (ii), we get
5x+ 5 y = 15
⇒ x + y = 3 …… (iii)
Also, subtracting (ii) from (i), we get
x – y = 7 ……. (iv)
Adding (iii) and (iv), we get
2x = 10
⇒ x = 5
Putting the value of x in equation (iii), we get
5 + y = 3
⇒ y = – 2
∴ x = 5 and y = – 2.

Question 27.
Find the coordinates of the points of trisection of the line segment joining the points (2, – 2) and (- 7, 4). [3]
Answer:
Let, the given points be A and B. Then, A (2, -2) and B (-7, 4).
Let, P and Q be the two points of trisection of $\overline{\mathrm{AB}}$ such that, P divides $\overline{\mathrm{AB}}$ in the ratio 1 : 2; and Q divides $\overline{\mathrm{AB}}$ in the ratio 2 : 1.

Question 28.
In the figure, DEFG is a square and ∠BAC = 90°. Prove that [3]

(A) ∆AGF ~ ∆DBG
(B) ∆AGF ~ ∆EFC
(C) ∆DBG ~ ∆EFC
OR
Determine the A.P. whose 3^rd term is 5 and the 7^th term is 9.
Answer:
(A) Consider As AGF and DBG.
Here, GF || BD and AB is a transversal
So, ∠AGF = ∠DBG
(corresponding angles are equal)
Also, ∠GAF = ∠GDB (each is 90°)
So, by AA similarity criteria,
∆AGF ~ ∆DBG

(B) Consider A’s AGF and EFC.
Here FG || CE and AC is a transversal
So, ∠AFG = ∠FCE
(corresponding angles are equal)
Also, ∠FAG = ∠FEC (each is 90°)
So, by AA similarity criteria,
∆AGF ~ ∆EFC

(C) Since ∆DBG ~ ∆AGF [by (i)]
and ∆EFC ~ ∆AGF, [by (ii)]
we get ∆DBG ~ ∆EFC
OR
Let a be the first term and d be the common difference of A.P.
Then, a₃ = a + 2d = 5
and a₇ = a + 6d = 9
Solving these simultaneously, we get:
a = 3 and d = 1
Thus, the required A.P. is 3, 4, 5, 6, ………….

Question 29.
A quadrilateral ABCD circumscribe a circle. Prove that AB + CD = BC + DA. [3]
Answer:
We know, length of tangents, drawn from a same external points are equal.
So here,
AP = AS; BP = BQ; CQ = CR and DS = DR. ……….. (i)

Now, AB + CD = (AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS) [By eqn. (i)]
= (AS + DS) + (BQ + CQ)
= AD + BC, or BC + DA.

Question 30.
A automobile has wheels that are each 80 cm in diameter. When the car is moving at a speed of 66 km/h, how many complete rotations do each wheel complete in 10 minutes? [3]
Answer:
Diameter of wheel = 80 cm
⇒ Radius of wheel (r) = 40 cm
Distance covered by wheel in one revolution
= 2πr = 2 × $\frac{22}{7}$ × 40 = $\frac{1760}{7}$ cm
∵ Distance covered by wheel in 1 hour = 66 km = 66000 m = 6600000 cm
Distance covered by wheel in 10 minutes
= $\frac{6600000}{60}$ × 10 = 1100000 cm
No. of revolutions
= $\frac{\text { Total distance }}{\text { distance of one revolution }}$ = $\frac{1100000 \times 7}{1760}$
= 4375

Question 31.
A girl of height 90 cm is standing near a lamp-post. Now, she starts walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, then what is the length of her shadow after 4 seconds? [3]
Answer:
Speed of girl = 1.2 m/s
∴ In 4 seconds, travels

distance = 12 × 4 = 4.8 m
∴ After 4 seconds, she reaches at D.
∴ BD = 4.8 m
Let CD be the length of her shadow
Now, ∠ABD = ∠EDC = 90°
∴ AB || ED
Hence, by BPT
$\frac{A B}{E D}=\frac{B C}{D C}$
$\frac{3.6}{0.9}=\frac{4.8+x}{x}$
⇒ 4x = 4.8 + x
⇒ x = 1.6 m

Section – D (20 marks)
(Section – D consists of 4 questions of 5 mark each)

Question 32.
Solve for x:
$\frac{1}{x+4}-\frac{1}{x-7}$ = $\frac{11}{30}$,
OR
Using quadratic formula, solve for x:
3x² + 2√5x – 5 = 0. [5]
Answer:
We have

⇒ (x + 4) (x – 7) = – 30
⇒ x² – 3x – 28 = – 30
⇒ x² – 3x + 2 = 0
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1 (x – 2) = 0
⇒ (x – 1) (x – 2) = 0
⇒ x = 1 or 2

Using the quadratic formula, we have:

Question 33.
Izmir Clock Tower is a historic clock tower in Konak Square in the center of Izmir, Turkey. The French architect Raymond Charles Pere designed the Izmir Clock Tower.
Let us assume that the height of the tower AB = 14 m, height of tree CD = 5 m and BD – BC = 1 m. As the tower is vertical ∠ABC = 90°. Further, let us denote ∠CBD by ‘a’ and ∠BAD by ‘b’.

Find the value of sin a and tan b.
OR
The angle of elevation of the top of a tower from o point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower. [5]
Answer:
To find sin a, we will first find BD. It is given that BD – BC = 1 m and CD = 5 m. Therefore, applying Pythagoras theorem in triangle BCD, we get:
BD² = BC² + CD²
⇒ BD² = (BD – 1)² + 5²
⇒ BD² = BD² – 2BD + 1 + 25
Solving further, 2BD = 26, or BD = 13 m
Therefore, BC = 12 m.
In ∆BCD, sin a = $\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
= $\frac{C D}{B D}=\frac{5}{13}$
To find tan b, we will find AE and DE (drawn parallel to BC).

We construct DE || BC and we get a rectangle.
∴ AB = BE + AE
or 14 = 5 + AE
Therefore, AE = 9 m
and DE = BC = 12 m
Therefore, tan b = $\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{12}{9}$

Let P be the point of observation on the ground and TQ be the tower.
So, PQ = 30 m and ∠TPQ = 30°.
Let TQ = h metres
From rt. ∠d ∆PQT,
$\frac{\mathrm{TQ}}{\mathrm{PQ}}$ = tan 30°

⇒ h = $\frac{30}{\sqrt{3}}$
= 10√3 metres

Question 34.
As shown in the figure, two hemispheres of a solid metal ball ore divided in half and joined. The solid is positioned in a water-filled, cylindrical tub such that it is completely submerged. The cylindrical tub has a radius of 4 cm and a height of 11 cm, respectively. Additionally, a spherical ball has a 3 cm radius. Calculate how much water is left in the cylindrical tub. [5]

Answer:
Radius of spherical ball r = 3 cm
∴ Radius of each hemisphere, r = 3cm
Radius of cylinderical tub, R = 4 cm
Height of cylinderical tub, H = 11 cm
Volume of water left in tub = volume of water in cylinderical tub – Volume of the solid
= πR²H – 2 × $\frac{2}{3}$ πr³
= $\frac{22}{7}$ × 4 × 4 × 11 – $\frac{4}{3} \times \frac{22}{7}$ × 3 × 3 × 3
= $\frac{22}{7}$ (176 – 36)
= $\frac{22}{7}$ × 140
= 22 × 20
= 440 cm²

Question 35.
Find the mean, median and mode of the following frequency distribution: [5]

Answer:
Given, distribution is:

Mean X̄ = $\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{4225}{25}$
For median, n = 25
$\frac{n}{2}$ = 12.5
Then, median class is 150 – 200.
we have, l = 150, c.f. = 10, f = 6, h = 50

For mode
modal class = 150 – 200
f₀ = 5, f_i = 6, f₂ = 5, h = 50, l = 150
Mode = l + $\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)$ × h
= 150 + $\left(\frac{6-5}{12-5-5}\right)$ × 50
= 150 + $\frac{50}{2}$
= 150 + 25
= 175

Section – E (12 marks)
(Case Study-Based Questions)
(Section – E consists of 3 questions. All are compulsory.)

Question 36.
Due to ongoing COVID-19 crises, Surbhi Medical store has started stocking up and sell masks of decent quality as sourced from a disposable medical device manufacturer. The owner of Surbhi Medical store is selling two types of masks currently – A and B. The cost of one type A mask is ₹ 10 and of one type B mask is ₹ 12. In the month of April, 2020, the store sold 100 masks for total sales of ₹ 1082.

Due to great demand and short supply, the store has increased the price of each type by ₹ 1 from May 1, 2020 . In the month of May, 2020, the store sold 250 masks for total sales of ₹ 2920.

On the basis of the above information, answer the following questions:
(A) How many masks of each type were sold in the month of April? [1]
OR
How many masks of each type were sold in the month of May?
(B) If the store had sold 125 masks of each type, what would be its sale in month of May? [1]
(C) What percent of masks of each type sale was increased in the month of May, compared with the sale of month April? [2]
Answer:
(A) Let, the mask of type A sold in April be x and type of mask B sold in April be y.
Then, x + y = 100 …(i)
and 10x + 12y = 1082 …(ii)
Multiply equation (i) by 10 and subtract (ii) from (i).
10x + 10y = 1000
10x + 12y = 1082
-2y = – 82
y = 41
Then, x = 100 – 41 = 59

For May, Let, the mask of type A sold be x and type B be y.
Then, x + y = 250 …(i)
and 11x + 13y = 2920 …(ii)
Multiply equation (i) by 11 and subtract it from equation (ii), we get
11x + 11y = 2750
11x + 13y = 2920
– 2 y = – 170
y = 85
and x = 250 – 85 = 165

(B) 11 × 125 + 13 × 125 = 1375 + 1625
= ₹ 3000

Increase in type B = $\frac{85-41}{41}$ × 100
= 107.31%
= 110%

Question 37.
A game at a stall in Diwali fare involves using a spinner first as a pre-cursor to complete the game with certain rules. If the spinner stops at a particular number, then the player is allowed to roll a 6-faced unbiased die.

Rules:
(1) If the spinner stops at a particular number, then the player is allowed to roll a 6- faced unbiased dice.
(2) If the spinner stops at any other number, you get to try again and only two tries allowed maximum.
(3) If you reach the next stage and roll a dice, the shopkeeper will open a chit to disclose the number if it matches, the player gets a prize.

On the basis of the above information, answer the following questions:
(A) What is the probability of getting an odd number on the spinner? [1]
(B) If getting an even number on the spinner allows a player to roll the die, then find the probability of his rolling the die. [1]
(C) If the player is allowed to roll the die and getting a prime number entitles him to get prize, then find the probability of his winning the prize and if getting a square number on the spinner allows a player to roil the die, then find the probability of his rolling the die.
OR
If the player is allowed to roll the die and getting a number greater than 5 entitles him to get prize, then find the probability of his winning the prize. [2]
Answer:
(A) Total number of cases = 6
Favourable outcomes = (1, 9, 15) i.e., 3
P (dice will be thrown) = $\frac{3}{6} = \frac{1}{2}$

(B) Even number = 4, 8, 12
∴ P(getting in even number) = $\frac{3}{6} = \frac{1}{2}$

(C) Prime no. on dice = 2, 3, 5, i.e., (3) outcomes
∴ Total outcomes = 6
∴ P(getting a prime no.) = $\frac{3}{6} = \frac{1}{2}$
Total outcomes = 6
Favourable outcomes = {1, 4, 9} i.e., 3
P (dice will be thrown) = $\frac{3}{6} = \frac{1}{2}$
OR
Total outcomes = 6
Favorable outcomes is 6 i.e., 1
P(getting a no. greater than 5) = $\frac{1}{6}$

Question 38.
Radio towers are typically tall structures designed to support antennas for telecommunications and broadcasting, including television. There are 2 main types: guyed and self-supporting structures.
They are among the tallest human-made structures. Masts are often named after the broadcasting organizations that originally built them or currently use them.

On a similar concept, a radio-station tower was built in two sections A and B. From a point 24 m from the base of the tower, the angle of elevation of the top of section A is 30° and the angle of elevation of the top of section B is 45°.

On the basis of the above information, answer the following questions:
(A) Find the height of the section A. [1]
(B) Find the height of the section B. [1]
(C) Find the length of the wire structure from the point O to the top of section A.
OR
Find the length of the wire structure from the point O to the top of section B. [2]
Answer:
(A) In ∆AOC,
tan 30° = $\frac{A C}{O C}$
⇒ AC = 24 × $\frac{1}{\sqrt{3}}$
= 8√3
= 13.84 m

(B) In ∆BOC,
tan 45° = $\frac{B C}{O C}$
⇒ BC = OC
⇒ BC = 24 m
Now, AB = 24 – 13.84
= 10.16 m

(C) In ∆OAC,
cos 30° = $\frac{O C}{O A}$
⇒ $\frac{\sqrt{3}}{2}=\frac{24}{\mathrm{OA}}$
⇒ OA = $\frac{48}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
= 16√3
= 27.68 ≃ 27.7 m

OR
In ∆OBC,
cos 45° = $\frac{O C}{O B}$
⇒ $\frac{1}{\sqrt{2}}=\frac{24}{\mathrm{OB}}$
⇒ OB = 24√2
= 33.84 m

CBSE Sample Papers for Class 10 Social Science Set 10 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Social Science Set 10 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
Section A – From question 1 to 20 are MCQs of 1 mark each.
Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
Which of the following most statements appropriately describes underemployment?
(a) Workers are not paid as per their work.
(b) Worker are working less than what they are capable of doing.
(c) Workers are not skilled.
(d) Workers are not willing to work. [1]
Answer:
(b) Workers are working less than what they are capable of doing.
Explanation: When the work can be done with similar efficiency even if some workers are removed, the workers are said to be underemployed- which means they are working less than they are capable of, due to Lack ofjob opportunities.

Question 2.
Identify the country in which the principle of majoritarianism led to civil war.
(a) Belgium
(b) Sri Lanka
(c) Netherlands
(d) Germany [1]
Answer:
(b) Sri Lanka
Explanation: The majoritarian decisions taken by democratically elected Sinhala government against the minority- Sri Lankan Tamils-led to the development of sentiments of alienation amongst them and a civil war was thus caused where Tamils fought for a separate state EeLam.

Related Theory:
Belgium worked out its ethnic complications democratically and emerged out to be a harmonious state, ruled by law which is equal for all.

Question 3.
Which of the following is regulated by the institution featured in the picture?

(a) Formal Sector loans
(b) Informal Sector loans
(c) Domestic loans
(d) Loans for Marriage [1]
Answer:
(a) Formal Sector loans
Explanation: The Reserve Bank of India supervises the functioning of formal sources of loans. The RBI sees that the banks give loans notjust to profit-making businesses and traders but also to small cultivators, small scale.

Related Theory:
Banks have to submit information to the RBI on how much they are lending, to whom, at what interest rate, etc

Question 4.
Overgrazing is the main reason behind land degradation in which of the following states?
(a) Maharashtra
(b) Chhattisgarh
(c) Punjab
(d) Western Uttar Pradesh [1]
Answer:
(a) Maharashtra
Explanation: In states like Gujarat, Rajasthan, Madhya Pradesh and Maharashtra, overgrazing is one of the main reasons for land degradation.

Related Theory:
Mining sites are abandoned after excavation work is complete leaving deep scars and traces of over burdening. In states like Jharkhand, Chattisgarh, Madhya Pradesh and Odisha, deforestation due to mining has caused severe land degradation.

Question 5.
Identify this political party with the help of the following information.
Formed in 1999 following a split in the Congress party. Espouses democracy, Gandhian secularism, equity, social justice and federalism. Wants that high offices in government be confined to natural born citizens of the country.
(a) INC
(b) BJP
(c) NCP
(d) BSP [1]
Answer:
(c) NCP
Explanation: The NCP Nationalist Congress Party is a major party in Maharashtra and has a significant presence in Meghalaya, Manipur and Assam. It has been a coalition partner in the state of Maharashtra in alliance with the Congress in recent past. Since 2004, it has been a member of the United Progressive Alliance.

Caution:
Important details of most popular and important political parties should be memorised by heart because they can be often asked in form of identification questions

Question 6.
How can we increase the proportion of women in the important political and social institutions of the country?
(a) by educating women to participate more
(b) by declaring separate electorates for them
(c) by reserving seats for them at high level institutions
(d) by making these institutions open only for females [1]
Answer:
(c) by reserving seats for them at high level institutions
Explanation: One way to solve this problem is to make it legally binding to have a fair proportion of women in the elected or constituted bodies. This is what the Panchayati Raj Institutions have done in India. One-third of seats in local government bodies – in panchayats and municipalities – are now reserved for women.

Related Theory:
There are more than 10 lakh elected women representatives in rural and urban local bodies in India.

Question 7.
Match the items in column A with those of column B and find the most appropriate code which reflects the correctly matched pairs.

	Column A		Column B
(A)	Coir Industry	(I)	Odisha
(B)	Aluminium smelting plants	(II)	Singrauli
(C)	Software Park	(III)	Kerala
(D)	Thermal Power plant	(IV)	Noida

(a) A-(I), B-(III), C-(IV), D-(II)
(b) A-(III), B-(IV), C-(II), D-(I)
(c) A-(III), B-(I), C-(IV), D-(II)
(d) A-(IV), B-(III), C-(II), D-(I) [1]
Answer:
(c) A-(III), B-(I), C-(IV), D-(II)
Explanation: Coir industry is found in regions where coconut is easily available, i.e. in Kerala.

Question 8.
Which of the following statements related to Dandi March organised by Mahatma Gandhi is incorrect?
(a) Mahatma Gandhi started the Dandi March from Sabarmati Ashram.
(b) Dandi March is also known as Salt March
(c) Dandi March was started on 11 March, 1930
(d) Mahatma Gandhi was accompanied by 72 of his trusted members. [1]
Answer:
(d) Mahatma Gandhi was accompanied by 72 of his trusted members.
Explanation: Gandhiji was followed by 78-80 of his followers and disciples in his march to break the infamous Salt law.

Related Theory:
Dandi March began on 11th-12th March, 1930 and lasted for 24 days, culminating on 6th April, 1930 in Dandi.

Question 9.
Fill in the blank marked by A by choosing the most appropriate option:

Name of the Book	Sacchi Kavitayen	Istri Dharam Vichaar
Author	Sudarshan	A

(a) Kashibaba
(b) Ram Chaddha
(c) Bal Gangadhar Tilak
(d) Raja Ravi Varma [1]
Answer:
(b) Ram Chaddha
Explanation: Ram Chaddha published the fast-selling Istri Dharm Vichar to teach women how to be obedient wives. The Khalsa Tract Society published cheap booklets with a similar message.

Question 10.
What is also known as the economic barometer of the country?
(a) Communication
(b) International trade
(c) Waterway transport
(d) Spice trade [1]
Answer:
(a) International trade
Explanation: Advancement of international trade of a country is an index to its economic prosperity. It is, therefore, considered the economic barometer for a country.

Related Theory:
As the resources are space bound, no country can survive without international trade. Export and import are the components of trade.

Question 11.
According to the classification by the World Bank, based on per capita income, choose the correctly matched pair.
(a) Middle income group – USA
(b) Low income group – Saudi Arabia
(c) High income group – Bangladesh
(d) Low middle income group – India [1]
Answer:
(d) Low Middle Income group
Explanation: World Bank uses Per-Capita Income to compare development in various countries. Countries with per capita income- 12056 $ and above are Rich Countries. Countries with 955$ or less as their per capita income are called Low Income Countries.

Related Theory:
India has a per capita income of 1820 $m in 2017 and hence is better than low income group but not as good as middle or high income countries.

Question 12.
There are two statements marked as Assertion (A) and Reason (R). Mark your answer as per the codes provided below:
Assertion (A): It was felt that if the power to decide is dispersed, it would not be possible to take quick decisions and to enforce them. Reason (R): Unstable governments cannot take decisions easily and quickly.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
Explanation: In a good democratic government, due respect is given to diverse groups and views that exist in a society. Everyone has a voice in the shaping of public policies. Getting every person to agree and come to a common conclusion takes a Long time and therefore decision making in democracies is a long process.

The idea of power sharing has emerged in opposition to the notions of undivided political power.

Question 13.
Which institution is headquartered in Brussels, Belgium?
(a) African Union
(b) WHO
(c) UNICEF
(d) European Union [1]
Answer:
(d) European Union
Explanation: The Belgium model of Administration has been able to keep Brussels away from all kinds of social and political conflicts due to its accomodation and inclusiveness.

Related Theory:
The accomodation in Belgium helped to avoid civic strife between the two major communities and a possible division of the country on linguistic lines.

Question 14.
Which of the following acts is NOT applicable to in the unorganised sector?
(I) Factories Act
(II) Minimum Wages Act
(III) Shops and Establishments Act
(IV) Equal Remuneration Act Codes:
(a) I, II and IV
(b) II and III
(c) III and IV
(d) I, II, III and IV [1]
Answer:
(d) I, II, III and IV
Explanation: Organised sectors registered by the government and have to follow its rules and regulations which are given in various laws such as the Factories Act, Minimum Wages Act, Payment of Gratuity Act, Shops and Establishments Act etc.

Related Theory:
Organised sectors is called so because it has some formal processes and procedures.

Question 15.
Choose the incorrect pair.
(a) Eldorado-City of Diamond
(b) Europe-Centre of world trade
(c) London-Financial centre
(d) Punjab-Canal colonies [1]
Answer:
(d) Eldorado- City of Diamond
Explanation: Eldorado was a fictional city of gold. Legends spread in seventeenth-century Europe about South America’s fabled wealth where El Dorado was supposed to be.

Related Theory:
From the fifteenth century, China’s reduced role and the rising importance of the Americas gradually moved the centre of world trade westwards. Europe now emerged as the centre of world trade.

Question 16.
Fill in the blank by choosing the most appropriate option.
Entry of an MNC in a domestic market may prove harmful for ………………..
(a) large scale producers
(b) all domestic producers
(c) all international producers
(d) all sub-standard small-scale producers [1]
Answer:
(d) all sub-standard small-scale producers
Explanation: Batteries, capacitors, plastics, toys, tyres, dairy products, and vegetable oil are some examples of industries where the sub-standard small manufacturers have been hit hard due to competition

Question 17.
Which one of the following is a major benefit of joint production between a local company and an MNC?
(a) MNCs supply new technology through such mergers
(b) MNCs can control the increase in the price through joint production of goods
(c) The local company can buy all the technology from the MNC
(d) MNC can sell the products under their brand name. [1]
Answer:
(a) MNCs supply new technology through such mergers
Explanation: The benefit to the local company of such joint production is two-fold. First, MNCs can provide money for additional investments, like buying new machines for faster production. Second, MNCs might bring with them the latest technology for production.

Related Theory:
MNC invests to buy up local companies and then to expand production

Question 18.
The 1929 Lahore Session is famous for which of the following decisions/declarations?
(a) Declaration of Non-cooperation movement
(b) Declaration of Independence
(c) Declaration of Purna Swaraj Resolution
(d) Calling off of Non-Cooperation Movement [1]
Answer:
(c) Declaration of Purna Swaraj Resolution
Explanation: Lahore Congress formalised the demand of ‘PurnaSwaraj’ after continuously being dejected by the multiple vague concessions offered by British to call off the on-going revolutionary movements.

Related Theory:
Non-Cooperation movement was adopted in 1920 while it was called off in 1922. India’s attainment of independence was declared in 1947.

Question 19.
Identify the institution mentioned through the given hints:
(1) They raise and highlight issues.
(2) They have lakhs of members and activists spread all over the country.
(3) Many of the pressure groups are the extensions of these institutions among different sections of society.
(a) Banks
(b) Political porties
(c) Interest group
(d) Cooperative societies [1]
Answer:
(b) Political Parties
Explanation: Parties shape public opinion. They raise and highlight issues. Parties have lakhs of members and activists spread all over the country.

Related Theory:
Many of the pressure groups are the extensions of political parties. Parties sometimes also launch movements for the resolution of problems faced by people.

Question 20.
Which of the following statements about the early entrepreneurs from India is (are) True?
(I) The history of many business groups goes back to trade with America.
(II) The British in India began exporting opium to China and took tea from China to England.
(III) In Bengal, Dwarkanath Tagore made his fortune in the China trade before he turned to industrial investment, setting up six joint-stock companies in the 1830s and 1840s.
(IV) Seth Hukumchand set up the first Indian jute mill in Calcutta in 1917.
Codes:
(a) I and IV
(b) I, II and III
(c) I and III
(d) II, III and IV [1]
Answer:
(d) II, III and IV
Explanation: The history of many business groups goes back to trade with China. The British in India began exporting opium to China and took tea from China to England. Dwarkanath Tagore made his fortune in the China too.

Related Theory:
Till the First World War, European Managing Agencies regulated and controlled a sector of Indian industries. Three of the biggest ones were Bird Heiglers & Co., Andrew Yule, and Jardine Skinner & Co. These agencies mobilised capital, set up joint-stock companies and managed them. Indian financiers provided the capital while the European Agencies made all investment and business decisions.

Section – B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
What is per capita income? Can per capita income be considered the real income of a citizen? [2]
Answer:
To calculate income of every citizen in a country is difficult therefore an average income is calculated by dividing the total income of the country by its total population.

This average income is also known as per capita income or average income of every citizen in a country.
Per capita income however is not the real income of a citizen but an estimate. Per capita income doesn’t prove to be a reliable unit to measure whether or not a citizen is developing in a country.

Question 22.
Why did Ambedkar establish the Depressed Class Association in 1930? [2]
Answer:
Ambedkar organised the Dalits into the Depressed Classes Association in 1930 because:

To help the disadvantaged community to organise rallies, debates and open conversations to attain appropriate rights and opportunities
To support his demand for Separate Class electorates over which he had been clashing with Gandhi.

Question 23.
Mention two characteristics of the textiles industry.
OR
How are industrial locations chosen? What factors are kept in mind? [2]
Answer:
Two characteristics of the textile industry are:

It is self reliant and compliant in its value chain.
It contributes significantly in employment generation and revenue collection.

OR
Industrial locations are complex to find.

They are influenced by availability of raw materials, labour, capital, transportation, power and proximity to the market, etc.
It is rarely possible to find all these factors available at one place. Factories are located where all the factors of production are available at the lowest cost.

Question 24.
How has the creation of linguistic states affected the unity of the country? [2]
Answer:
The formation of linguistic states has actually made the country more united because it has helped recognise the differences based on geography, language or culture of the states.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
The Second World War caused an immense amount of economic devastation and social disruption. Reconstruction promised to be long and difficult. What are the two factors which influenced this reconstruction? [3]
Answer:
Two crucial influences shaped post-war reconstruction:

The first was the emergence of the United States as the dominant economic, political and military power in the Western world.
The dominance of the Soviet Union also influenced the post war reconstruction. Russia was a backward agricultural country which suddenly developed itself into a world power.
The dynamics between these countries affected the post war world reconstruction as it divided the world into two/three blocs. Countries who supported the US on one side, parts of the Soviet Union on the other side and the Non-aligned bloc on the third side.

Question 26.
How is transparency ensured in a democracy? [3]
Answer:
To ensure Transparency in a democracy, the Constitution and our law makers have made several provisions.
(1) Democracy ensures that decision making process is based on strict rules and norms written in the constitution. The process is transparent at each step for any citizen to inquire and understand.

(2) A citizen has a right through the Right to Information Act to inquire about any political process or person involved unless the information being asked is hidden for the safety and maintenance of integrity of the country.

(3) A citizen can also scrutinize all the decision along with the actions of their decision makers and criticise the same through multiple platforms. He or she can also use the same analysis to vote for the said candidate in the next election.
OR
Yes, the three sectors namely, Primary, Secondary and Tertiary are highly interdependent on each other.
(1) The secondary sector which basically manufactures or processes goods produced by the primary sector- for example, makes biscuit, bread by grounded wheat(flour), constructs buildings by cement, tar, stones is heavily dependent on primary sector. The Primary Sector is dependent on Secondary sector for the processing of its raw material into finished products.

(2) The tertiary sector, that advertises, sells, transports products of both primary and secondary sector is thus dependent on these sectors as well.

(3) Both Primary and Secondary sectors are dependent on Tertiary sector for capital to invest and continue their production. Hence all three sectors are interdependent on each other to work successfully.

Question 27.
What are the differences between formal and informal sources of credit?
OR
Economic activities, though, grouped into three different categories, are highly interdependent. Do you agree? Elaborate. [3]
Answer:
The difference between formal and informal sources of credit are as follows:

Basis of Distinction	Formal Sources of Credit	Informal Sources of Credit
(1) Need of Collateral	Collateral is essential for borrowing money.	Most informal sources lend without collateral or proper agreement.
(2) Rate of Interest	Reasonably Low	Exorbitantly High
(3) Safety of Borrower	Backed by the Government, monitored by RBI. Extremely safe.	No monitoring agency backs the loan. The lenders turn exploitative most of the times.

OR
Yes, the three sectors namely, Primary, Secondary and Tertiary are highly interdependent on each other.
(1) The secondary sector which basically manufactures or processes goods produced by the primary sector- for example, makes biscuit, bread by grounded wheat(flour), constructs buildings by cement, tar, stones is heavily dependent on primary sector. The Primary Sector is dependent on Secondary sector for the processing of its raw material into finished products.

(2) The tertiary sector, that advertises, sells, transports products of both primary and secondary sector is thus dependent on these sectors as well.

Question 28.
Write a note on Greek war of Independence. [3]
Answer:
Constructing sophisticated hydraulic structures like dams built of stone rubble, reservoirs or lakes, embankments and canals for irrigation was a common practise during ancient India. Three of them are:

In the first century B.C., Sringaverapura near Allahabad had sophisticated water harvesting system channelling the flood water of the river Ganga.
During the time of Chandragupta Maurya, dams, lakes and irrigation systems were extensively built in his reign.
In the 11th Century, Bhopal Lake was one of the largest artificial lakes of its time was built.

Question 29.
Mention three hydraulic structures found in Ancient India. [3]
Answer:
Greek war of Independence mobilised nationalist feelings among the educated elite across Europe. It had been a part of the Ottoman Empire since the fifteenth century. Fuelled by the growing revolutionary nationalism and liberalism in the 19th century a spark for independence was lit among Greeks in 1821. Nationalist Greeks were rendered huge supports from all ends. To invoke nationalism against a Muslim rule, contemporary literature was used. After a long struggle and fierce war, the treaty of Constantinople in 1832 recognised Greece as an independent state.

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
How did the officials of East India Company procure regular supplies of cotton and silk textiles from the Indian weavers?
OR
‘British rule in India would have collapsed within a year, if Indians would have refused to cooperate.’ Elaborate upon this statement said by Gandhi ji. [5]
Answer:
These are the following ways in which officials of East India Company procures regular supplies of cotton and silk textiles from the Indian weavers:
(1) After the East India Company established political power, it asserted a monopoly right to trade.

(2) The company proceeded to develop a system of management and control that would eliminate competition, control costs, and ensure regular supplies of cotton and silk goods.

(3) The company tried to eliminate the existing traders and brokers connected with the cloth trade, and appointed a paid servant called the Gomostha who supervised the weavers, collect supplies, and examine the quality of cloth.

(4) To have a direct control over the weavers, the company started the system of advances. After the advances were given to farmers, raw material was supplied to them and they were bound to supply everything produced to the Gomastha. They could not take it to any other trader.

(5) The places where the weaver refused to cooperate with the Company, it used its police and power. Weaver were often beaten and flogged for delays in supply
OR
(1) Mahatma Gandhi understood that Britishers were only able to rule Indian people because of Indians silent cooperation and their incapability to United we stand against the exploitation meted out to them.

(2) He declared that if Indians refused to cooperate, British rule in India would collapse within a year and thus proposed Non cooperation movement.

(3) As the first step, surrendering of titles that the government had awarded to the Indians, complete boycott of government controlled institutions like civil services, army, police, courts and legislative assemblies, schools and foreign goods began.

(4) Mahatma Gandhi suggested Indians to promote native craftsmen and artisans and to wear Indian clothes and use Indian words instead of important goods from Britain.

(5) Boycott of elections, titles and other important offices, silent protests by marches were the next steps taken in this direction. This discovery led to a pan- India movement which later became the foundation to the birth of nationalism in the country.

Question 31.
Name the most significant ferrous metal which can be also called as the backbone of industrial development. Mention its uses and availability.
OR
Underline the significance of the role played by Wildlife Protection Act of 1972 in protecting wildlife resources.[5]
Answer:
Iron ore is the most significant and basic mineral and the backbone of industrial development.
Its feature and uses are:

India is endowed with fairly abundant resources of iron ore.
Magnetite with an extremely valuable iron content is heavily used in the electrical industry. Haematite is the most commonly used type of iron-ore. It is also called industrial iron ore. Limonite and Sidrite are used for various other purposes.
Iron ore is mostly used for steel production which again forms the basis of any construction in the country.
Almost the entire production of iron ore (97%) is accrued from Odisha, Chhattisgarh, Karnataka and Jharkhand.
Iron ore is used for manufacturing of machines, agricultural equipment and multiple other goods.

OR
The contribution of WPA of 1972 in protecting wildlife resources is:

The act has enabled the creation of a pan-India list of protected species.
Central and State governments have established multiple wildlife sanctuaries and national parks to protect wildlife resources.
Government has also announced various pan-Indian protection programmes to accord legal sanction and protection to animal species.
This act has also helped in giving legal sanction to the protection campaigns run by local communities and rural people.
The act has helped in conserving various species under threat protecting their habitats and ecological niches.

Question 32.
Explain the factors which facilitate Globalisation.
OR
Describe any five factors that promote multinational corporations (MNCs) to set up their production units at a particular place. [5]
Answer:
The factors which facilitate Globalisation are:
(1) Rapid improvement in Technology which has stimulated the process of Globalisation largely.

(2) Liberalisation of foreign trade and foreign investment policies has helped more traders and investors to come up and invest, boosting the economy of various countries together simultaneously.

(3) Pressure from international organizations like WTO and World Bank has led to countries opening up their markets to foreign investors and companies. This has made such markets accessible.

(4) Improvement in transportation and communication facilities has helped in easy transportation of goods, final products or raw materials easily to the farthest distances.

(5) Dependence of developing countries upon developed countries for goods, technology and money has stimulated the developed countries to explore and invest in the markets of developing countries in exchanged of goods and services provided buy them- thus making it a win-win situation for both countries.
OR
Factors that promote multinational corporations (MNCs) to set up their production units at a particular place are:
(1) Proximity to the markets as it helps them to sell their products easily.

(2) Availability of skilled and unskilled labourers at cheap rate which help them in cutting costs and aiding in production work.

(3) Presence of favourable government policies looking after their interest or which could support their functioning.

(4) Availability of other factors of production such as raw materials, water, electricity and transport, as it eases the production and transportation of finished goods.

(5) Presence of standard safety measures for assured production as it helps them to grow and stay assured.

Question 33.
Decentralisation can be taken as means to “Sexual division of labour is not based on minimise conflicts. Explain.
OR
“Sexual division of labour is not based on biology but on social expectations and stereotypes.” Support the statement. [5]
Answer:
(1) Decentralisation helps to settle various of problems and issues at the local level because more local people are involved in this tier of Government.

(2) It provides a platform for the direct participation of people in decision-making.

(3) In another way, decentralisation in the form of “local self-government’ is the best way to realise the principles of democracy.

(4) It saves from accumulation of power in the hands of one community, caste or economic group and thus is more representative. The minorities do not feel ill-represented or neglected.

(5) Since Local people become Leaders and representatives of the area, it makes it simpler for people to convey their aspirations and grievances to them, resolving conflicts in the initial stages only.
OR
Sexual division is not based on biology. This can be asserted as follows:
(1) Women are believed to do housework and look after children and take important decisions keeping family wellbeing in mind but they are stopped from assuming decision making roles because they are believed to be emotional.

(2) Traditional roles of women have been performed by men in recent times.

(3) Role of women in public life, especially in politics is minimal because it is believed that they are weak and cannot handle the stress.

(4) Only men were allowed to work in public affairs, women are not given equal opportunities to prove themselves outside of their physiological differences.

Section – E
Case Based Questions (4 x 3 = 12)

Question 34.
Read the given source and answer the following questions:
The contiguous stretch of Sahyadri could be crossed only through gaps or passes (Ghats). It has also faced a number of problem such as sinking of track in some stretches and landslides. Today, they have become more important in our national economy than all other means of transport put together. However, this type of transport suffers from certain problems as well. Many passengers travel without tickets. Thefts and damaging of property has not yet stopped completely. People stop the vehicles, pull the chain unnecessarily and this causes heavy damage to the transports.
(A) Identify the type of transport mentioned in the given source. [1]
(B) Mention two problems associated with this type of transport. [1]
(C) Mention two benefits of this type of transport vis-a-vis Airways. [2]
Answer:
(A) The type of transport mentioned in the given source is Railways. Railways includes tracks and streches.

(B) Two problems being faced by this type of transportation are:

People travel without tickets and misbehave in the trains with authorities and other passangers.
Chain pulling, destruction of property and tracks are also important issues which affect railways.

Railway transportation is cheap and affordable.
Railways connect the remotest regions in the country. The initial capital and maintenance is not very expensive.

Question 35.
Read the given source and answer the following questions:
Most of the major parties of the country are classified by the Election Commission as ‘State parties’. These are commonly referred to as regional parties. Yet these parties need not be regional in their ideology or outlook Some of these parties are all India parties that happen to have succeeded only in some states.. over the last three decades, the number and strength has expanded.

This made the Parliament of India politically more and more diverse. As a result, the national parties are compelled to form alliances with State parties. Since 1996, nearly every one of the state parties has got an opportunity to be a part of one or the other national level coalition government. This has contributed to the strengthening of federalism and democracy in our country.
(A) What is a State party? [1]
(B) Name four state parties popular in India. [1]
(C) What has contributed to the strengthening of federalism and democracy in our country? [2]
Answer:
(A) A party which is active in a particular region or state is called a State party. For example, Shiv Sena is an example of a state party, active in the state of Maharashtra.

Related Theory:
State parties are also known as regional political parties.

(B) Four State parties are:

Trinamool Congress Party
Akalidal
Shiv Sena
AIADMK

(C) Every political party has got an opportunity to be a part of one or the other national level coalition government. This has led to the strengthening of democracy in our country. Citizens have been able to join regional parties and bring up their local demands into friction. Since 1996, there has been a trend of coalition governments in India in the centre particularly.

Related Theory:
When more than two political parties join hand together to win the election and form the government after the election is called a coalition government. For example, the Modi government at the centre in India was a coalition government in 2014-2019.

Question 36.
Read the given source below and answer the following questions:
We have seen that money is something that can act as a medium of exchange in transactions. Before the introduction of coins, a variety of objects was used as money. For example, since the very early ages, Indian used grains and cattle as money. Thereafter came the use of metallic coins—gold, silver, copper coins-a phase which continued well into the last century.
(A) Mention four forms of modern currency. [1]
(B) Mention an advantage of using the faa’lity of cheque against demand deposits. [1]
(C) Why can money act as a medium of exchange? [2]
Answer:
(A) Four forms of modern currency are Cheque, Credit Cards, Digital money, notes and coins.
Related Theory:
A cheque is a paper instructing the bank to pay a specific amount from the person’s account to the person in whose name the Cheque has been issued while deposits in the bank that can be withdrawn on demand and used are called demand deposits.

(B) Payments can directly be settled without the use of cash by using a cheque since demand deposits are accepted widely as a means of payment, along with currency, they constitute money in the modern economy.

(C) Since money acts as an intermediatry in the exchange process, introduction on money has solved the problem of double coincidence of wants-the main problem of barter system. Money has been accepted as the medium of exchange because the currency is authorized by the government of the country.

Section – F
Map Based Questions (2 + 3 = 5)

Question 37.
(a) On the given political map of India, identify the places marked as A and B with the help of the following information and write their correct names on the lines drawn near them.
(A) The place where the peasants struggled against the Indigo Plantation system.
(B) The place where a session of Indian National Congress was held in September 1920. [2]

(b) On the same outline map of India locate and label any three of the following with suitable symbols.
(a) Tarapur-Nuclear power plant
(b) Tungabhadra – Dam
(c) Kandla – Seaport
(d) Bengaluru – Software Technology Park [3]

Answer:
(a) (A) Champaran (B) Nagpur
(b) Located and Labelled on the map

CBSE Sample Papers for Class 10 Maths Basic Set 3 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 3 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A, B, C, D, and E.
Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
Section C has 6 Short Answer-ll (SA-II) type questions carrying 3 marks each.
Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section
Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section – A (20 marks)
(Section – A consists of 20 questions of 1 mark each.)

Question 1.
The value of $\frac{\cot A+\tan B}{\cot B+\tan A}$ is: [1]
(a) cot A tan B
(b) cot B tan A
(c) cot A
(d) tan B
Answer:
(a) cot A tan B

Explanation:

Question 2.
For any two positive integers ‘a’ and ‘b’, what is the value of HCF (a, b) x LCM (a, b)? [1]
(a) ab²
(b) a²b
(c) a × b
(d) a + b
Answer:
(c) a × b

Explanation:
We know the relation,
HCF (a, b) × LCM (a, b) = Product of a and b.

Question 3.
The mean of first 10 odd natural numbers is: [1]
(a) 20
(b) 40
(c) 30
(d) 10
Answer:
(d) 10

Explanation:
First ten odd numbers are: 1, 3, 5, ………. 19.
Their sum = $\frac{10}{2}$ (2 × 1 + 9 × 2) = 100
Mean = $\frac{100}{10}$ = 10

Question 4.
Using prime factorisation method, the HCF and LCM of 210 and 175 is: [1]
(a) 35, 1000
(b) 34, 1050
(c) 35, 1050
(d) 35, 1010
Answer:
(c) 35, 1050

Explanation: The prime factorisations of 210 and 175 are:
210 = 2 × 3 × 5 × 7
175 = 5 × 5 × 7
So, HCF (210, 175) = 5 × 7 = 35; and
LCM (210, 175) = 2 × 3 × 5 × 7 × 5 = 1050

Question 5.
If 2x, x + 10, 3x + 2 are in A.P., the value of x is: [1]
(a) 6
(b) 7
(c) 9
(d) 10
Answer:
(a) 6

Explanation:
Since 2x, x + 10, 3x + 2 are in A.P.,
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 = 5x + 2
⇒ 3x = 18
⇒ x = 6

Question 6.
By taking A = 90° and B = 30°, sin A cos B – cos A sin B is: [1]
(a) 0
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) $\frac{1}{\sqrt{2}}$
Answer:
(b) $\frac{\sqrt{3}}{2}$

Explanation:
sin A cos B – cos A sin B
= sin 90° cos 30° – cos 90° sin 30°
= 1 × $\frac{\sqrt{3}}{2}$ – 0 × $\frac{1}{2}$ = $\frac{\sqrt{3}}{2}$

Question 7.
The degree of the polynomial (x + 1) (x² – x – x⁴ + 1) is: [1]
(a) 4
(b) 2
(c) 5
(d) 3
Answer:
(c) 5

Explanation:
(x + 1)(x² – x – x⁴ + 1) = (x³ – x² – x⁵ + x) + (x² – x – x⁴ + 1)
= – x⁵ – x⁴ + x³ + 1
So, the degree of the polynomial is 5.

Question 8.
The ratio of the volume of a right circular cone to that of the volume of right circular cylinder, of equal diameter and height is: [1]
(a) 4 : 3
(b) 1: 4
(c) 2 : 3
(d) 1: 3
Answer:
(d) 1 : 3

Explanation:
Volume of a right circular cone = $\frac{1}{3}$ ≠ r²h
Volume of a right circular cylinder = πr²h
So, required ratio is 1: 3.

Question 9.
The value of (cos 0° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°) is: [1]
(a) $\frac{1}{4}$
(b) $\frac{3}{4}$
(c) $\frac{5}{4}$
(d) $\frac{7}{4}$
Answer:
(d) $\frac{7}{4}$

Explanation:
(cos 0° + sin 45° + sin 30°)
(sin 90° – cos 45° + cos 60°)

Question 10.
A quadratic polynomial sum of whose zeros is 3 and product is – 6 is: [1]
(a) x² – 3x – 6
(b) 2x²2 + 3x + 6
(c) x² + 3x + 6
(d) x² – 6x – 3
Answer:
(a) x² – 3x – 6

Explanation:
Here, sum of zeros = 3
Product of zeros = – 6
∴ Quadratic polynomial is
x² – (sum of roots) x + product of roots
⇒ x² – 3x – 6, is the required quadratic polynomial

Question 11.
The value of ‘k’ for which the pair of linear equations kx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is: [1]
(a) 4
(b) 3
(c) 2
(d) None of these
Answer:
(d) None of these

Explanation:
The given pair of equations will have infinitely many solutions when
$\frac{k}{6}=\frac{-1}{-2}=\frac{2}{3}$
No value of k satisfy the above relation.
So, no value of k exist.

Question 12.
The probability of getting a number which is neither prime nor composite in a single throw of a fair dice is: [1]
(a) $\frac{1}{5}$
(b) $\frac{2}{5}$
(c) $\frac{1}{6}$
(d) $\frac{1}{4}$
Answer:
(c) $\frac{1}{6}$

Explanation:
Total number of outcomes on rolling a fair dice = 6
∵ 1 is only the number which is neither prime nor composite.
So, required probability is $\frac{1}{6}$.

Question 13.
The mean of the following data is: [1]
1, 7, 9, 3, 4, 5, 6
(a) 4
(b) 2
(c) 3
(d) 5
Answer:
(d) 5

Explanation:
We know,
Mean = $\frac{\text { Sum of observation }}{\text { Total no. of observation }}$
= $\frac{1+7+9+3+4+5+6}{7}$
= $\frac{35}{7}$ = 5

Question 14.
One card is drawn at random from a well shuffled deck of 52 cards. What is the probability to get a face card ? [1]
(a) $\frac{1}{13}$
(b) $\frac{3}{13}$
(c) $\frac{2}{13}$
(d) $\frac{4}{13}$
Answer:
(b) $\frac{3}{13}$

Explanation:
In a deck of 52 cards, number of face cards is 12.
So, the required probability = $\frac{12}{52}$, i.e. $\frac{3}{13}$

Question 15.
A solid cube is cut into two cuboids of equal volumes. The ratio of surface areas of the given cube and one of the resulting cuboid is: [1]
(a) 2 : 3
(b) 1: 3
(c) 3 : 2
(d) 3 : 1
Answer:
(c) 3 : 2

Explanation:
Since the cube is cut into two cuboids of equal volumes, so the two cuboids are equal.

Let the edge of the cube be 2a units and the cube be cut along its length.
∴ Dimensions of each cuboid formed are a × 2a × 2a.
So, Surface area of cuboid
= 2 (a × 2a + 2a × 2a + 2a × a)
= 16a²
Also, Surface area of cube
= 6(2a)² = 24a²
∴ Required ratio = 24a² : 16a² = 3 : 2.

Question 16.
An unbiased dice is rolled once. What is the probability of getting a prime number? [1]
(a) $\frac{3}{2}$
(b) $\frac{5}{2}$
(c) $\frac{1}{3}$
(d) $\frac{1}{2}$
Answer:
(d) $\frac{1}{2}$

Explanation:
When a dice is rolled, the total outcomes are 1, 2, 3, 4, 5 and 6.
∴ Total outcomes = 6
Favourable outcomes = 2, 3, 5 i.e. 3
∴ P (getting a prime number) = $\frac{3}{6}=\frac{1}{2}$
Hence, required probability is $\frac{1}{2}$

Question 17.
If the ratio of the Length of a rod to its shadow ¡s 1 : √3, then, angle of elevation of the sun is: [1]
(a) 30
(b) 60°
(c) 45°
(d) 900
Answer:
(a) 30°

Explanation:
Let θ be the angle of elevation of the sun.

Then, tan θ = $\frac{BC}{AB}$
= $\frac{1}{\sqrt{3}}$
θ = 30°

Question 18.
A rolling pin is made by joining three cylindrical pieces of wood, as shown in the figure: [1]

Assuming that there is no wastage of wood, the volume of wood used in making the rolling pin is:
(a) 64π cm³
(b) 280π cm³
(c) 480π cm³
(d) 544π cm³
Answer:
(d) 544 π cm³

Explanation:
For the bigger cylinder:
r₁ = 4 cm
h₁ = 30 cm
For the two smaller cylinders:
r₂ = $\frac{4}{2}$ = 2 cm,
h₂ = 8 cm
Now, Volume of wood used = Volume of the pin
= Volume of bigger cylinder + 2x volume of smaller cylinder
= πr₁²h₁ + 2 × πr₂²h₂
= π × (4)² × 30 + 2 × π × (2)² × 8
= 480π + 64π
= 544 π cm³

Question 19.
Assertion (A): The base radii of two right circular cylinders of the same height are in the ratio 3 : 5. The ratio of their curved surface area is 3 : 5.
Reason (R): CSA of right circular cylinder is 2πr²h. [1]
Answer:
(c) Assertion (A) is true but reason (R) is false.

Explanation:
Let r₁ h₁ and r₂, h₂ be the radii and the heights of the first and second cylinders respectively.

Question 20.
Assertion (A): If the probability of the occurence of an event is $\frac{3}{7}$, then the probability of its non-occurence is $\frac{5}{7}$.
Reason (R): P(A) + P(Ā) = 1, when P(Ā) is a complement. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
For an event A, the P(A) + P(Ā) = 1
Here, P(A) = $\frac{3}{7}$
Then, P(Ā) = 1 – $\frac{3}{7}$
= $\frac{4}{7}$

Section – B (10 marks)
(Section – B consists of 5 questions of 2 marks each.)

Question 21.
Determine the A.P. whose 3^rd term is 5 and the 7^th term is 9. [2]
Answer:
Here, a₃ = a + 2d = 5
and a₇ = a + 6d = 9
Solving the two equations, we get:
a = 3, d = 1
So, required A.P. is 3, 4, 5, 6, …………

Question 22.
Find the zeros of 3x² – x – 4.
Answer:
Let p(x) = 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)
So, the two zeros of 3x² – x – 4 are $\frac{4}{3}$ and – 1.

Question 23.
Find the coordinates of the point which divides the line joining (1, -2) and (4, 7) internally in the ratio 1: 2.
OR
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin. [2]
Answer:
Let P(x, y) divide AB in the ratio 1 : 2. Then,

Let the third vertex be (x, y). Then,
$\left(\frac{x-3+0}{3}, \frac{y+1-2}{3}\right)$ = (0, 0)
⇒ $\frac{x-3}{3}$ = 0 and $\frac{y-1}{3}$ = 0
⇒ x = 3 and y = 1
Thus, the third vertex is (3, 1).

Question 24.
Find the area of the shaded region:

OR
Two dice are thrown simultaneously and the outcomes are noted. Find the probability that: [2]
(A) doublets are obtained
(B) sum of numbers on the two dice is 5.
Answer:
Area of the shaded region = Area of semi-circle of radius 14 cm + 2 × Area of semi-circle of radius 7 cm.
= ($\frac{\pi}{2}$(14)² + 2 × $\frac{\pi}{2}$(7)²) sq.cm
= ($\frac{1}{2} \times \frac{22}{7}$ × 14 × 14 + $\frac{22}{7}$ × 7 × 7) sq.cm
= (308 + 154) sq.cm
= 462 sq.cm

On throwing two dice together, Total number of outcomes = 36
(A) Favourable outcomes = {(1, 1), (2, 2), (3, 3), (4, 4). (5, 5), (6, 6)}.
⇒ Number of favourable outcomes = 6 6 1
∴ P (a doublet) = $\frac{6}{36}$ i.e., $\frac{1}{6}$

(B) Favourable outcomes = {(1, 4), (2, 3), (3, 2), (4,1)}.
⇒ Number of favourable outcomes = 4
∴ P (sum of 5 on two numbers)
= $\frac{4}{36}$ i.e., $\frac{1}{9}$

Question 25.
Amrish wakes up in the morning and notices that his digital clock reads 07: 25 am. After noon, he looks at the clock again. [2]

What is the probability that:
(A) the number in column A is 4?
Answer:
The number in column A can be 0,1, 2, 3, 4 and 5.
So, P(4) = $\frac{1}{6}$

(B) the number in column B is 8?
Answer:
The number in column B can be 0, 1, 2, 3, …….. 9.
So, P(8) = $\frac{1}{10}$

SECTION – C (18 marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
On a morning walk, three girls step off together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? [3]
Answer:
Required minimum distance
= LCM (40, 42, 45)
∵ 40 = 2 × 2 × 2 × 5 i.e., 2³ × 5
42 = 2 × 3 × 7
∴ 45 = 3 × 3 × 5 i.e., 3² × 5
LCM (40, 42, 45) = 2³ × 3² × 5 × 7
= 2520

Question 27.
There is a circular park of radius 24 m and there is a pole at a distance of 26 m from the centre of the park as shown in the figure. It is planned to enclose the park by planting trees along line segments PQ and PR tangential to the park. [3]

(A) Find the length of PQ and PR;
Answer:
In right triangle PRO,
We have PR = $\sqrt{P O^2-R O^2}$

= $\sqrt{26^2-24^2}$
= $\sqrt{676-576}$
= √100
= 10 m
⇒ PR + PQ =10 m

(B) If six trees are to be planted along each tangential line segments at equal distances, find the distance between any two consecutive trees.
Answer:
As six trees are to be planted along PQ and PR each.

Let’s assume the distance between consecutive trees is x.
Total trees are at 5 equal distances.
Hence, 5x = 10
x = 2

Question 28.
In the figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. [3]

Find the area of the shaded region.
Answer:
Area of the shaded region
= Area of the sector of radius 7 cm – Area of the sector of radius 3.5 cm
= ($\frac{30^{\circ}}{360^{\circ}}$π(7)² – $\frac{30^{\circ}}{360^{\circ}}$π(3.5)²)sq.cm
= $\frac{30}{360} \times \frac{22}{7}$ × (49 – 12.25) sq.cm
= 9.625 sq.cm

Question 29.
The sum of reciprocals of a child’s age (in years) 3 years ago and 5 years from now is $\frac{1}{3}$. Find his present age. [3]
OR
Solve for x and y:
7x – 4y = 49; 5x – 6y = 57
Answer:
Let the present age of the child (in years) be x. Then,
$\frac{1}{x-3}+\frac{1}{x+5}$ = $\frac{1}{3}$
⇒ $\frac{(x+5)+(x-3)}{(x-3)(x+5)}$ = $\frac{1}{3}$
⇒ 3(2x + 2) + (x – 3) (x + 5)
⇒ 6x + 6 = x² + 2 x + 15
⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 or x + 3 = 0
⇒ x = 7 or x = – 3
(x ≠ – 3, as age cannot be negative)
Thus, present age of child is 7 years.

Given equations are:
7x – 4y = 49 ….. (i)
5x – 6y = 57 ….. (ii)
Multiplying eq. (i) by 5 and eq. (ii) by 7, we get:
35x – 20y = 245 …..(iii)
35x – 42y = 399 …… (iv)
Subtracting equation (iv) from equation (iii), we get
⇒ 22y = – 154
⇒ y = – 7
Substituting y = – 7 in eq. (i), we get
7x + 28 = 49
⇒ 7x = 21
⇒ x = 3
Thus, x = 3, y = – 7 is the required solution.

Question 30.
ABP and ACQ are two tangents to the circle with O as its centre in the given figure. If ∠TCQ = 60° and ∠TBP = 50°, then find the measure of ∠BTC. [3]

Answer:

Join OB, OT and OC
We know, tangent is perpendicular to radius at the point of contact.
∴ OB ⊥ AP and OC ⊥ AQ
∴ ∠OBP = 90°
⇒ ∠OBT + ∠TBP = 90°
⇒ ∠OBT + 50° = 90°
⇒ ∠OBT = 90° – 50° = 40°
Similarly, ∠OCQ = 90° and ∠TCQ = 60°
∴ ∠OCT = 30°
Now, in ∆OBT
OB = OT (Radii)
∠OTB = ∠OBT
[Equal angles opposite to equal sides]
⇒ ∠OTB = 40°
Similarly in AOTC
OT = OC
⇒ ∠OCT = ∠OTC = 30°
So, ∠BTC = ∠OTB + ∠OTC
= 40° + 30° = 70°

Question 31.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
OR
Find two numbers whose sum is 27 and product is 182. [3]
Answer:
Let the natural number be x. Then,
x + 12 = 160 × $\frac{1}{x}$
⇒ x² + 12x – 160 = 0
⇒ x² + 20x – 8x – 160 = 0
⇒ x(x + 20) – 8 (x + 20) = 0
⇒ (x + 20) (x – 8) = 0
⇒ x + 20 = 0
⇒ x = – 20
(Rejected, as x is a natural number)
or x – 8 = 0
⇒ x = 8
Thus, the required natural number is 8.

Let first number be x and let second number be (27 – x)
According to given condition, the product of two numbers is 182.
Therefore,
x(27 – x) = 180
⇒ 27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14)= 0
⇒ (x – 14) (x – 13)= 0
⇒ x = 14, 13
Therefore, the first number is equal to 14 or 13
And, Second number is = 27 – x = 27 – 14 = 13 or Second number = 27 – 13 = 14
Therefore, two numbers are 13 and 14.

Section – D (20 marks)
(Section – D consists of 4 questions of 5 marks each.)

Question 32.
A number consists of two digits. When it is divided by the sum of the digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number. [5]
Answer:
Let the number be ab, i.e., 10a + b
As per the question,
$\frac{10 a+b}{a+b}$ = 6 or 10a + b = 6a + 6b
⇒ 4a = 5b ….. (1)
(10a + b) – 9 = 10b + a
or 9a = 9b + 9
or a = b + 1 ….. (2)
Solving Eq. (1) and Eq. (2), we get
b = 4 and a = 5.
So, the required number is 54.

Question 33.
Amit and Prem were very good cricketers and also represented their school team at district and even state level. One day, after their match, they measured the height of the wickets and found it to be 28 inches. They marked a point P on the ground as shown in the figure below:

(A) If cot P = $\frac{3}{4}$, then find the length of P
(B) Find the value of cosec P.
(C) Find the value of $\frac{1+\sin \mathrm{P}}{1+\cos \mathrm{P}}$.
(D) Find the value of sec R.
(E) Find The value of coses² R – cot² R.
OR
Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road , the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the two poles.
Answer:
(A) It is given that QR = 28 inches and
cot P = $\frac{3}{4}$
We know that cot P = $\frac{\text { Base }}{\text { Perpendicular }}$
= $\frac{P Q}{Q R}=\frac{P Q}{28}$

Therefore, $\frac{P Q}{28}=\frac{3}{4}$
⇒ PQ = $\frac{28 \times 3}{4}$ = 21 inches

(B) To evaluate cosec P, we will first find PR
by applying Pythagoras theorem ∆PQR.
PR² = PQ² + QR²
= 21² + 28²
= 441 + 784
= 1225
⇒ PR = 35 inches
∴ cosec P = $\frac{\text { Hypotenuse }}{\text { Perpendicular }}$ = $\frac{\mathrm{PR}}{\mathrm{QR}}=\frac{35}{28}=\frac{5}{4}$

(C)

(D) sec R = $\frac{\text { Hypotenuse }}{\text { Base }}$ = $\frac{\mathrm{PR}}{\mathrm{QR}}$
= $\frac{35}{28}=\frac{5}{4}$

(E)

In the figure. AB and XY are two poles of equal height, say ‘h’ metres.

Here, AX represents the width of the road and P is a point on the road.
In ∆ BAP,

⇒ h = 20√3 metres
Thus, the height of each pole is 20√3 metres.
From eqn. (i) and (ii), we aLso have,
AP = $\frac{20 \sqrt{3}}{\sqrt{3}}$ i.e. 20 metres
and XP = √3 × 20√3, i.e. 60 metres.
Thus, distance, of the point from the two poles are 20 metres and 60 metres.

Question 34.
State and prove Basic Proportionality Theorem. [5]
Answer:
Statement:
If a line is drawn parallel to one-side of a triangle intersecting the other sides, the other two sides are divided in the same ratio.

Proof:
Consider a ∆ABC is which DE || BC
To prove: $\frac{A D}{B D}=\frac{A E}{E C}$

Construction: Draw EM ⊥ AB, DN ⊥ AC and join BE and CD.
Proof: We know,
Area of triangle = $\frac{1}{2}$ × AD × EM
Also, (∆ADE) = $\frac{1}{2}$ × AE × DN
Similarly, ar (ABDE) = $\frac{1}{2}$ × BD × EM
And, ar (ADEC) = $\frac{1}{2}$ × EC × DN
Now, triangles BDE and DEC are on same base DE and lying between some parallels DE and BC.
∴ ar (∆BDE) = ar (∆DEC)

Question 35.
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequencies fi and

OR
A test tube has a hemisphere-shaped lower portion and a cylindrical upper portion with the same radius. The test tube fills up to the point of being exactly full when $\frac{4554}{7}$ cu cm of water is poured, but when only 396 cu cm is added, 9 cm of the tube is left empty. Calcualte the test tube’s radius and the cylindrical part’s height. [5]
Answer:
The frequency distribution for calculating the mean, for the given data is:

We know that
30 + f₁ + f₂ = 50
⇒ f₁ + f₂ = 20
⇒ f₂ = 20 – f₁
Now, mean = A + $\frac{\Sigma f_i d_i}{\Sigma f_i}$
⇒ 62.8 = 50 + $\frac{560+20 f_2-20 f_1}{50}$
⇒ 12.8 = 50 + $\frac{560+20\left(20-f_1\right)-20 f_1}{50}$ [Using (i)]
⇒ 640 = 960 – 40 f₁
⇒ 40 f₁ = 320
f₁ = 8
∴ f₂ = 20 – 8 = 12
∴ f₁ = 8, f₂ = 12.

Volume of water that can fill the test tube = $\frac{4554}{7}$ cm³

Let r’ be the radius of cylinder and hemispherical part.
And let ‘h’ be height of cylinderical part

= 21 cm
Hence, radius of the test tube is 3 cm and height of cylinderical part is 21 cm.

Section – E (12 marks)
(Case Study-Based Questions)
(Section – E consists of 3 questions. All are compulsory.)

Question 36.
Rishu is riding in a hot air balloon. After reaching a point P. he spots a car parked at B on the ground at an angle of depression of 30°. The balloon rises further by 50 metres and now he spots the same car at an angle of depression of 45° and a lorry parked at S’ at an angle of depression of 30°. (Use √3 = 1.73)

The measurement of Rishu facing vertically is the height. Distance is defined as the measurement of car/lorry from a point in a horizontal direction. If an imaginary line is drawn from the observation point to the top edge of the car/lorry, a triangle is formed by the vertical, horizontal and imaginary line.
On the basis of the above information, answer the following questions:
(A) If the height of the balloon at point P is ‘h’m and distance AB is’x’ m, then find the relation between ‘x’ and ‘h’: [1]
(B) What is the relation between the height of the balloon at point P and distance AB. [1]
(C) Find the height of the balloon at point P and the distance AB on the ground.
OR
Find the distance B’B on the ground. [2]
Answer:

(A) In ∆APB,
tan 30° = $\frac{\mathrm{AP}}{\mathrm{AB}}$
⇒ $\frac{1}{\sqrt{3}}=\frac{h}{x}$
⇒ x = √3h

(B) In ∆AP’B,
tan 45° = $\frac{A P^{\prime}}{A B}$
⇒ AB = AP’
⇒ x = h + 50

In ∆AP’B’
tan 30° = $\frac{A P^{\prime}}{\mathrm{AB}^{\prime}}$
⇒ $\frac{1}{\sqrt{3}}=\frac{68.25+50}{\mathrm{AB}^{\prime}}$
⇒ AB’ = 118.25 × 1.732
= 204.809
BB’ = AB’ – AB
= 204.809 – 118
= 86.80 = 87 m

Question 37.
A factory is using an inclined conveyor belt to transport its product from level 1 to level 2 which is 3 m above level 1 as shown in the figure below. The inclined conveyor is supported from one end to level 1 and from the other end to a post located 8 m away from level 1 supporting point.

The factory wants to extend the conveyor belt to reach at a new level 3 which is 9 m above level 1 while maintaining the inclination angle.
On the basis of the above information, answer the following questions:
(A) Find the distance at which a new post is to be installed to support the conveyor belt at level 3. [1]
(B) How much distance is extended from D to B? [1]
(C) Find the length of the conveyor belt up to level 3.
OR
Find the length of the conveyor belt up to level 2. [2]
Answer:
(A) In ∆ADE and ∆ABC,
Since, both triangles are similar, then, their corresponding sides will be proportional
Then $\frac{A D}{A B}=\frac{D E}{B C}$
⇒ $\frac{8}{A B}=\frac{3}{9}$
⇒ AB = 24 cm

(B) Distance extended = AB – AD
= 24 – 8 = 16 m

(C) Since, ∆ABC is a right-angled at B.
∴ AC² = AB² + BC² (by Pythagoras theorem)
AC² = (24)² + 9²
AC = $\sqrt{676+81}$ = 25.63
≃ 26 m
Then, distance need to be travelled to reach new level is 26 m

In ∆ADE, by pythagoras theorem
AD² + DE² = AE²
⇒ 8² + 3² = AE²
⇒ AE² = 64 + 9 = 73
⇒ AE = √73 = 8.5 m

Question 38.
To conduct sport day activities in the rectangular school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each along DB.
100 flower pots have been placed at a distance of 1 m from each other along DA as shown in the figure below.

Radha runs $\frac{1}{4}$th of the distance DA on 2nd line and post a green flag at X . Preeti runs $\frac{1}{5}$th of the distance DA on other line post a red flag at Y.
on the basis of the above information, answer the following questions:
(A) Treating DB as x-axis and DA as y-axis, find the position of green flag. [1]
(B) Treating DB as x-axis and DA as y-axis, find the position of red flag. [1]
(C) Find the distance (in complete metres) between the two flags.
OR
Find the perimeter (in complete metres) of the triangular region OXY. [2]
Answer:
(A) Radha’s distance on x-axis is 2 and on y-axis she is at $\frac{1}{4}$ × 100 = 25 4
Green flag coordinates are (2, 25)

(B) X-coordinate = 8
Y-coordinate = $\frac{1}{5}$ × 100 = 20
∴ Coordinates of red flag (8, 20)

(C) Coordinates of green flag is (2, 25)
∴ Coordinates of red flag is (8, 20)
∴ By distance formula

= √61 = 7.8 m
≃ 8 m (approx)

The distance formula $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$ = $\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}$. It gives the same answer.

Perimeter = OX + OY + YY
= 25.07 + 21.54 + 7.81
= 54.42
= 55 m

CBSE Sample Papers for Class 10 Maths Basic Set 8 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 8 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

This Question Paper has 5 Sections A, B C D. and E
Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
Section Chas 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
Section D has 4 Long Answer (LA) type questions carrying S marks each.
Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
AU Questions are compulsory. However an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2
Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions
of Section E
Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
Two positive integers p and q are expressible as p = a³b and q = ab². The HCF (p, q) and LCM (p, q) is: [1]
(a) ab²
(b) a³b²
(c) a²b³
(d) a³b
Answer:
(b) a³b²

Explanation:
HCF (p, q) = ab;
LCM (p, q) = a³b²

Question 2.
If given A.P. is 11, 8, 5, 2, …………………, then the sum of 10th term is: [1]
(a) -23
(b) 28
(c) 24
(d) -25
Answer:
(d) -25

Explanation:
Given AP is 11, 8, 5, 2, ……..
a = 11
d = 8 – 11 = -3
n = 10
S₁₀ = $\frac{10}{2}$(2 × 11 + (10 – 1) × – 3)
= 5(22 – 27)
= 5 × -5
= -25

Question 3.
The ideal times of year to have chilLed shakes and ice creams are during the summer. During Lockdown, Saumyo was eager to try the watermeLon sharbat she had recently Learned to make from her online culinary classes. She cut a watermelon slice with a semi-circuLar cross section. If the perimeter of a semi- circuLar portion Is 36cm, then its radius is: [1]
(a) 12 cm
(b) 15 cm
(c) 7 cm
(d) 14 cm
Answer:
(c) 7 cm

Explanation:
Perimeter of semicircular = 36 cm
πr + 2r = 36
⇒ (π + 2)r = 36
⇒ ($\frac{1}{2}$ + 2)r = 36
⇒ $\frac{36}{7}$r = 36
⇒ r = $\frac{36 \times 7}{36}$
⇒ r = 7

Question 4.
The solution of the pair of equations: [1]
2x + 3y = 9; 3x + 4g = 5 is
(a) 21, -17
(b) 20, 14
(c) -21, 17
(d) 20, 3
Answer:
(c) -21, 17

Explanation:
2x + 3y = 9 …(i)
3x + 4y = 5 …(ii)
Multiplying eq. (i) and (ii) by 3 and 2, respectively and then subtracting them, we get

From eq. (i)
2x + 51 =9
⇒ 2x = -42
⇒ x = -21
Thus, x = -21, y = 17 is the required solution.

Question 5.
Two vertices of a triangle are (4, -5) and (-5, -2). If the centroid of the triangle is the origin, the third vertex of the triangle is: [1]
(a) (1, 5)
(b) (2, 4)
(c) (1,4)
(d) (1,7)
Answer:
(d) (1,7)

Explanation:
Let the third vertex be (x, y). Then,
$\left(\frac{4-5+x}{3}, \frac{-5-2+y}{3}\right)$ = (0, 0)
⇒ $\frac{x-1}{3}$ = 0, $\frac{y-7}{3}$ = 0
⇒ x = 1, y = 7
Thus, the third vertex is (1, 7).

Question 6.
In the adjoining figure, if PA = 10 cm, then the perimeter of ΔPCD is: [1]

(a) 16 cm
(b) 21 cm
(c) 18 cm
(d) 20 cm
Answer:
(d) 20 cm

Explanation:
We know lengths of tangents drawn from an external point to a circle are equal.
∴ From the figure, we have
PA = PB, CA = CE and DE = DB.
Now,
Perimeter of ΔPCD = PC + CE + ED + DP
= (PC + CE) + (ED + DP)
= (PC + CA) + (BD + DP)
= PA + PB
= 2 PA .
= 2 × 10 cm = 20 cm

Question 7.
What is mid-point of the line segment AB where A (- 5, 0) and B (0, 5) ? [1]
(a) $\left(-\frac{5}{2}, \frac{5}{2}\right)$
(b) (3, 5)
(c) $\left(\frac{5}{2}, \frac{3}{2}\right)$
(d) (2, 4)
Answer:
(a) $\left(-\frac{5}{2}, \frac{5}{2}\right)$

Explanation:
The mid-point of AB is
$\left(\frac{-5+0}{2}, \frac{0+5}{2}\right)$ i.e, $\left(-\frac{5}{2}, \frac{5}{2}\right)$

Question 8.
If x sec 45° = 2, then what is the value of x ? [1]
(a) $\frac{\sqrt{3}}{2}$
(b) 2√2
(c) √2
(d) $\frac{1}{\sqrt{2}}$
Answer:
(c) √2

Explanation:
Given, x sec 45° = 2
⇒ x(√2) = 2
⇒ x = √2

Question 9.
If tan θ + cot θ = 4, then the value of tan⁴θ + cot⁴θ is: [1]
(a) 196
(b) 200
(c) 194
(d) 198
Answer:
(c) 194

Explanation:
Given, tan θ + cot θ = 4
Squaring on both sides,
We get,
tan² θ + cot² θ + 2 tan θ cot θ = 16
tan² θ + cot² θ + 2 = 16
tan² θ + cot² θ = 14

Again, squaring on both sides, we get
tan⁴ θ + cot² θ + 2 tan² θ cot² θ = 196
⇒ tan⁴θ + cot⁴θ + 2 = 196
⇒ tan⁴θ + cot⁴θ = 194

Question 10.
In an A.P., if a = 3.5, d = 0, n = 101, then the value of a_n is: [1]
(a) 2.5
(b) 3
(c) 4
(d) 3.5
Answer:
(d) 3.5

Explanation:
In the given A.P., d = 0
So, all its terms are same as a = 3.5,
∴ a₁₀₁ = 3.5

Question 11.
If A = 900, If Σf_id_i = – 400 and Σf_i = 100, then what is the value of x? [1]
(a) 890
(b) 986
(c) 465
(d) 896
Answer:
(d) 896

Explanation:
We know that
x̄ = A + $\frac{\Sigma f_i d_i}{\Sigma f_i}$
= 900 + $\frac{(-400)}{100}$
= 900 – 4 = 896

Question 12.
A 6 faced cube has letters A, B, C, D, A and C on its six faces. This cube is rolled once. What is the probability of getting B or C? [1]
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{1}{4}$
Answer:
(a) $\frac{1}{2}$

Explanation:
Total number of outcomes = 6
∴ Number of favourable outcomes = 3
P (B or C) = $\frac{3}{6}$ i.e. $\frac{1}{2}$

Question 13.
Which criterion of similarity will be used in proving that ΔABD – ACE ? [1]

(a) SAS
(b) AA
(c) RHS
(d) SSS
Answer:
(b) AA

Explanation:
In As ABD and ACE, we have AD = AE (given)
So, ∠D = ∠E (Angles opposite to equal sides are equal)
∠A = ∠A (Common)
So, by AA similarly criterion, ΔABD ~ ΔACE.
Hence, AA similarity criteria will be used.

Question 14.
A letter is chosen from the letters of the word MAINTENANCE The probability that it is N is: [1]
(a) $\frac{1}{11}$
(b) $\frac{2}{11}$
(c) $\frac{3}{11}$
(d) $\frac{4}{11}$
Answer:
(c) $\frac{3}{11}$

Explanation:
In the given word, there are in all eleven letters, of which three are N.
So, the required probability is $\frac{3}{11}$

Question 15.
If the equation x² + 4x + k = 0 has real and distinct roots, then the value of ‘k’ is: [1]
(a) k > 4
(b) k > 4
(c) k < 4
(d) k = 4
Answer:
(c) k < 4

Explanation:
As the given equation has real and distinct roots,
∴ Discriminant = (4)² – 4(1)(k) > 0
i.e. 16 – 4k > 0
or k < 4

Question 16.
Which term of the A.P.: -2, -7, -12,… will be -77? [1]
(a) 16th
(b) 10th
(c) 15th
(d) 12th
Answer:
(a) 16th

Explanation:
Let, the nth term of the A.P. be -77.
Then, a + (n – l)d = -77
For the given A.P., a = -2 and d = -5.
So, a + (n – 1)d = (-2) + (n – 1)(-5) = -77
-5(n – 1) = -75
or n- 1 = 15 or n = 16
So, the 16th term of the A.P. is (-77).

Question 17.
What type of lines are represented by the pair of equations 10x + 6y = 9 and 5x + 3y + 4 = 0 ? [1]
(a) Straight lines
(b) Intersecting lines
(c) Parallel lines
(d) None of these
Answer:
(c) Parallel lines

Explanation:
The pair of equations satisfy the relation
$\frac{a_1}{a_2}=\frac{b_1}{b_2} \uparrow \frac{c_1}{c_2}$
as $\frac{10}{5}=\frac{6}{3} \neq \frac{9}{-4}$
Hence, these equations represent parallel lines.

Question 18.
If an event is sure to occur, then what is its probability of occurrence ? [1]
(a) 0
(b) 1
(c) 2
(d) $\frac{1}{2}$
Answer:
(b) 1

Explanation:
The probability of occurrence of a sure event is 1.

Direction for questions 19 and 20: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Assertion (A): The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. The radius of the circle is 17 cm
Reason (R): Tangent is perpendicular to radius at the point of contact. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
OQ is perpendicular to PQ.

∴ In ΔPOQ,
PQ² + OQ² = OP²
25² = OQ² + 24²
OQ² = 625 – 576 = 49
OQ = 7 cm

Question 20.
Assertion (A): If circumference of two circles are equal, then their areas are also equal.
Reason (R): Two circles are congruent if their radii are equal 1. [1]
Answer:
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)

Explanation:
Let’s consider 2 circles of radii r₁ and r₂.
Then, 2πr₁ = 2πr₂
r₁ = r₂ = r
Then, A₁ = πr₂² = πr²
A₁ = A₂

SECTION – B (10 marks)
(Section – B consists of 5 questions of 2 marks: each)

Question 21.
Assuming that √2 is irrational, show that 5√2 is an irrational number. [2]
Answer:
Let 5√2 be rationaL Then,
5√2 = $\frac{p}{q}$, where p and q are co-prime and q ≠ 0
⇒ √2 = $\frac{p}{5q}$
Here, $\frac{p}{5q}$ is rational which implies √2 is rational which is a contradiction, as it is given than √2 is irrational
⇒ 5√2 is irrational.

Question 22.
Find the greatest number that divides 45 and 210 completely. [2]
Answer:
The greatest number that divides 45 and 240 completely is the HCF of 45 and 210.
Now, 45 = 3 × 3 × 5, or 3² × 5¹
210 = 2 × 3 × 5 × 7,
So, HCF (45, 210) = 3¹ × 5¹, i.e. 15
Hence, the required number is 15.

Question 23.
If x = a cos³θ and y = b sin³θ, then prove that: [2]
$\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}$ = 1
OR
Prove that $\sqrt{\sec ^2 \theta+{cosec}^2 \theta}$ = tan θ + cot θ.
Answer:

Question 24.
The largest possible sphere is carved out of wooden solid cube of side 7 cm. What is the radius of this sphere?
OR
A line intersect the y-axis and x-axis at the points P and Q respectively. If (2, – 5) is the mid point of PQ, then find the coordinates of the points P and Q. [2]
Answer:
The largest possible sphere that can be carved out of a sphere, is equal to diameter of the sphere.
∴ Diameter of sphere = 7 cm.
So, the radius of the largest possible sphere is 3.5 cm.
OR
Let the coordinates of P and Q be (0, y) and (x, 0) respectively.
Here, the mid-point of PQ is (2, – 5)
So, $\left(\frac{x+0}{2}, \frac{0+y}{2}\right)$ = (2, -5)
⇒ x = 4, y = -10

Thus, the coordinates of P and Q are (0, – 10) and (4, 0) respectively.

Question 25.
In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [2]

Answer:
OB = $\sqrt{O A^2+A B^2}=\sqrt{O A^2+O A^2}$
= √2 OA = √2 × 20 = 20√2 cm
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
= $\frac{90^{\circ}}{360^{\circ}}$ × 3.14(20√2)² – 20 × 20
= $\frac{1}{4}$ × 3.14 × 800 – 400
= 200 × 3.14 – 400
= 228 cm²

SECTION – C (18 marks)
(Section – C consists of 6 questions of 3 mark each)

Question 26.
Solve for x and y:
x + $\frac{y}{4}$ = 11; $\frac{5 x}{6}-\frac{y}{3}$ = 7
OR
A 2-digit number is such that the product of the digits is 20. If 9 is subtracted from the number, the digits interchange their places. Find the number. [3]
Answer:
The given equations are rewritten as:
4x + y – 44 = 0 ;
5x – 2y – 42 = 0
From 4x + y – 44 = 0, we have
y = 44 – 4x …..(i)
Substituting this value of y in 5x – 2y – 42 = 0, we have:
5x – 2 (44 – 4x) – 42 = 0
=> 13x – 88 – 42 = 0
=> x = 10
From (i) y = 44 – 40 = 4 Thus, x = 10, y = 4 is the required solution.
OR
Let ten’s digit and one’s digit of the two digit number be a and b respectively. Then, the number is 10a + b.
Here, ab = 20 ……..(i)
and (10a + b) – 9 = 10b + a
i.e 9a – 9b = 9
or a – b = 1
Solving (i) and (ii) simultaneously, we get
a = 5, b = 4
Thus, the number is 54.

Question 27.
Show that ΔABC with vertices A (- 2, 0), B (2, 0) and C(0, 2) is similar to ΔDEF with vertices D (- 4, 0), E (4, 0) and F (0, 4). [3]
Answer:
Let’s say that ΔABC ~ ΔDEF.
If $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}$

Now, using distance formula,
AB = 4, BC = √8 = 2√2
CA = √8, 2√2
DE = 8, EF = √32 = 4√2
FD = √32 , i.e 4√2
Here $\frac{A B}{D E}=\frac{B C}{D E}=\frac{C A}{F D}=\frac{1}{2}$
So, ΔABC ~ ΔDEF.

Question 28.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
In the figure, PQ and RS are the common tangents to two circles intersecting at O. [3]

Prove that PQ = RS
Answer:
Let AP and AQ be the two tangents drawn
to the circle from an external point A.
We need to show that AP = AQ.
Join OA, OP and OQ.
We know that tangent is perpendicular to radius at the point of contact.
∴ OP ⊥ AP and OQ ⊥ QA.

Consider ΔOPA and ΔOQA.
Here, OQ = OP (radii of the circle)
OA = OA (common)
∠OPA = ∠OQA
So, ΔOPA ≅ ΔOQA
⇒ PA = QA or AP = AQ
OR
We know that lengths of tangent drawn from an external point to a circle are equaL

So, from the figure, OP = OR and OS = OQ
Now, PQ = PO + OQ
= OR + OS
= RS

Question 29.
A number x is selected from the numbers 1, 2, 3 and then a second number y is selected randomly from the numbers 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9? [3]
Answer:
Total number of possible outcomes of
product xy = {1, 4, 9, 2, 8,18, 3,12, 27}.
Of these, 5 are less than 9.
So, P(xy < 9) = $\frac{5}{9}$

Question 30.
Find the value of:
$\frac{5 \sin ^2 30^{\circ}+\cos ^2 45^{\circ}-4 \tan ^2 30^{\circ}}{2 \sin 30^{\circ} \cdot \cos 30^{\circ}+\tan 45^{\circ}}$ + cos 0° [3]
Answer:

Question 31.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there in the A.P. ?
Answer:
Let the A.P. contains ‘n’ terms. Then,
a_n = l = 350
⇒ a + (n – 1)d = 350
⇒ 17 + (n – 1)(9) = 350
⇒ 9(n – 1) = 333
⇒ n – 1 = 37
⇒ n = 38
Thus, A.R contains 38 terms.

SECTION – D (20 marks)
(Section – D consists of 4 questions of 5 mark each)

Question 32.
From the top of a building 60 m high, the angle of depression of the top and bottom of a vertical lamp-post are observed to be 30° and 60° respectively. Find the height of the lamp-post, and the distance between the top of building and the top of lamp-post. [5]
Answer:
Let AB be the building and XY be the lamp post.
∴ AB = 60 m, ∠BYM = 30° and ∠BXA = 60°
Let ‘h‘ metres be the height of the lamp post and ‘d’ metres be the horizontal distance between the lamp post and the building.
Then,
From right ΔBMY, we have:
$\frac{B M}{Y M}$ = tan 30°
⇒ $\frac{60-h}{d}=\frac{1}{\sqrt{3}}$

From (i) and (ii)
√3(60 – h) = 20√3
⇒ 60 – h = 20
⇒ h = 40
Hence the height of lamp post is 40 m
Now, BY = $\sqrt{Y M^2+B M^2}$
= $\sqrt{d^2+20^2}=\sqrt{1200+400}$
= $\sqrt{1600}$ = 40 m
Thus, the distance between the top of the building and the top of lamp-post is 40 m.

Question 33.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Prove it. [5]
OR
BL and CM are medians of ΔABC, right¬angled at A.
Prove that: 4(BL² + CM²) = 5 BC²

Answer:
ABC is a triangle in which DE is a line parallel to side BC which cuts AB at D and AC at E.
We need to prove that:

Join BE and CD and draw EF ⊥ AB and DN ⊥ AC.
Now,

But ΔBDE and ΔCDE are on the same base DE and between the same parallels DE and
So, ar(ΔBDE) = ar(ΔCDE) … (iii)
⇒ $\frac{{ar}(\triangle \mathrm{ADE})}{{ar}(\triangle \mathrm{BDE})}=\frac{{ar}(\mathrm{CDE})}{{ar}(\mathrm{CDE})}$
Hence, by (i), (ii) and (iii), we have
$\frac{A D}{B D}=\frac{A E}{E C}$
or
$\frac{A D}{D B}=\frac{A E}{E C}$
OR
Since BL and CM are medians, So L is the mid-point of CAç and M is the mid-point of AB.
i.e. AL = $\frac{1}{2}$CA and AM = $\frac{1}{2}$AB
Using Pythagoras theorem in right triangle LAB, we have:
RL² = LA² + AB²
BL² = $\frac{\mathrm{CA}^2}{4}$ + AB² ………..(i)
SimilarLy, in right triangLe CAM, we have:
CM² = CA² + AM²
⇒ CM² = CA² + $\frac{\mathrm{AB}^2}{4}$

Adding (i) and (ii), we get

(Using Pythagoras theorem in ΔABC)
⇒ 4(BL² + CM²) = 5 BC²

Question 34.
Find the median marks for the following frequency distribution : [5]

Answer:

Marks	Frequency	Cumulative Frequency
0-20	7	7
20-40	12	19
40-60	23	42
60-80	18	60
80-100	10	70

Here, N = 70, ∴ $\frac{N}{2}$ = 35
Cumulative frequency just greater than 35 is 42 which belongs to class 40 – 60.
So, the median class is 40-60.
For this class,
l = 40,h = 20, c.f. = 19 and f = 2
So, Median = l + $\frac{\frac{N}{2}-c . f}{f}$ × h
= 40 + $\frac{35-19}{23}$ × 20
= 40 + 13.9 (approx)
= 53.9 (approx)

Question 35.
Between Mysore and Bangalore, 132 km apart, an express train travels in 1 hour less time than a passenger train (without taking into consideration the time they stop at intermediate stations). Find the average speed of the two trains if the express train’s average speed is 11 km/h higher than the passenger train’s.
OR
Form a pair of linear equations for the following problems and find their solution by substitution method.
(A) The cost of a taxi in a city consists of a fixed charge and a charge for the distance travelled. The cost for a 10 km travel is ₹ 105, while for a 15 km journey, the cost is ₹ 155. What are the fixed charges and the km charged? How much will it cost someoneto drive 25 Km?
(B) For ₹ 3800, the cricket team’s coach purchases 6 balls and 7 bats, he then spends 1750 for 3 bats and 5 balls. Find out how much each ball and bat costs. [5]
Answer:
Let average speed of passenger train = x km/h
Let average speed of express train = (x + 11) km/h
Time taken by passenger train to cover 132 km = $\frac{132}{x}$hours
Time taken by express train to cover 132 km = $\left(\frac{132}{x+11}\right)$hours

According to the given conditions,
$\frac{132}{x}=\frac{132}{x+11}$ + 1
⇒ $\left(\frac{1}{x}-\frac{1}{x+11}\right)$ = 1
⇒ 132$\left(\frac{x+11-x}{x(x+11)}\right)$ = 1
⇒ 132 (11) = x(x + 11)
⇒ 1452 = x² + 11x
⇒ x² + 11x – 1452 = 0
Comparing equation x² + 11x – 1452 = 0 with general quadratic equation ax² + bx + c = 0, we get a = 1, b = 11 and c = -1452

Applying Quadratic Formula

As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h And, speed of express train = x + ll = 33 + ll = 44 km/h
OR
(A) Let fixed charge = ₹ x and let charge for every km = ₹ y
According to given conditions, we have
x + 0y = 105 …(i)
x + 15 = 155 …(ii)
Using equation (i), we can say that
x = 105 – 10y
Puttinmg this in equation (ii), we get
105 – 10y + 15y = 155
⇒ 5x = 50
⇒ y = 10
Putting value of y in equation (i), we get x + 10(10) = 105
⇒ x = 105 – 100
⇒ x = 5
Therefore, fixed charge = ₹ 5 and charge per km = ₹ 10
To travel distance of 25 km, person will have to pay
= ₹ (x + 25y)
= ₹ (5+ 25 × 10)
= ₹ (5 + 250)
= ₹ 255

(B) Let cost of each bat = ₹ x
and let cost of each ball = ₹ y
According to given conditions, we have 7x + 6y = 3800 _(i)
And, 3x + 5y = 1750 …(ii)
Using equation 0), we can say that
7x = 3800 – 6y
x = $\frac{3800-6 y}{7}$

Putting this in eq. (ii), we get

⇒ 17y = 850
⇒ y = 50
Putting value of y in (ii). we get 3x + 250 = 1750
⇒ 3x = 1500
⇒ x = 500
Therefore, cost of each bat = ₹ 500 and cost of each ball = ₹ 50.

SECTION – E (12 marks)
(Case Study-Based Questions)
(Section – E consists of 3 questions. All are compulsory)

Question 36.
‘Origami’ is the art of paper folding, which is often associated with Japanese culture. Gurmeet is trying to learn Origami using paper cutting and folding technique. A square base is sometimes referred to as a “preliminary” base or preliminary fold.

Here ¡s a 20 cm × 20 cm square. Gurmeet wants to first cut the squares of integral Length from the corners and by foLding the flaps along the sides.

On the basis of the above information, answer the following questions
(A) How many different sizes of boxes Gurmeet can make? [1]
(B) How many different sizes of boxes Gurmeet can make if sides of the squares are not integral length? [1]
(C) Find the equation relating the size of the square cut out and volume of the box.
OR
Find the dimension of the box with maximum volume and minimum volume. [2]
Answer:
(A) Different size of squares are = 18 × 18 × 1,16 × 16 × 2,14 × 14 × 3 ,12 × 12 × 4, 10 × 10 × 5, 8 × 8 × 6
(B) As the side length of any value could be cut out from the square and it could be infinite in number.
(C) Let the width of square of each side be
Then, sides of box are 20 – 2x, 20 – 2x and x
Volume = lbh
= (20 – 2x) (20 – 2x)x
= (400 – 40x – 40x + 4x²)x
= 4x³ – 80x² + 400x
OR
On calculating the volume of the boxes given in the options. The box with dimension 14 × 14 × 3 has maximum volume as 588.
On calculating the volume of the boxes given in the option. The box with dimensions 18 × 18 × 1 has minimum volume.

Question 37.
The students of a school decided to beautify the school on the Annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag.

Ruchi was given the responsibility of placing the flags Ruchi kept her books where the flag were stored She could carry only one flag at a time.
On the basis of the above information, answer the following questions
(A) What is the position of the middle flog? [1]
(B) Find totaL distance travelled for placing all the flogs. [1]
(C) Find total distance travelled for placing 13 flags on left
OR
Find the maximum distance she travelled carrying a flag. [2]
Answer:
(A) There are 27 flags. So the middle most flag is 14th flag.

(B) Total distance travelled = 13 flags on left side +13 flags on right side
= 364 + 364
= 728 m

(C) For placing first placing she go 2 m and come back 2 m. Then for second flag, she goes 4 m and come back 4 m and so one….
Distance travailed = 4,8,12
Then it forms an A.P. with a = 4, d = 4 and n = 13
Then S₁₃ = $\frac{13}{2}$ [8 + 12 × 4]
= $\frac{13}{2}$ (56) = 364 m

Then, a₁₃ = a + (n – 1)d
= 4 + (13 – 1) × 4
= 4 + 48 = 52
∴ From carrying the flag to its position 52 she covers distance = $\frac{52}{2}$
= 26 m

Question 38.
Soumya throws a ball upwards, from a rooftop, 80 m above. It will reach a maximum height and then fall back to the ground. The height of the ball from the ground at time ‘t’ is ‘h’ which is given by h = -16t² + 64t + 80.

On the basis of the above information, answer the following questions:
(A) What is the height reached by the ball after 1 second? [1]
(B) How long will the ball take to hit the ground? [1]
(C) What are the two possible times to reach the ball at the same height of 128 m?
OR
What is the maximum height reached by the ball? [2]
Answer:
(A) In the basis of given equation, h = – 16t² + 64t + 80
when, t = 1 second
h = – 16(1)² + 64(1) + 80
= -16 + 144 = 128 m

(B) When ball hits the ground, h = 0
-16t² + 64t + 80 = 0
∴ t² – 4t – 5 = 0
(t – 5) (t + 1) = 0
t = 5 or t = – 1
Since, time cannot be negative, so the time = 5 seconds.

(C) Since, h = -16t² + 1612 + 80
⇒ 128 = -16t² + 64t² + 80
⇒ 16t² + 64t + 80 – 128 = 0
⇒ 16t² + 64t – 48 = 0
⇒ t² – 4t + 3 = 0
⇒ t² + 3t – t + 3 = 0
⇒ (t – 3) (t – 1) = 0
⇒ t = 3,1
OR
Rearrange the given equation, by completing the square, we get
h = -16 (t² – 4t – 5)
= -16[(t – 2)² – 9]
= – 16(t- 2)² + 144
Height is maximum, when t = 2
∴ Maximum height = 144 m

CBSE Sample Papers for Class 10 Social Science Set 9 with Solutions

October 26, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Social Science Set 9 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
Section A – From question 1 to 20 are MCQs of 1 mark each.
Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
Which of the following is NOT applicable for a worker called Shyama working in the organised sector?
(a) Shyama gets a regular salary at the end of the month for her work.
(b) She is not paid for a leave she takes when she falls ill.
(c) She gets medical allowance to handle emergency medical situations like surgery.
(d) She got an appointment letter stating the terms and conditions of work when she joined her workplace. [1]
Answer:
(b) She is not paid for a leave she takes when she falls ill.
Explanation: In an organised sector, employees are given medical leaves. They are paid despite being absent from work if the reason behind their absence is medical in nature. For example- Maternity leaves given to pregnant women who receive their pay for a fixed period even when they are on leave during child birth and after that.

Question 2.
Observe the picture below carefully and answer the following question:

Which of the following challenges is being highlighted in this cartoon?
(a) the challenge of growing role of money and muscle power in parties.
(b) the challenge of lack of internal democracy within parties
(c) the challenge of dynastic succession
(d) the challenge of external democracy and party politics [1]
Answer:
(a) the challenge of growing role of money and muscle power in parties
Explanation: The woman can be seen bribing the person in authority in the party with money. Influential people use money and muscle power to sway political parties in their favour and to make them work according to their interests. This is one of the most fundamental challenges faced by political parties.

Related Theory:
Political parties are focused on winning elections. They nominate rich candidates. Rich people and big companies who fund the parties, tend to have an influence on the policies and decisions of the party. Parties can support criminals too if they can win the elections for them. This increasing role of rich people and big companies in democratic politics is a challenge to democrats all over the world.

Question 3.
Which of the following options can be an ideal developmental goal for an unemployed urban youth?
(a) An opportunity to travel across the world and learn new languages.
(b) A stable job appropriate to his qualifications.
(c) An opportunity to study for other degrees.
(d) Both (a) and (c) [1]
Answer:
(b) A stable job appropriate to his qualifications
Explanation: Development Goals are decided according to immediate requirements of people for development. For an unemployed youth who is qualified and well educated, a stable job which meets his qualifications is the most appropriate development goal.

Question 4.
Which type of government was functioning in France before the revolution of 1789?
(a) Dictatorship
(b) Military Rule
(c) Monarchical Rule
(d) Council of French Citizen was ruling [1]
Answer:
(c) Monarchical Rule
Explanation: France was a full-fledged territorial state in 1789 under the rule of an absolute monarch. The political and constitutional changes that came in the wake of the French Revolution led to the transfer of sovereignty from the monarchy to a body of French citizens.

Question 5.
Which of the following countries can provide a cheap manufacturing location to MNCs?
(a) Australia
(b) India
(c) United States of America
(d) Japan [1]
Answer:
(b) India
Explanation: Developing countries provide a cheap manufacturing location because they have abundance of labour and raw material. They have cheaper transportation and are well connected because of the newly developing infrastructure.

Related Theory:
India has highly skilled engineers who can understand the technical aspects of production. It also has educated English speaking youth who can provide customer care services.

Question 6.
Decisions in democracy are generally delayed because of which of the following features of democracy?
(a) Democratic government is formed by the choice of the people of the country.
(b) Democratic government is based on the idea of deliberation and negotiation.
(c) Democratic government listens to the voice of both minority and majority.
(d) Democratic government promotes the dignity and harmony of each individual. [1]
Answer:
(b) Democratic government is based on the idea of deliberation and negotiation.
Explanation: Since Democracy involves general consensus among members of legislature both at Central and State level- which means lengthy discussion and negotiation among the hundreds of members of these bodies. It becomes time consuming. However this deliberation and negotiation represents the idea of taking into consideration interests of each and every citizen of the country while deciding something that affects them all. This makes decisions legitimate.

Question 7.
What is a vellum?
(a) a kind of vehicle
(b) a kind of animal
(c) a kind of stationary material
(d) a kind of musical instrument [1]
Answer:
(c) a kind of stationary material
Explanation: A vellum is a parchment made from the skin of animals. Luxury editions were still handwritten on very expensive vellum, meant for aristocratic circles and rich monastic libraries which scoffed at printed books as cheap vulgarities.

Related Theory:
By 1448, Gutenberg perfected the Press system. The first book he printed was the Bible.

Caution:
Terms mentioned in the chapter must be written separately in a list. Each of such terms can be asked in the exam in form of definitions. Students must pay attention to these terms and memorise a short definition of each one of them in their own words.

Question 8.
Which deposits are formed by the decomposition of a wide variety of rocks rich in aluminium silicates?
(a) Mica
(b) Limestone
(c) Bauxite
(d) Manganese [1]
Answer:
(c) Bauxite
Explanation: Bauxite, a clay-like substance that alumina and later aluminium is obtained. Bauxite deposits are formed by the decomposition of a wide variety of rocks rich in aluminium silicates.

Related Theory:
Aluminium combines the strength of metals such as iron, with extreme lightness and also with good conductivity and great malleability.

Question 9.
Choose the correctly matched pair
(a) Life expectancy at birth – weight
(b) BMI – height
(c) Infant mortality rate – height
(d) Literacy rate – attendance in school [1]
Answer:
(b) BMI – height
Explanation: Body mass index is calculated by measuring weight and height and thereafter dividing them. Other pairs are not inter-related. Correctly matched pairs are: Life expectancy at birth-age and nutrition Infant mortality rate-age and nutrition literacy rate academic skills.

Question 10.
Following are some suggestions to reform parties and leaders in India. Identify which of them is helpful to curb the misuse of money and muscle power in politics. Suggestions:
(I) There should be state funding of elections.
(II) Parties should give one-third of its tickets to women candidates.
(III) Parties should look for leaders from humble background.
(IV) A law should be made to regulate the internal affairs of political parties.
Options:
(a) (I) and (IV)
(b) (I) and (II)
(c) (I), (II) and (III)
(d) (I), (III) and (IV) [1]
Answer:
(b) (i) and (ii)
Explanation: State funding of elections means the government should give parties money to support their election expenses. This support could be given in kind (petrol, paper, telephone etc.) or it could be given in cash on the basis of the votes secured by the party in the last election.

Similarly, there should be a quota for women in the decision making bodies of the party other than giving one-third of party tickets to women candidates.

Related Theory:
The Supreme Court of India passed an order to reduce the growing role of money and muscle power which make it mandatory for every candidate who contests elections to file an affidavit giving details of his property and criminal cases pending against him. This has made a lot of information available to the public.

Question 11.
Chinkara is the other name given to:
(a) a dolphin
(b) a Gharial
(c) a Black Buck
(d) a Kashmiri Stag [1]
Answer:
(c) A Black Buck

Question 12.
How does the private sector contribute in the development of a nation?
(a) It increases the demand of the products by advertisements.
(b) It funds the welfare scheme for the disadvantaged classes.
(c) It raises the revenue of the country massively.
(d) It provides luxurious services to the rich. [1]
Answer:
(c) It increases the productivity of the country in the manufacturing of industrial goods.
Explanation: Activities in the private sector are guided by the motive to earn profits. However, through the sale of goods and services, the activities of the private sector exponentially increases the revenue of the country.

Caution:
Reading between the lines is an extremely important device to understand and use. For students, every statement given in the chapter is important but rather than rote learning, it is important for them to understand what a statement implies and then relate it to other sub-topic. Most conceptual questions are based on such topics.

Question 13.
Identify the personality with the help of given hints.
(I) He headed the Awadh Peasant movement along with Baba Ramchandra.
(II) He was the President of Lahore Congress Session in December 1929.
(III) He addressed the peasants in United Provinces after the Rae Bareli farming in 1921.
(a) M. R. Jayakar
(b) Mahatma Gandhi
(c) Jawahar Lai Nehru
(d) Sardar Vallabh Bhai Patel [1]
Answer:
(c) Jawahar Lal Nehru
Explanation: In June 1920, Jawaharlal Nehru began going around the villages in Awadh, talking to the villagers, and trying to understand their grievances. By October, the Awadh Kisan Sabha was set up headed by Jawaharlal Nehru, Baba Ramchandra and a few others.

Question 14.
Rnd the odd one out:
(a) Nehru, Sardar Patel, Tilak
(b) Gandhi, C. R. Das
(c) M. R. Jaykar, Iqbal, Irwin
(d) Motilal Nehru, Bose, Ghaffar Khan [1]
Answer:
(c) M.R. Jaykar, Iqbal, Irwin
Explanation: All the other options include leoders which were a part of Indian National Congress at one point in their careers.

Question 15.
There are two statements marked as Assertion (A) and Reason (R). Mark your answer as per the codes provided below:
Assertion (A): An industrial society based on mass production cannot be sustained without mass consumption.
Reason (R): The goal of full employment could only be achieved if governments had power to control flows of goods, capital and labour.
(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but the Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Explanation: An industrial society based on mass production cannot be sustained without mass consumption because people need incomes to sustain lifestyles. If not for demand, the supply wouldn’t rise.

Related Theory:
To ensure mass consumption, there was a need for high and stable incomes. Incomes could not be stable if employment was unstable. Thus stable incomes also required steady, full employment which can only be provided if the government ensures this.

Question 16.
There are two statements marked as Assertion (A) and Reason (R). Mark your answer as per the codes provided below:
Assertion (A): The cost to the borrower of informal loans is much higher.
Reason (R): The principals of these kinds of loans are really large.
(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but the Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(c) Assertion (A) is true but Reason (R) is false.
Explanation: The cost to the borrower of informal loans is much higher because the rates of interest are really high and the borrower has to almost pay double take principal borrowed in the name of interest. This happens because informal loans are lent without collateral and arbitrarily.

Question 17.
Observe the table and answer the following question:

	Column A		Column B
(A)	Ninety five theses	(I)	Martin Luthur King
(B)	Kesari	(II)	Tilak
(C)	Ramcharitmanas	(III)	……………..
(D)	Amar Jiban	(IV)	Rashsundar Devi

Based on your understanding of the relationship between items in Column A and B, choose the option that fills the blank most appropriately.
(a) jyotiba Phule
(b) Valmiki
(c) Tulsidas
(d) Tarabai Shinde [1]
Answer:
(c) Tulsidas
Explanation: Tulsidas wrote Ramcharitmanas in the sixteenth century. The first printed edition of this book came out in 1810 in Culcutta

Question 18.
Identify the sanctuary mentioned in the hints.
In this sanctuary, villagers have fought against mining by citing the Wildlife Protection Act. In many areas, villagers themselves are protecting habitats and explicitly rejecting government involvement.
(a) Sacred Groves
(b) Sariska Tiger Reserve
(c) Buxa Tiger Reserve
(d) Bhairodev Dakav Sonchuri [1]
Answer:
(b) Sariska Tiger Reserve
Explanation: In Sariska Tiger Reserve, Rajasthan, villagers have fought against mining by citing the Wildlife Protection Act.

Related Theory:
The inhabitants of five villages in the Alwar district of Rajasthan have declared 1,200 hectares of forest as the Bhairodev Dakav ‘Sonchuri, declaring their own set of rules and regulations which do not allow hunting, and are protecting the wildlife against any outside encroachments.

Question 19.
The Central government has the power of making laws on which of the following subjects?
(a) Subjects in the Union list.
(b) Subjects in the state list.
(c) Subjects in the concurrent list.
(d) Central govt cen make laws on all the mentioned lists. [1]
Answer:
(d) Central government can make laws on all the mentioned lists.
Explanation: Central Government generally makes laws on subjects in Union and Concurrent list but can also make laws on subjects involved in State list of requested by two or more states in form of writing. Thus, Central Government can make laws on all the three lists with certain restrictions due to the federal setup of Indian administration.

Question 20.
King Victor Emmanuel II belonged to which of the following countries?
(a) England
(b) Spain
(c) France
(d) Italy [1]
Answer:
(d) Italy
Related Theory:
In 1861 Victor Emmanuel II was proclaimed king of united Italy.

Section – B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
Why are metalled roads are called all weather roads?
OR
What is an agglomeration economy? [2]
Answer:
Metalled Roads are made of cement, concrete and bitumen of coal which makes them strong enough to handle all kinds of weather. This makes these roads accessible even in the roughest rain or snow storms. Thus they are known as all weather roads.
OR
Co-operation of multiple industries together at a place (generally urban centres) which also offers other tax, market and raw material related benefits are called Agglomeration Economics.

Question 22.
How are the modern forms of money different from the early forms of money? [2]
Answer:
Facts that show modern forms of money are different from the early forms of money:

Modern forms of money include currency- paper, notes and coins, while the older form included gold or silver coins and even grains and objects.
Unlike the things that were used as money earlier, modern currency is not made of precious metal such as gold, silver and copper, grains or cloth.
Value of the new currency is not calculated by the material it is made of. Paper is cheaper than metal yet still larger in value in form notes.
Modern currency also includes debit cards, online transfers and other forms in which currency is not present in physical form. It is transferred virtually to the seller.

Question 23.
What were people’s apprehension’s weather roads? associated with the printed books? [2]
Answer:
People were apprehensive of the impact printed books had on the minds of people who read them. It was feared that if there was no control over what was printed and read then rebellious and irreligious thoughts might spread.

Question 24.
How do Political parties encourage public participation in a democracy? Elaborate. [2]
Answer:
Political Parties encourage public participation in the following ways:

Political parties raise issues keeping interests of various classes and caste groups in mind. This encourages participation of the involved communities in taking a decision.
They also stage various protests against dissatisfying policies and procedures which gives the public a platform to come ahead and speak for themselves.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
What was the aim of the business class in supporting the Civil Disobedience movement and how did they help? [3]
Answer:
During the First World War, Indian merchants and industrialists had made huge profits. They wanted relief against colonial policies that restricted business activities.

They wanted protection against imports of foreign goods along with a rupeesterling foreign exchange ratio that would discourage imports.
They formed various organisations to organise their interests and supported the Civil Disobedience Movement with financial assistance and refused to buy or sell imported goods.
After the movement was called off once, they lost interest. Later they were apprehensive of the spread of militant activities and withdrew support.

Question 26.
What is Bank? Illustrate the functions of a Bank. [3]
Answer:
Banks are institutions which accept the deposits from the public. they allow withdrawal of money by cheques and by demand deposits and provide advance loans of various types to the borrowers.
Following are the functions of a bank:

Depositor of Money: Banks accepts deposits from customers in various forms. It can be in the form of saving account deposits, current account and fixed deposits.
Interest on Savings: It gives interests on the savings to the customers. This can add on to their principal amount deposited earlier and can give a handsome return to the depositors.
Withdrawal Facilities: It provides withdrawal facilities to the customers as and when needed.
Providing Loan: It gives loans and advances to the borrowers and charge certain interests on them.
Agency Function: It provides agency function like transfer of funds, collection of funds, payment of various items, purchase and sale of shares and securities. (Mention any 3 points)

Question 27.
How is the sharing of power between the Union and the State Government basic of the structure of the constitution of India? Explain. [3]
Answer:
Power sharing is a basic structure of constitution. This can be asserted in the following ways:

It is not easy to make changes in power sharing arrangement. The process is long and tedious.
Parliament has not been given the sole power to change this arrangement.
Any change to it has to be first passed by both the houses of parliament, with at least 2/3 majority. It has to be ratified by legislature of at least half of the total states.

Question 28.
What are millets? Why are millets a very important food crop in India? [3]
Answer:
Millets are a group of highly variable small- seeded grasses, widely grown around the world as cereal crops or grains for fodder and human food. Millets are very important food crop in India because:

They are crops with very high nutritional value. These are used as food crops as well as fodder crops also.
These crop is favoured due to its productivity and short growing season under dry, high temperature conditions. They are also known as coarse grains.
Jowar, Bajra and Ragi are the important millets grown in India.

Question 29.
“There is overwhelming support for the idea of democracy all over the world.” Support the statement.
OR
Democracy is more effective than its other alternatives. Justify the statement. [3]
Answer:
There is an overwhelming support for the idea of democracy all over the world because:

A democratic government is people’s own government.
The evidence from South Asia shows that the support exists in countries with democratic regimes.
People wished to be ruled by representatives elected by them.
People believe that democracy is suitable for their country.
Democracy has the ability to generate its own support which in itself is an outcome that cannot be ignored.

OR
Democracy is more effective than its other alternatives in the following ways:

Democracy promotes equality among citizens.
It enhances the dignity of the individual.
It also improves the quality of decision-making.
It provides a method to resolve conflicts.
It gives room to correct mistakes.
Democratic government is a legitimate government.
Democracy’s ability to generate its own support is itself an outcome that cannot be ignored. (Mention any 3 points)

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
What are Multinational Corporations? Describe any three features of multinational corporations.
OR
Describe the role of technology in promoting globalisation process. [5]
Answer:
Multinational corporations are the companies that own or control the manufacturing of goods in more than one country. They are based in multiple countries and thereby control and integrate markets through the same. They base their operations in suitable countries to earn the most profit.

Some of the main features of a multi-national Cooperation are:

They set up factories and offices in more than one country.
Multi-National corporations run through Foreign and domestic investments and employ thousands of employees in different countries on both contract and permanent basis.
Multi-National corporations help in integrating markets and transfer of culture from one country to another.

For example-PUMA, Ranbaxxy, Asian Paints, Hero Honda, Google etc.
OR
The contribution of technology in globalization is as follows:

There have been many improvements in the transport sector, which helps in the export and import of goods. This increases trade relations between countries.
Since technology improves efficiency, the process of exchange has become faster and cheaper.
Development in information and communications technology has been the most beneficial since information can now sent across the world.
Developments in IT have also led to the production of services through outsourcing like call centres, online teaching, etc.
Telecommunications have improved contact between people. People living in two different countries can easily remain in touch with minimal prices.
Now, people can place an order for anything from any part of the world and it is at there doorstep with in the next few days. (Any 5 of 6 points can be written to get full marks)

Question 31.
Railways are the principal mode of transportation in India. Explain.
OR
What is the manufacturing sector? Why is it considered the backbone of development? Interpret the reason. [5]
Answer:
Railways are the principal mode of transportation for freight and passengers in India. There are various reasons behind it.
(1) Railways make it possible to conduct multifarious activities like business, sightseeing, and pilgrimage along with transportation of goods over longer distances.

(2) Apart from an important means of transport, Indian Railways has been a great integrating force for more than 150 years.

(3) Railways in India bind the economic life of the country as well as accelerate the development of the industry and agriculture.

(4) The Indian Railways have a network of 7,031 stations spread over a route Length of 63,221 km. with a fleet of 7817 locomotives, 5321 passenger services vehicles, 4904 other coach vehicles and 228,170 wagons as on 31 March 2004.

(5) Railways are the cheapest mode of transportation. They can be used by rich and poor classes alike. For long distances, they can transfer both goods and people easily.
OR
Production of goods in large quantities after processing from raw materials to more valuable products is called manufacturing. The sector that carries this process out is known as Manufacturing Sector.
It is considered the backbone of development because:
(1) It not only helps in modernising agriculture but also forms the backbone of our economy by employing a large number of people. It also reduces the heavy dependence of people on agricultural income by providing them jobs in secondary and tertiary sectors.

(2) It provides new modern techniques for farmers to apply to increase their agricultural produce.

(3) Industrial development is a pre-requisite for eradication of unemployment and poverty from a country.

(4) Export of manufactured goods expands trade and commerce. It helps to bring in more foreign reserves and strengthens the economy of our country.

(5) Establishing industries in backward and tribal areas helps them progress and mix better with other developed areas. Thus, it helps in regional development

Question 32.
Highlight the relationship between democracy and economic development?
OR
Suggest and explain any five ways to reform Political Parties in India. [5]
Answer:
The relationship between democracy and economic development is as follows:

Democracies promote dignity and progress and thus are expected to produce development.
Democracy is not directly related or proportional to economic development.
Between Democracies and dictatorships, dictatorships have been known to produce better higher rates of economic growth.
Economic development of any country depends on country’s population and size, global market, demand and supply, economic policies pursued by the government, the technology etc.
Multiple variables affect the relationship between democracy and economic development, so even when democracies are able to achieve redistribution of economic resource, economic growth is still not guaranteed.

A law should be made to regulate the internaL affairs of poLiticaL parties.
It should be made compulsory for the political parties to maintain an attendance register of its members.
It should be made mandatory for political parties to reserve 1/3rd seats for its women candidates.
There should be a quota to choose and appoint women to the decision-making bodies of the party.
There should be state funding of elections. Vote casting should be made compulsory in each election.

Question 33.
Highlight the origin and role of the Hindustan Socialist Republic Army.
OR
“Print Revolution in sixteenth century Europe transformed the lives of people.” Support the statement with suitable arguments. [5]
Answer:
In 1928, the Hindustan Socialist Republican Army (HSRA) was founded at a meeting in Ferozeshah Kotla ground in Delhi. Amongst its leaders were Bhagat Singh, Jatin Das and Ajoy Ghosh to fight against British colonial rule in India and achieve independence for the country through an armed rebellion if necessary. The Army was made of zealous individuals who believed moderate and constitutional methods would not be able to achieve Independence and Indians will have to gather together to fight British out of their land.

The HSRA targeted different symbols of British power in India in several instances, some of which included throwing a bomb at the legislative assembly. The army stood for revolution in a society but used violence as a method to achieve that.

Several revolutionary members of this party were hanged and executed due to their notorious activities against the British Government.
OR
Print revolution in sixteenth century Europe transformed the lives of people in the following ways:
(1) Print culture also affected the life of poor people and women in many ways. The print gave birth to a new form of popular literature. Very small books were brought out. They were sold at crossroads.

(2) The poor people brought these books and read with great interest. Books were cheap so that the poor people could also afford them.

(3) Reading by women increased enormously in middle class homes. Liberal husbands and fathers began educating their women folk at home and sending them to schools. Women schools were also set up.

(4) Print popularized the ideas of the Enlightenment thinkers. Collectively, their writings provided a critical commentary on tradition, superstition and despotism.

(5) The ideas of scientists and philosophers now became more accessible to the common people.

Related Theory:
The Print Revolution transformed the lives of people by opening the door of knowledge to a vast literate population. It also changed people’s relationship to information and knowledge and with institutions and authorities.

Section – E
Case Based Questions (3 x 4 = 12)

Question 34.
Read the source given below and answer the following questions:
Political parties are easily one of the most visible institutions in a democracy. For most ordinary citizens, democracy is equal to political parties. If you travel to remote parts of our country and speak to the less educated citizens, you could come across people who may not know anything about our Constitution or about the nature of our government.

But chances are that they would know something about our political parties. At the same time this visibility does not mean popularity. Most people tend to be very critical of political parties. They tend to blame parties for all that is wrong with our democracy and our political life. Parties have become identified with social and political divisions.
(A) What is a political party? [1]
(B) Define partisanship. [1]
(C) Analyse the reason for blaming political parties. [2]
Answer:
(A) A political party is a group of people who come together to contest elections and hold power in the government. Explanation: All those who come together to contest elections and hold power in the government agree on some policies and programmes for the society with a view to promote the collective good.

(B) A person who is strongly committed to a party, group or faction is called partisan. Partisanship is marked by a brandency to take sides and inability to be natural about a decision.

(C) People blame political parties for all that is wrong with democracy and our political life because they are the most visible institutions in a democracy.

However, it is wrong to blame political parties for the loopholes in democracy because political parties are only instruments to implement the pronciples of democracy. They are not a constituent of democracy.

Question 35.
Read the given source and answer the following questions:
Pipeline transport network is a new arrival on the transportation map of India. In the past, these were used to transport water to cities and industries. Now, these are used for transporting crude oil, petroleum products and natural gas from oil and natural gas fields to refineries, fertilizer factories and big thermal power plants.

Solids can also be transported through a pipeline when converted into slurry. The far inland locations of refineries like Barauni, Mathura, Panipat and gas based fertilizer plants could be thought of only because of pipelines.
(A) Mention the names of goods that can be generally transferred through pipelines. [1]
(B) How is pipelines an environment-friendly mode of transport? [1]
(C) Mention two problems associated with the use of pipelines. [2]
Answer:
(A) Pipelines can transfer slurry, liquid fuels, gasses and water from one place to another.

(B) Pipelines is an environment-friendly mode of transport because:

It doesn’t release any gasses or fumes while the transfer is taking place leading to almost zero wastage or leakage.
It doesn’t leave behind any residues to be dumped or disposed off anywhere.

Initial cost of laying pipelines is very high.
Maintenance might obstruct the working of the pipelines for a long time. Also, pipelines cannot be laid in all terrains. Suitable conditions are necessary.

Question 36.
Read the source given below and answer the following questions:
Suppose for the present that a particular country is quite developed. We would certainly like this level of development to go up further or at least be maintained for future generations. This is obviously desirable. However, since the second half of the twentieth century, a number of scientists have been warning that the present type, and levels, of development are not sustainable.

For example, recent evidence suggests that the groundwater is under serious threat of overuse in many parts of the country. About 300 districts have reported a water level decline of over 4 metres during the past 20 years. ‘Nearly one-third of the country is overusing their groundwater reserves. In another 25 years, 60 percent of the country would be doing the same if the present way of using this resource continues.

Groundwater overuse is particularly found in the agriculturally prosperous regions of Punjab and Western up, hard rock plateau areas of central and south India, same coastal areas and the rapidly growing urban settlements.
(A) Mention two examples of renewable resources? [1]
(B) Analyse the reason behind overuse of groundwater resources. [1]
(C) How can ground water be conserved? [2]
Answer:
(A) Groundwater and Solar Power
Explanation: Groundwater is a renewable resource because this is replenished by nature as in the case of crops and plants.

(B) Groundwater is a source of freshwater. It is exploited in Punjab and Uttar Pradesh because:

The regions are heavily dependent upon rainfall and external water supply for irrigation. Groundwater resources are utilized to fulfill this requirement.
These regions are densely populated and groundwater is their primary source of potable drinking water. Rising population puts greater pressure on these resource.

Recycle, reuse and treat the used water so it can be further upcycled for other purposes.
Minimise surface run-off and cultivate crops with a lower water footprint.

Section – F
Map Based Questions (2 + 3 = 5)

Question 37.
(a) Two places A and B have been marked on the given political outline map of India. Identify them with the help of the following information and write their correct names on the lines drawn near them.
(A) A place where cotton mill workers organised their Satyagraha.
(B) A place where Congress Session was held in December 1927 [2]

(b) Locate and label any three of the following with appropriate symbols on the same given outline political map of India.
(a) Singrauli – Thermal Power Plant.
(b) Mumbai – Chhatrapati Shivaji International Airport.
(c) Nagarjuna Sagar Dam
(d) Gandhinagar STP [3]

Answer:
(a) (A) Ahmedabad (B) Nagpur
(b) Located and labelled on the map.

CBSE Sample Papers for Class 10 Social Science Set 8 with Solutions

October 25, 2022

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Social Science Set 8 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
Section A – From question 1 to 20 are MCQs of 1 mark each.
Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
Why did industries like gas works, breweries, book-binders prefer hand labour over machines in Victorian Britain?
(a) Machines were not available for most of the work.
(b) Demand was seasonal so maintaining machines for the entire year didn’t seem economic.
(c) The products of these industries were not in demand.
(d) The work was intricate and could not be done through machines. [1]
Answer:
(b) Demand was seasonal so maintaining machines for the entire year didn’t seem economic.
Explanation: In these industries, the demand was seasonal. Industrialists did not want to introduce machines to get rid of human labour because maintenance for the rest of the year was expensive and unproductive. They preferred to employ workers only for a limited period of time during peak demand.

Related Theory:
Other such industries where demand fluctuated with season include printing and ship maintenance.

Question 2.
Which of the following movements for Independence was never called off by Gandhi?
(a) Non-Cooperation movement
(b) Khilafat movement
(c) Quit India movement
(d) Civil Disobedience movement [1]
Answer:
(c) Quit India movement
Explanation: Non-Cooperation movement was called off by Gandhi in 1922 after violence ensued in Chauri Chaura at Gorakhpur, Uttar Pradesh. Civil Disobedience movement was called off by Gandhi in December 1931 when he decided to attend the Second Round Table Conference in London.

Related Theory:
Khilafat Movement was a part of the Non¬cooperation movement. It was started to protest against the ill-treatment meted out to the Khalifa of Turkey by the British.

Question 3.
Identify this crop through the given hints: Sometimes, the new crops could make the difference between life and death. In Ireland, hundreds of thousands people died of starvation due to their dependency on a crop in mid 1840s.
(a) Tomato
(b) Groundnuts
(c) Maize
(d) Potato [1]
Answer:
(d) Potato
Explanation: After potato was discovered in America by the Spaniards, it quickly became popular in the Europe. People’s diet centred around the crop so much that hundreds and thousands died later during the Irish potato famine where a pest disease affected the crop resulting in its failure.

Related Theory:
Many of our common foods come from the native inhabitants of America- American Indians. America had been cut off from the rest of the world before it was discovered by Columbus.

Question 4.
Why did Mahatma Gandhi manufacture salt to violate a law?
(P) Salt tax bothered the rich and the poor alike. It joined people in a common struggle against their oppressors.
(Q) Breaking a law by manufacturing salt was the least violent and most effective rebellion.
(R) People were not. interested in breaking the law in any other form.
(a) Both P and R are correct
(b) Only P is correct
(c) P and Q are both correct
(d) P, Q and R are correct [1]
Answer:
(c) P and Q are both correct.
Explanation: Mahatma Gandhi was confident that salt was a common link between the rich and the poor. Both classes used them everyday. The tax and British monopoly on salt production hurt both these classes. Hence, he chose salt as a symbol to unite Indians. It was
also important for Gandhi, as a supporter of Non-violence, to choose a battle which would unite Indians without shedding any blood. Hence, breaking this law was a safe bet.

Question 5.
Identify the political party using the given picture and hints:
(1)

(2) It was formed in 1999 following a split in the Congress party.
(3) The party wants the high offices in the country to be restricted to natural born citizens of India.
(4) It is a major party in Maharashtra.
(a) NCP
(b) INC
(c) BJP
(d) BSP [1]
Answer:
(a) NCP
Explanation: The Nationalist Congress Party was formed in 1999 after INC went through a split. The party espouses democracy, Gandhian secularism, social justice and freedom. Apart from Maharashtra, the party also has significant presence in North-east India.

Question 6.
Which of the following characteristics is NOT true about commercial farming?
(a) Use of higher doses of modern inputs, high yielding variety (HYV) seeds is prominent in this type of farming.
(b) Use of chemical fertilisers, insecticides and pesticides is prominent in this type of farming.
(c) The degree of commercialisation of agriculture remains the same across various regions.
(d) Crops are grown for business and not sustenance. [1]
Answer:
(c) The degree of commercialisation of agriculture remains the same across various regions.
Explanation: The degree of commercialisation of agriculture varies from one region to another. At some place, commercialisation is greater while in other states, crops are still grown for sustenance more than for commercial purposes.

Related Theory:
Rice is a commercial crop in Haryana and Punjab, but in Odisha, it is a subsistence crop.

Question 7.
Match the items in column A with those of column B and choose the most appropriate code which reflects the correctly matched pairs.

	Column A		Column B
(A)	Khadins	(I)	Bikaner
(B)	Kuls	(II)	Jaisalmer
(C)	Drip Irrigation	(III)	Western Himalayas
(D)	Tankas	(IV)	Meghalaya

Codes:
(a) (A)-(II) (B)-(III) (C)-(IV) (D)-(I)
(b) (A)-(I) (B)-(IV) (C)-(II) (D)-(III)
(c) (A)-(III) (B)-(I) (C)-(II) (D)-(IV)
(d) (A)-(II) (B)-(l) (C)-(IV) (D)-(III) [1]
Answer:
(a) A-(II). B-(III), C-(IV), D-(I)
Explanation: In hill and mountainous regions, people built diversion channels like the ‘Guls’ or ‘Kuls’ of the Western Himalayas for agriculture. ‘Khadins’ in Jaisalmer and ‘Johads’ are agricultural fields converted into rain-fed storage structures in Rajasthan.

Drip Irrigation is practiced in North-Eastern States because of water scarcity prevalent in the region. Moisture retention of the soil is low in North-East India.

Question 8.
By the seventeenth century, as urban culture bloomed in China, the uses of print diversified. Which of the given statements supports this assertion?
(a) Reading became a monotonous activity.
(b) People started reading less books and more pamphlets.
(c) Merchants began using print everyday as they collected trade information.
(d) Women started reading more books. [1]
Answer:
(c) Merchants began using print everyday as they collected trade information.
Explanation: The usage of print diversified. It was no longer only used for publishing books. It was also used by merchants for book keeping.

Caution:
It is important for the students to read the chapter and understand each line rather than trying to remember specific details. Once the students grasp the concept, it will be easier to answer indirect questions like these which are given as an assertion.

Question 9.
How does the practice of decentralisation help a democracy?
(a) It increases voter bank
(b) It diversifies caste expression.
(c) It increases democratic participation.
(d) It gives birth to new political leaders. [1]
Answer:
(c) It increases democratic participation.
Explanation: At the local level, it is possible for the people to directly participate in decision making. Thus, decentralisation helps to inculcate a habit of democratic participation.

Related Theory:
The local government is the best way to realise one important principle of democracy, namely local self-government.

Question 10.
Which of the following Indian companies has emerged as an MNC lately?
(a) Nykaa
(b) The man company
(c) Asian Paints
(d) Puma [1]
Answer:
(c) Asian Paints
Explanation: Globalisation has enabled some large Indian companies to emerge as multinationals themselves. Some of these are Tata Motors (automobiles), Infosys (IT), Ranbaxy (medicines), Asian Paints (paints), Sundaram Fasteners (nuts and bolts) etc.

Related Theory:
An MNC has operation and production set up in more than one country. They cater to consumers based in different parts of the world.

Caution:
For students to answer questions based on real- life situations, it is important that they be aware of current affairs and recent events. For example, for this question, it is important for the students to recognise these companies to be able to decide if they are MNCs or not

Question 11.
Choose the correctly matched pair:

Column A	Column B
(a) Automobile Industry	Manganese
(b) Cement Industry	Silica
(c) Aluminium Smelting Industry	Bauxite
(d) Fertiliser Industry	Alumina

Answer:
(c) Aluminium Smelting Industry-Bauxite
Explanation: Bauxite is the raw material used in the smelters. It is a very bulky, dark reddish coloured rock.

Related Theory:
Aluminium smelting plants in the country are located in Odisha, West Bengal, Kerala, Uttar Pradesh, Chhattisgarh, Maharashtra and Tamil Nadu.

Question 12.
There are two statements marked as Assertion (A) and Reason (R). Mark your answer as per the codes provided below :
Assertion (A): Per Capita Income can adequately convey the status of development of a community.
Reason (R): High income helps them to afford most basic necessities including food, water, education and medical care.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(d) Assertion (A) is false but Reason (R) is true.
Explanation: Per Capita Income cannot adequately convey the status of development of a community because income can only buy materialistic necessities. It cannot buy the citizens their rights, justice, equality and liberty which are equally important for holistic development.

Question 13.
Which country has a special arrangement for accommodation of linguistic diversity in form of a community government?
(a) Sri Lanka
(b) France
(c) Germany
(d) Belgium [1]
Answer:
(d) Belgium
Explanation: Belgium has accommodated its linguistic diversity by electing a special community government which is chosen by individual linguistic groups.

Related Theory:
The community government has powers regarding cultural, educational and other language related issues.

Question 14.
Sources of Credit per Rs. 10,000 for rural households have been tabulated below :

Sources of Credit	Percentage
Government	1%
Cooperative Society Banks	25%
Commercial Banks	25%
Landlords	1%
Relatives and Friends	8%
Other non-lnstitutional agencies	2%
Money lenders	33%
Other Institutional agencies	5%

Which of the following inferences can be made after reading the given table?
(a) Rural Indians were mainly borrowing from informal sources of credit in 2012.
(b) In India, Banks are the largest source of credit for rural people.
(c) Cooperative societies/Banks lend to common people more than commercial Banks.
(d) Relatives and friends are the go-to sources for rural people when they require credit. [1]
Answer:
(b) In India, Banks are the largest source of credit for rural people.
Explanation: In India, Banks are the largest source of credit for rural people. Here, Banks include both the cooperative and commercial banks which together form the source of half of the total credit borrowed by the rural households. According to the given table, rural Indian households were mainly borrowing from formal sources of credit in 2012 (25%+25%+5%+1= 56%).

Question 15.
Read the given btable.
Some data regarding India and its neighbours for 2017:

Country	Gross national income (GNI) per capita	Life Ex-pectancy at birth (2017)	Mean years of school¬ing of peo¬ple aged 25 and above	HDI Rank in the World (2018)
Srilanka	11326	75.5	10.9	76
India	6653	68.8	6.4	130
Mayan-mar	5567	66.7	4.9	148
Pakistan	5331	66.6	5.2	150
Nepal	2471	70.6	4.9	149
Bangla-desh	3677	72.8	5.8	136

Based on your reading of the given table, choose the country the citizens of which have a comparatively better chance of holistic development.
(a) India
(b) Bangladesh
(c) Pakistan
(d) Sri Lanka [1]
Answer:
(d) Sri Lanka
Explanation: Since, almost all the indicators in the given table show better values for Sri Lanka- namely the per capita income, life expectancy at birth and mean years of schooling, it provides better opportunities for holistic development than the rest of the countries.

Related Theory:
This report has been released by the UNDP which considers not just per capita income but also educational and medical facilities available for the citizens of the country.

Question 16.
Find the odd one out:
(a) Farmer, Shopkeeper, Civil Engineer
(b) Manufacturing technician, Fisherman, Doctor
(c) Food processor, Construction worker, Worker in Textile mill
(d) Miner, Hotelier, Oilfield worker [1]
Answer:
(c) Food processor, Construction worker, Worker in textile mill.
Explanation: Other options have a mixture of employees and workers from all sectors of economic activities primary, secondary and tertiary sector.
Food processor, construction workers, workers in textile mills all belong to the secondary sector of economic activties. They process goods obtained from the primary sector.

Question 17.
Which of the following industries has received an immense boost after the Green Revolution?
(a) Automobile industry
(b) Chemical industry
(c) Textile industry
(d) Fertiliser industry [1]
Answer:
(d) Fertiliser industry
Explanation: Green Revolution was introduced to make India self-sufficient in producing food and other crops for its sustenance by employing new hybrid variety seeds, equipment, chemical fertilisers etc This led to a boost in the expansion of the fertiliser industry as a consequence.

Related Theory:
Gujarat, Tamil Nadu, Uttar Pradesh, Punjab and Kerala contribute towards half of the fertilizer production. Other significant producers are Andhra Pradesh, Odisha, Rajasthan, Bihar, Maharashtra, Assam, West Bengal Goa, Delhi, Madhya Pradesh and Karnataka.

Question 18.
What were the contents of the letter written by Mahatma Gandhi?
(a) It included the new Constitution of India. Mahatma Gandhi wanted the British to run India based on that.
(b) The letter contained 11 demands from the Indian citizens. They included fair treatment, removal of salt duty, freedom to revolutionaries and other prisoners etc.
(c) They letter included a contract from the Indian National Congress to rule India as partners.
(d) The letter included a request to form a new political party apart from the Indian National Congress. [1]
Answer:
(b) The letter contained 11 demands from the Indian citizens. They included fair treatment, removal of salt duty, freedom to revolutionaries and other prisoners, etc.
Explanation: On 31st January, 1930, he sent a ^letter to Viceroy Irwin stating eleven demands-of general and specific interests. These demands addressed the requests of different classes, from industrialists to peasants.

Related Theory:
Mahatma Gandhi found salt a powerful symbol that could unite the nation. Mahatma Gandhi believed that the tax on salt and the government monopoly over its production revealed the most oppressive face of the British rule.

Question 19.
Which continent was monumentally affected by Rinderpest in the nineteenth century?
(a) Asia
(b) Africa
(c) America
(d) Australia [1]
Answer:
(b) Africa
Explanation: Rinderpest arrived in Africa in the late 1880s. It was carried by infected cattle imported from British Asia to feed the Italian soldiers invading Eritrea in East Africa. Entering Africa in the east, rinderpest moved west ‘like forest fire’, reaching Africa’s Atlantic coast in 1892. Rinderpest killed 90 per cent of the cattle.

Related Theory:
This disease destroyed the livelihoods of Africans. Control over the scarce resource of cattle enabled European colonisers to conquer and subdue Africa.

Question 20.
Which of the following statements stand(s) true about the primary sector?
(I) Almost half of the workers in the country are working in the primary sector.
(II) Primary sector is the most important contributor in total production of goods and services in the country.
(III) Primary sector is dependent on tertiary sector.
Codes:
(a) Only I
(b) I and II
(c) I and III
(d) I, II and III [1]
Answer:
(c) I and III
Explanation: With time, the primary sector has not been able to retain its position as the most important contributor to the GDP. Tertiary sector has replaced the primary sector from that position because it generates more revenue.

Section – B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
Why is it significant to release reports like the Human Development Report by the UNDP or the World Bank? Mention any two points. [2]
Answer:
It is significant to release such reports because:
(1) Countries in different corners of the world develop at different pace. It is extremely important to know which countries are still facing issues like poverty, hunger, illiteracy, malnutrition or diseases to be able to help them through grants and transfer of technology.

(2) Every country has different standards of calculating poverty and development. Such reports set common standards for comparison.

Question 22.
Differentiate between the Eastern and Western world based on a picture called Two Magicians which was published in the 20th century. [2]
Answer:
The Two magicians was a picture published on the cover of a trade magazine in 1901. It distinguished between the Eastern and the Western world in the following ways:
(1) It signifies East through the figure of Aladdin, a fictional character, who built a palace through his magic lamp. This shows how the Eastern world was still steeped in mysticism, antiquity and fantasy.

(2) It shows the Western world through a modern mechanic who built bridges, ships and high rise building which indicates that the West was developing, advancing and learning about newer technology unlike the poor but pompous East

Question 23.
Mention any two reasons why one should prefer democracy over any other form of government.
OR
How has the constitutional status granted to local government units helped democracy? [2]
Answer:
Two reasons why one should prefer Democracy over other forms of government are:

Democracy promotes and enhances the dignity of an individual It guarantees them a platform to put forth their opinion.
Democracy provides methods to solve social and political conflicts which in turn makes expression of social, economic and political differences easier.

OR
The Constitutional status granted to local governments has helped in deepening the democracy immensely. This can be understood by the following points:

It has increased minority and women’s representation in the government of institutions.
It has helped in imparting power to the most elemental unit of a society- a common citizen.

Question 24.
Mention any two features of Alluvial soil. [2]
Answer:
Alluvial soil is the most widespread soil in India. It is spread across the entire Gangetic plains. Some of its features are:

Alluvial soils are very fertile. They contain adequate content of potash, phosphoric acid and lime.
Alluvial soils are classified as old alluvial soil (Bangar) and new alluvial soil (Khadar).

Caution:
To answer such questions, it is important for the students to memorise and recall the features of different classes and types discussed in each chapter. In this case, for example, students should be able to recall different features of each type of soil. To achieve that students should draw individual tables or helpful figures to understand the features and memorise them efficiently.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
How did the birth of nation-states in the nineteenth century rearrange the world into new groups and factions? [3]
Answer:
This can be understood as:

Before the birth of nation-states, the world was mostly divided into multi-national dynastic empires and provinces. The criterion of identification was not based on territory.
After nation-states were born, people began to identify with one another based on common language, political and economic struggle, common territory and values. This united them into different groups and factions than before.
For example, multi-national empires in Europe broke down into multiple nation-states, each with their own language, belief system, flags, lifestyle and struggles.

OR
France was truly considered to be the harbinger of change in Europe. This can be evidenced through these points:

France has always been the first nation to ignite a spark of rebellion among other European nations. French revolution led to the birth of nation states. It popularised values like equality, liberty and fraternity.
The July upheaval of 1830 which dethroned Bourbon kings and allowed the liberal revolutionaries to rule caused a revolution in Brussels which in turn led to Belgium breaking away from the United Kingdom of the Netherlands.
French revolutionaries were also responsible for liberating the people of Europe from despotism. They carried this message of nationalism and patriotism across Europe.

Caution:
It is important for the students to read the questions carefully before attempting to write it. The keywords used by the examiners set the tone for the answer. For example, in this case, the question only asks the students to support Metternich’s statements. Only those incidents which involve France and are relevant to support the assertion should be included in the answer.

Question 26.
In general, MNCs set up production where it is close to the markets. Do you agree? What other factors do MNCs consider while setting up a factory? [3]
Answer:
Yes, MNCs mostly set up production units or factories in the proximity of a well functioning market This is to reduce the cost of transportation and thus gain greater profits. Other factors considered by MNCs include:
(1) Availability of cheap labour – Cheap and abundant labour is extremely important for any factory to function without any obstructions. Rising labour wages can exponentially increase the cost of production.

(2) Availability of power, water and other factors of production at low costs – Cheap inputs without any supply obstructions help to produce efficiently.

(3) MNCs also set up factories at places with loosely implemented labour laws and regulations.

Question 27.
How can power-sharing reduce the fear of despotism in a country? Mention any three points. [3]
Answer:
Power sharing reduces the fear of despotism in the following ways:
(1) Power-sharing, by definition, can be understood as the process of distributing the responsibility and power to administer a state among different factions, institutions, communities and representatives.

(2) It provides representation to all the different communities residing the country. As a consequence, resources are equitably distributed and in most cases, a social conflict is avoided.

(3) Power sharing reduces the possibility of a majority community oppressing a minority community by using political power.

Question 28.
Mention three benefits of airway transportation. [3]
Answer:
Airways is an extremely beneficial mode of transportation. This can be explained through the following points:
(1) Airways is the fastest and most comfortable form of transportation.

(2) It makes difficult terrains like hills, mountains, deserts and forests extremely accessible. Airways is extremely important and useful for the citizens of the north-eastern states.

(3) Air transport also helps in providing relief to disaster-struck regions in the country. Helicopters help drop food packets and planes are used to evacuate people out of areas under immense stress.

Question 29.
Suggest a few measures to reform political parties in India. [3]
Answer:
Three measures to reform political parties are:
(1) Proper laws should be made to regulate the internal functioning of a party. Party workers should be made to maintain attendance registers and follow a constitution to administer these parties.

(2) Political parties should reserve seats for women and people belonging to different minority communities at positions of influence. This will enable parties to be more representative and fair.

(3) Political parties should be held accountable for their decisions and actions. They should be funded by the state to contest elections.

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
How did the attitude of the East India Company towards the Indian press change after 1857? Support your argument using five points.
OR
Indian trade had played a crucial role in the late nineteenth century world economy. Analyze the statement. [5]
Answer:
After the revolt of 1857, the attitude of the East India Company towards the Indian press changed drastically in the following ways:
(1) The British believed that the native press was mainly responsible for instigating and spreading the rebellion and demanded severe restrictions on its freedom.

(2) Vernacular newspapers were becoming nationalist in their fervour. British officers discussed implementing stringent measures to control what was being published in these papers and pamphlets.

(3) To censor reports and editorials in such newspapers, the Vernacular Press Act was passed in 1878. It was based on the Irish Press laws.

(4) Papers were rigidly tracked. The reports were judged and if suspected to be seditious, the newspapers were warned and press machines were confiscated.

(5) Prosecution measures were made even more rigorous after some protests were observed.
OR
The role of Indian trade was:
(1) Trade Surplus: Britain had a trade surplus with India which it used to balance its trade deficit with other countries.

(2) Home charges : Britain’s trade surplus in India also helped to pay ‘home charges’ which included private remittances by British officials and traders, interest payments on India’s external debts and pensions of the British officials in India.

(3) Major supplier of cotton : India remained a major supplier of raw cotton to Britain. This helped British textile industries to grow.

(4) Supplier of indentured workers: Many indentured workers from Bihar. Uttar Pradesh, Central India and Tamil Nadu migrated to other countries to work in mines and plantations.

(5) India traded in opium with China, the proceeds of which financed British ventures and imports.

Question 31.
What is the benefit of using non-conventional sources of energy? Mention three non- conventional sources of energy.
OR
Describe the mechanism of Rainwater Harvesting. [5]
Answer:
Using non-conventional sources of energy is beneficial for the environment They reduce humanity’s dependence upon coal, oil and gas which are not just finite and non-renewable but are also mainly responsible for polluting the environment. Unconventional sources of energy provide a viable alternative to such sources.

Some non-conventional sources of energy are as follows:

Solar Energy: Using photovoltaic technology, solar energy can be directly converted into electricity. It also helps rural households to reduce their dependency upon firewood and other fuels.
Nuclear Energy: It is the energy obtained by altering the structure of atoms. The energy released is used to generate electricity.
Wind Energy: Nagarcoil and Jaisalmer are well known for effective use of wind energy in the country. It is clean and viable. India has immense potential for it.

Rainwater harvesting systems include capturing of rainwater by directing it from large, flat surfaces (e.g. roofs) to underground or over-ground holding tanks.
The collected rainwater is later filtered and pumped directly to the appliances or to a header tank.
Rainwater from the first shower is not collected and used because it washes the dirt away from the collecting surfaces.
The rainwater from subsequent showers is collected. This roof water is an extremely reliable source of drinking water when all other sources are dried up.
It is commonly referred to as Palar-Pani and is considered as the purest form of natural water.

Question 32.
What does the term ‘Secular’ mean? How does the Indian Constitution ensure that India is a secular country?
OR
What are feminist movements ? How have they brought improvement in the condition of women? [5]
Answer:
Secularism entails separation of religion and the state. The principle of secularism means giving equal respect to all religions and cultural beliefs without any prejudice and discrimination.

To ensure India is a secular country, the Indian Constitution laid down the following rules:
(1) Indian state has no officially recognised religion. It practises total non-interference in religious matters unless it is to upkeep the peace and harmony in the country.

(2) The Constitution provides to all individuals and communities, freedom to profess, propogate, practise their beliefs and religions or not to follow any.

(3) The Constitution prohibits any form of discrimination against any citizen because of any cultural, religious or personal beliefs.

(4) Indian Constitution allows the state to interfere in religious matters only to ensure equality, harmony and mutual peace.

(5) The Constitution also allows all religions to establish religious and educational institutions for minority communities. It establishes all forms of equality through the Fundamental Rights.
OR
A woman or man who believes in equal rights and opportunities for both women and men is a feminist. Thus, these feminist movements aimed at achieving equality in personal and family life.
As a result of these feminist movements, the condition of women has improved in the following ways:
(1) Their role in public life was improved. They can now vote, contest elections, drive, travel, establish a business all by themselves. Earlier they required permission from their male counter parts.

(2) They are working as scientists, doctors, engineers, lawyers, managers, college and university teachers etc. These positions were not considered suitable for women earlier.

(3) In Scandinavian countries such as Sweden, Norway and Finland, the participation of women in public life is very high. In India and other developing countries, their position and role as political participants is rising.

Question 33.
Mention any five reasons or situations in which a person might need to borrow money.
OR
Why is sustainability an important component of development? Discuss a situation where one without the other would be extremely disastrous for the humanity. [5]
Answer:
One might need to borrow money on credit in the following five situations:
(1) Farmers require credit to sow or harvest crops in every growing season. They borrow at the beginning of growing seasons to buy HYV seeds, advanced equipment and tools.

(2) People belonging to the middle class often require to borrow credit for personal purposes- for marriage, medical care etc.

(3) Students require to borrow money for educational purposes. Students require educational loans to migrate or pay their tuition fees.

(4) Entrepreneurs and Businessmen constantly require credit to invest back in their ventures to increase their profits.

(5) People suffering from medical emergencies might often require credit on short notices.
OR
Sustainability is an extremely integral component of development today. This is because technological advancement has driven the world towards instability and mass destruction of finitely available resources.
(1) Sustainability can be understood as the first step in the right direction. Sustainable development entails no compromise with the pace of development but keeps the quality of development and the cost of development at par with that of its pace.

(2) Sustainable development keeps in mind the needs of the present as well as coming generations at the same pedestal.

(3) A situation where sustainability without development and vice versa would prove to be disastrous may be the development and usage of biological weapons.

(4) Bio-weapons have been used in the past to ruin settlements and communities for the purpose of war and capturing authority.

(5) If sustainability was not kept in mind, one could create a lethal weapon capable of destroying the humanity but if development (including vaccines required to keep such weapons in check) was not thought of, more destruction would be guaranteed. The same can be said in terms of utilisation of resources like coal and petroleum.

Section – E
Case Based Questions (4 x 3 = 12)

Question 34.
Read the given source and answer the following questions:
Nature worship is an age old tribal belief based on the premise that all creations of nature have to be protected. Such beliefs have preserved several virgin forests in pristine form called Sacred Groves (the forests of God and Goddesses). These patches of forest or parts of large forests have been left untouched by the local people and any interference with them is banned. Certain societies revere a particular tree which they have preserved from time immemorial
(A) Which practise can nature worship be called an example of? [1]
(B) How are Sacred groves preserved? [1]
(C) Why are these forests called virgin? [2]
Answer:
(A) Nature worship can be understood as an example of a community conservation programme because the entire native community participates in the practise without any interference from the government.

Related Theory:
The decision to ascribe such values and preserve the flora and fauna is of the entire community by extension. Joint Forest Management is another form of Community Conservation programme. Social forestry also involves similar approach.

(B) Sacred groves have been preserved by native communities by ascribing religious meanings to the available flora and fauna in the region. They are called forests of Gods and Goddesses.

(C) Human interference apart from the native community is not allowed in such forests. This is why these forests are known as Virgin or untouched. They are preserved in their virgin or unadulterated state. No other alien or foreign species is introduced due to the fear of an ecological succession.

Question 35.
Read the given source and answer the following questions:
But it does not mean that there is only a one-way relation between caste and politics. Politics too influences the caste system and caste identities by bringing them into the political arena. Thus, it is not politics that gets caste-ridden, it is the caste that gets politicised. This takes several forms.
(A) How does caste influence politics? [1]
(B) How does the expression of caste help in reducing the social conflicts? [1]
(C) How is racism different from caste ism? [2]
Answer:
(A) When parties choose candidates in elections, they keep in mind, the caste composition of the electorate and nominate candidates from different castes, so as to muster necessary support to win elections. This is how caste influences politics.

Related Theory:
Voters keep in mind the caste of their candidates white choosing their representatives. People within the same caste or community have different interests depending on their economic condition. Rich and poor or men and women from the same caste often vote very differently.

(B) Expression of caste differences in politics gives many disadvantaged communities the space to demand their share of power which makes the institutions administrating the state more representative and fair.

Related Theory:
This reduces the possibility of sudden outbursts of anger caused due to exploitation and oppression.

(C) Racism basically distinguishes between people based on their skin colour. On the other hand, the Indian Varna system gave way to a distorted form of differentiation, the caste system, where one’s caste determines their position in the society. Racism doesn’t allow any mobility within the system. It is permanent. The caste system is rigid but permits mobility in specific situations.

Related Theory:
The Anti-Apartheid movement was kick-started against racism which was destroying the social structure of South Africa.

Question 36.
Read the given source and answer the following questions:
First as wheat, then as flour and finally as biscuits. The value of final goods and services produced in each sector during a particular year provides the total production of the sector for that year. And the sum of production in the three sectors gives what is called the Gross Domestic Product (GDP) of a country. It is the value of all final goods and services produced within a country during a particular year. GDP shows how big the economy is.
(A) What does the GDP measure? [1]
(B) What is meant by ‘Gross’ in the term ‘Gross Domestic Product’? [1]
(C) Which institution calculates the GDP of India? [2]
Answer:
(A) The GDP measures the value of the total production of a nation in one financial year. It includes the total value of production of final goods and services.

(B) The term ‘Gross’ means the value of total production without any tax deductions from the amount. The opposite to Gross is Net which is calculated after reducing all the tax deductions from the Gross amount.

Related Theory:
Other phrases are also meanings of Gross but they do not stand true when put in the context of the term, Gross Domestic Product

(C) (1) In India, the mammoth task of measuring GDP is undertaken by the Central Statistics Office under the Ministry of Statistics and Program.

(2) This ministry with the help of various government departments of all the Indian states and union territories, collects information relating to total volume of goods and services and their prices and then estimates the GDP.

Section – F
Map Based Questions (5 x 4 = 20)

Question 37.
(a) On the given outline Political Map of India, identify the places marked as A and B with the help of following information and write their correct name on the lines marked near it.
(A) The place where Non Cooperation Movement was called off due to violence.
(B) Mahatma Gandhi organized a Satgagraha Movement at this place for indigo planters. [2]

(b) On the same given map of India, locate and label any three of the following with suitable symbols:
(a) Rana Pratap Sagar-Dam
(b) Amritsar-Raja Sansi Airport
(c) Bengaluru-Software Technology Park
(d) Vishakhapatnam-Port [3]

Answer:
(a) (A) Chauri Chaura (B) Champaran

Category: CBSE Sample Papers

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