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 6th term and 8th term of an A.P. are 12 and 22 respectively, then its 2nd 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.
OR
(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 = \(\frac{7}{25} \times \frac{24}{25}=\frac{168}{625}\)
OR
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
OR
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
OR
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
Sn = \(\frac{n}{2}\)(a + l)
⇒ \(\frac{n}{2}\)(5 + 45) = 400
⇒ n = 16
As last term is the nth 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.
OR
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 Sn 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.
OR
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
OR
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
OR
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} = \(\frac{4}{16}\) = \(\frac{1}{4}\)
OR
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} = \(\frac{7}{17}\)