CBSE Sample Papers for Class 10 Maths Basic Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 2 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.
The common zero of the polynomial x³ + 1, x² – 1 and x² + 2x + 1 is: [1]
(a) 1
(b) -1
(c) 3
(d) -2
Answer:
(b) -1

Explanation:
x³ + 1 = (x + 1)(x² – x + 1)
x² + 1 = (x + 1)(x – 1)
x² + 2x + 1 = (x + 1)²
The common factor is (x + 1), i.e. the common zero is -1.

Question 2.
If p, 2p – 1, 2p + 1 are three consecutive terms of an A.P., then the value of ‘p’ is: [1]
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(a) 3

Explanation:
Since p, 2p – 1 and 2p + 1 are consecutive terms of an A.P.,
2(2p – 1) = p + (2p + 1)
⇒ 4p – 2 = 3p + 1
⇒ P = 3

Question 3.
In which quadrant does the point (-1, -2) lie? [1]
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III

Explanation:
Third quadrant since, both ‘x’ and ‘y’ are negative.

Question 4.
If radii of two concentric circles are 4 cm and 5 cm, then the length of the chord to one circle which is tangent to the other circle is: [1]
(a) 6 cm
(b) 5 cm
(c) 8 cm
(d) 9 cm
Answer:
(a) 6 cm

Explanation:
Here, OP ⊥ AB and
AP = PB = \(\frac{1}{2}\)AB
We know tangent is perpendicular to radious at the point of contact.
∴ OP ⊥ AB
Also, AB is a chord to smaller circle and perpendicular from centre to the chord bisects it

Question 5.
The tangent of a circle makes angle with radius at point of contact is: [1]
(a) 60°
(b) 30°
(c) 90°
(d) none of these
Answer:
(c) 90°

Explanation:
Tangent at any point of a circle is perpendicular to the radius at the point of contact So, the tangent makes a right angle with the radius at the point of contact

Question 6.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is: [1]
(a) 10 units
(b) 14 units
(c) 15 units
(d) 12 units
Answer:
(d) 12 units

Explanation:
Perimeter of ΔABC.
= OA + AB + OB

= 3 + \(\sqrt{(3-0)^2+(0-4)^2}\) + 4
= 3 + 5 + 4
= 12 units

Question 7.
What is the area of a circle inscribed in a square of side ‘a’ units ? [1]
(a) \(\frac{\pi a^2}{4} \)
(b) πa²
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi a}{2}\)
Answer:
(a) \(\frac{\pi a^2}{4} \)

Explanation:
The radius of the inscribed circle is \(\frac{a}{2}\).

So, its area is π\left(\frac{a}{2}\right)^2, i.e \frac{\pi a^2}{4}

Question 8.
A number from 11 to 30 was chosen at random. The probability of this chosen number being a multiple of 2 is: [1]

(a) 50-60
(b) 10-20
(c) 40-50
(d) 30-40
Answer:
(d) 30-40

Explanation:

Here, maximum frequency is 30, which belongs to class 30 – 40.
∴ Modal class = 30 – 40

Question 9.
A number from 11 to 30 was chosen at random. The probability of this chosen number being a multiple of 2 is: [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{3}{2}\)
Answer:
(a) \(\frac{1}{2}\)

Explanation:
Total numbers from 11 to 30 = 20
∴ Total number of outcomes = 20
Multiples of 2 from 11 to 30 = {12, 14, 16, … 28, 30}
⇒ Number of favourable outcomes = 10
∴ P (Multiple of 2) = \frac{10}{20}=\frac{1}{2}

Question 10.
A car travels 0.99 km in which each wheel makes 450 complete revolutions. The radius of the wheel is: [1]
(a) 30 cm
(b) 32 cm
(c) 35 cm
(d) 40 cm
Answer:
(c) 35 cm

Explanation:
The wheel covers a distance of 2πr, in one revolutions.
So, 450 × 2πr = 0.99 × 1000 × 100
450 × 2 × \(\frac{22}{7}\) × r = 99 × 1000
r = 35
Thus, the radius of the wheel is 35 cm.

Question 11.
If α and β are the zeros of the polynomial p(x) = x² – px + q, then the value of \frac{1}{\alpha}+\frac{1}{\beta} is: [1]
(a) p
(b) \(\frac{-p}{q}\)
(c) pq
(d) \(\frac{p}{q}\)
Answer:
(d) \(\frac{p}{q}\)

Explanation:
Since a and P are the zeros of p(x) = x² – px + q,
∴ α + β = p and αβ = q
Now, \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{p}{q}\)

Question 12.
The sum of first 20 even numbers is: [1]
(a) 420
(b) 320
(c) 140
(d) 450
Answer:
(a) 420

Explanation:
The list of first 20 even numbers is: 2, 4, 6, 8, ………., 40
It is an A.P. with a = 2, d = 2 20
So, the required sum = \(\frac{20}{2}\) [2 × 2 + (20 – 1)(2)]
= 10(4 + 38)
= 420

Question 13.
The discriminant of the quadratic equation (p + 3)x² – (5 – p)x + 1 = 0 is: [1]
(a) (p + 13)
(b) (p + 1)
(c) (P + 2)
(d) (p – 13) (p – 1)
Answer:
(d) (p – 13) (p – 1)

Explanation:
Discriminant = b² – 4ac
= [-(5 – p)]² – 4(p + 3)(1)
= 25 + p² – 10p – 4p – 12
= p² – 14p + 13
= p²– 13p – p+ 13
= p (p – 13) – 1 (p – 13)
= (p – 13) (p – 1)

Question 14.
In a ΔABC, DE ∥ BC with D on AB and E on AC. If \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{3}\), then \(\frac{B C}{D E}\) is: [1]
(a) \(\frac{5}{2}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{2}{3}\)
Answer:
(a) \(\frac{5}{2}\)

Explanation: Since DE is parallel to BC,
∴ ΔADE ~ ΔABC
⇒ \(\frac{A D}{A B}=\frac{D E}{B C}\) ………….(i)

Question 15.
The median of the following data is: [1]
5, 2, 7, 9, 3, 2, 4, 8.
(a) 5.5
(b) 4
(c) 4.5
(d) 5
Answer:
(c) 4.5

Explanation:
Arranging the data in ascending order,
2, 2, 3. 4, 5, 7, 8, 9

Here, number of terms (n) = 8 i.e., even
∴ median =

Question 16.
The discriminant 3x² + 2x = 0 is: of the equation [1]
(a) 4
(b) 2
(c) 1
(d) -5
Answer:
(b) 2

Explanation:
Here, equation is 3x² + 2x = 0 Then comparing it with ax² + bx + c = 0, we get
a = 3, b = 2, c = 0

Then, discriminant,
D = \(\sqrt{b^2-4 a c}\)
= \(\sqrt{(2)^2-4 \times 3 \times 0}\)
= √4 = 2

Question 17.
The 8th term from the end of A.P.: – 12, -7,-2, ………., 68 is: [1]
(a) 33
(b) 35
(c) 30
(d) 36
Answer:
(a) 33

Explanation:
Write the given A.P in reverse order, then series is,
68, 63, ……. – 2, – 7, – 12
Then, a = – 68, d = – 5
a8 = a + (8 – 1) d = 68 + 7 (- 5) = 33

Question 18.
In the given figure, ΔABC – ΔPQR. The value of x is: [1]

(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 5 cm

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.
Answer:
(c) 3 cm

Explanation:
Since, ΔABC ~ ΔPQR

Question 19.
Assertion (A): The length of the tangent will be 7 cm in a circle with a radius of 3 cm and a point’s distance from the centre of the circle is 5 cm. [1]
Reason (R) : (hypotenuse)² = (base)² + (height)²
(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.
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:

A tangent at any point of a circle is perpendicular to the radius through the point of contact
Therefore, DOBA = 90° and AOBA is a right-angled triangle.
By Pythagoras theorem,
OA² = OB² + AB²
5² = 3² + AB²
AB²= 25 – 9
AB² = 16
AB = ±4

Question 20.
Assertion (A): If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, then median is 21. [1]
Reason (R): Median \(\left(\frac{n+1}{2}\right)^{\text {th }}\) value, if n is odd.
(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.
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Arranging the terms in ascending order, 0, 5, 11, 19, 21, 30, 42, 50, 52.
Median = Value of (n + 1)/2^th observation
Median score = Value of (9 + 1)/2^th term
= Value of 5^th term
= 21

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

Question 21.
Show that 3 + √5 is an irrational number, assume that √5 is an irrational number.
OR
Solve algebraically: 4x + 3y = 14 and 3x – 4y = 23. [2]
Answer:
If possible, let us assume that 3 + √5 be a rational number. So, there exists positive integers a and b such that 3 + √5 = \(\frac{a}{b}\), where a and b are integers having no common foctor other than 1 and b ≠ 0.
⇒ √5 = \(\frac{a}{b}\) – 3
⇒ √5 = \(\frac{a-3 b}{b}\)
Since, \(\frac{a-3 b}{b}\) is a rational number, √5 is a rational number which is a contradiction to the fact that ”√5 is irrational”.
Hence, 3 + √5 is irrational
OR
Given, 4x + 3y = 14 …(i)
and 3x – 4 y = 23 ……(ii)
Adding (i) and (ii). we get
7x – y = 37
y = 7x – 37 …(iii)
Substituting the value of y in equation (i), we get
4x + 3 (7x – 37) = 14
⇒ 4x + 21x – 111 = 14
⇒ 25x = 125
⇒ x = 5
Putting x = 5 in eq. (iii).we get
y = 7 (5) – 37
= 35 – 37
= – 2
The solution is x = 5 and y = – 2.

Question 22.
Evaluate: (sin⁴ 60° + sec⁴ 30°) – 2 (cos² 45° – sin² 90°)
OR
Prove that \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ. [2]
Answer:
(sin⁴ 60° + sec⁴ 30°) – 2 (cos² 45° – sin² 90°)

Question 23.
A dice is thrown twice. Find the probability that 5 will not come up either time. [2]
Answer:
Total number of possible outcomes = 36
Number of favourable outcomes = 36 – 11 = 25
[Excluding (1, 5), (5, 1), (2, 5). (5, 2), (3, 5), (5, 3), (4, 5), (5, 4), (5, 5), (5, 6), (6, 5)]
So, required probability = \(\frac{25}{36}\)

Question 24.
Find a point on x-axis which is equidistant from A (-3, 4) and B (7, 6).
Answer:
Let P (x, 0) be a point on x-axis, equidistant from A (-3, 4) and B (7, 6), i.e.,
AP = BP or AP² = BP²
i.e. (-3 – x)² + (4 – 0)² = (7 – x)² + (6 – 0)²
i.e. 9 + x² + 6x + 16 = 49 + x² – 14x + 36
⇒ 6x + 25 = – 14x + 85
⇒ 20x = 60
⇒ x = 3
Thus, the required point is (3, 0).

Question 25.
If the circumfrence of a circle increases from 4π to 8π, then find the percentage increase in the area of the circle. [2]
Answer:
Let r and R be the radii of the initial circle and the increased circle, respectively. Then,
2πr = 4π and 2πR = 8π
⇒ r = 2 and R = 4
∴ Area of initial circle
= π(2)² i.e. 4π
and Area of increased circle
= π(4)² i.e. 16π

Thus, % increase = \(\frac{16 \pi-4 \pi}{4 \pi}\) × 100
= \(\frac{12 \pi}{4 \pi}\) × 100
= 300%

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

Question 26.
The sum of the squares of two consecutive multiples of 7 is 637. Find the two multiples.
OR
Find the last term of an AP having 9 terms whose last term is 28 and sum of all the terms is 144. [3]
Answer:
Let 7x and 7(x + 1) be two consecutive multiples of 7.
Then,
(7x)² + [7(x + 1)]² = 637
⇒ 49x² + 49 (x² + 2x + 1) = 637
⇒ 98x² + 98x + 49 = 637
⇒ 98x² + 98x – 588 =0
⇒ x² + x – 6 = 0
⇒ (x + 3)(x – 2) = 0
⇒ x + 3 = 0 or x-2 = 0
⇒ x = -3 or x = 2
Thus, two consecutive multiples of 7 are 14 and 21, or – 21 and – 14.
OR
Let ‘a’ be the first term of A.P. and‘d be the common difference.
Here, total number of terms of A.P. is 9, i.e. n = 9
nth term = last term = = an = a + 8d = 28 …(i)
Also, S_n = S₉ = \(\frac{9}{2}\)[2a + (9 – 1)d] = 144
⇒ 9 (a + 4 d) = 144
or 9a+ 36d= 144 …(ii)

Solving (i) and (ii) simultaneously, we get a = 4; d = 3
Thus, the required first term is 4.

Question 27.
The figure drawn on the graph paper shows a AABC with vertices A (-4, 1), B (-4, 6) and C (8, 1). [3]

(A) Find the length of BC;
Answer:
Length of BC = \(\sqrt{A B^2+A C^2}\)
= \(\sqrt{(1-6)^2+(8-(4))^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}=\sqrt{169}\) i.e 13

(B) Find sin ∠ABQ
Answer:
Here, AB = 5cm, AC = 12 cm and BC = 13 cm
So, sin ∠ABQ = \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{12}{13}\)

Question 28.
Prove that the length of two tangents drawn from an external point to a circle are equal. [3]
Answer:
Here, AP and AQ are two tangents drawn from an external point A, to the circle with centre O.
We need to prove that, AP = AQ
Join OP, OQ and OA.

Proof:
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OQ ⊥ QA and OP⊥ PA
So, ∠OQA = 90° = OPA
Now, in two Δs OQA and CPA. we have:
(j) QQ = OP(Rodii of the same cirde)
(ii) ∠OPA = ∠OQA (each 90°)
(iii) OA = OA (common)
So, by RHS congruence criterion,
ΔOPA = ΔOQA
This concludes that, AP = AQ
Thus, the lengths of tangents drawn from an external point to a circle are equal.

Question 29.
The minute hand of a clockis 2 cm long. Find the area of the face of the clock described by the minute hand between 7 am and 7:15 am.
OR
A cone of height 24 cm and radius of base 6 cm is made up from modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere. [3]
Answer:
We know that a minute hand sweeps by 6° angle in one minute.

So, it will sweep by an angle of 90° in 15 minutes.
So, area swept by the minute-hand in 15 minutes
= \(\frac{90}{360}\) × π(2)² sq cm
= π sq cm, or \(\frac{22}{7}\) sq cm.
OR
Height of the cone, h = 24 cm
Radius of base of the cone, r = 6 cm
Let, the radius of the sphere formed by reshaping cone be R.
As the cone is reshaped into sphere:
∴ Volume of cone = Volume of sphere
\(\frac{1}{3}\)πr²h = \(\frac{4}{3}\)πR³
r²h = 4R³
6 × 6 × 24 = 4 × R³
R³ = 6 × 6 × 6
R = 6 cm
∴ Radius (R) of the sphere formed = 6 cm
Hence, diameter of sphere = 6 × 2 = 12 cm

Question 30.
Determine the median for the following frequency distribution: [3]

Answer:

Class	Frequency	Cum. frequency
100-120	12	12
120-140	14	26
140-160	8	34
160-180	6	40
180-200	10	50

Here, N = 50, \(\frac{N}{2}\) = 25
Cumulative frequency just greater than 25 is 26, which belongs to class 120 – 140.
Hence, the median class is 120-140
For this class,
l = 120, f = 14, cf = 12, h = 20
The Median = l + \(\frac{\frac{N}{2}-c f}{f}\) × h
= 120 + \(\frac{25-12}{14}\) × 20
= 120 + 18.57
= 138.57
Thus, median of the given data is 138.57.

Question 31.
A bag contains 15 white balls and some black balls. If the probability of drawing a black ball from the bag is thrice that of a white ball, find the number of black balls in the bag. [3]
Answer:
Let the bag contains “x ’ number of black balls. Then, the total number of balls in the bag is (x+15).
It is given that,
P(a black ball) = 3 × P(a white ball)
⇒ \(\frac{x}{x+15}\) = 3 × \(\frac{15}{x+15}\)
⇒ x = 45
Thus, there are 45 black balls in the bag.

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

Question 32.
Prove the foLlowing identities, where the angles involved are acute angles for which the expressions are defined:
(A) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
(B) (sinA+ cosecA)² + (cosA + secA)² = 7 + tan²A + cot²A
OR
Two men on either side of a 75 m high building and in line with base of the building observe the angles of elevation of the top of the building as 30° and 60°. Find the distance between the two men (use √3 = 1.73).
Answer:

OR
Let AB be the building and, M₁ and M₂ be the positions of two men.

∴ Given, AB = 75 m,
∠AM₁B = 30°
and ∠AM₂B = 60°
Let AM₁ = x and AM₂ = y
From right triangle. M₁ AB.
\(\frac{A B}{M_1 A}\) = tan 30°
x = \(\frac{75}{x}=\frac{1}{\sqrt{3}}\)
x = 75√3

From right triangle. M₂AB.
\(\frac{A B}{M_2 A}\) = ton 60°
\(\frac{75}{y}\) = √3
y = \(\frac{75}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
y = 25√3
Thus. M₁M₂ = x + y
= (1√3 + 2√3)
= 100√3
= 100 × 1.73
= 173 metres.

Question 33.
Solve for x and y graphically: x – y + 1 = 0; 3x + 2y – 12 = 0. [5]
OR
The shadow of a vertical tower on level ground increases by 16 m, when the altitude of the Sun changes from angles of elevation 60° to 45°. Find the height of the tower, correct to one pLace of decimal (Use √3 = 1.73).
Answer:
Table for the value of x – y + 1 = 0

AS₁, AS₂ be is shadows when sun’s attitude is 60° and 45° respectively.
∴ S₁S₂ = 16 m
From S₁AB
\(\frac{\mathrm{AB}}{\mathrm{S}_1 \mathrm{~A}}\) = tan 60°

Question 34.
In the given figure, PA, QB and RC are perpendicuLars to AC such that PA = RC = y, QB = z, AB = a and BC = b. Prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\). [5]

Answer:
Given PA, QB and RC are perpendiculars to AC
To Prove: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\)
Proof:
Since PA, QB and RC are perpendicuLars, drawn on same tine segment AC.
∴ PA ∥ QB ∥ RC
In ΔARC, QB ∥ RC.
By Basic proportional Theorem,

Question 35.
If a + b = c + d where a, b, c, d are rational numbers, them prove that either a=c and b = d or b and d are squares of rational numbers. [5]
Answer:
Let a = c, then
a + √b = c + √d
⇒ √b = √d
⇒ b = d

So let a ≠ c, then, there exist a postive rational number x, such that a = c + x.
Now, a + √b = c + √d
⇒ c + x + √b = c + √d [∵ a = c + x]
⇒ x + √b = √d ……(i)
⇒ (x + √b)² = (√d)²
⇒ x² + 2 √b x + b = d
⇒ √b = \(\frac{d-x^2-b}{2 x}\)
⇒ √b is rational [∵ d, x, b are rationals, \(\frac{d-x^2-b}{2 x}\) is rational
⇒ b is a square of a rational number
From (i), we have
⇒ √d is rational
⇒ d is the square of a rational number
Hence, either a = c and b = d or b and d are the squares of rationats number.

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

Question 36.
Mr. Naik is a paramilitary Intelligence Corps officer who is tasked with planning a coup on the enemy at a certain date. Currently he is inspecting the area standing on top of the cliff. Agent Vinod is on a hot air balloon in the sky. When Mr. Naik looks down below the cliff towards the sea, he has Ajay and Maran in boats positioned to get a good vantage point.
The main goal is to scope out the range and angles at which they should train their soldiers.

On the basis of the above information, answer the following questions:
(A) Write a pair of ‘angles of elevation’ and ‘angle of depression’. [1]
(B) If the vertical height of the balloon from the top of the cliff is 12 m and ∠b = 30°, then find the distance between the Naik and vinod. [1]
(C) Ajay’s boat is 25 m away from the base of the cliff. If ∠d = 30°. What is the height of the cliff? (use √3 = 1.73).
OR
If the height of the cliff is 30 m , ∠c = 45° and ∠d = 30°, then find the horizontal distance between the two boats (use √3 = 1.73). [2]
Answer:
(A) A pair of ‘angle of elevation’ is ∠b°, ∠e and one pair of angle of depression is ∠c° and ∠d°

(B) Then, sin 30°
= \(\frac{\text { Vertical height }}{\text { Distance between Naik and Vinod }}\)
⇒ \(\frac{1}{2}=\frac{12}{D_{\text {Nand } V}}\)
⇒ Distance = 24 m

(C) Here. ∠d° = ∠f° = 30°
Then. \(\frac{\text { Height of cliff. }}{\text { Distance of Ajay s boat from the base of cliff }}\) = tan 30°
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{25}\)
⇒ h = \(\frac{25}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{25}{3} \times \sqrt{3}\)
= 14.45 m
OR
Here, height of cliff = 30 m
Then, ∠c = ∠e = 45°
⇒ tan 45° = \(\frac{\text { height of cliff }}{\text { Distance of Maran’s boat }}\)
⇒ 1 = \(\frac{30}{\text { Distance of Maran’s boat }}\)
⇒ Distance of morones boat = 30 m
And ∠d =∠f = 30°
tan 30° = \(\frac{\text { height of cliff }}{\text { Distance of Ajay’s boat }}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{30}{D_A}\)
⇒ D_A = 30√3

Distance between boats
= 30√3 – 30
= 30(√3 – 1)
= 30(173.-1)
= 30 × 0.73
= 21.9 m

Question 37.
Nimmi, a 10th class student makes a project on corona virus in Science for an exhibition in her schooL For this project, she picks a sphere which has volume 38808 cu. cm and 11 cylindrical shapes rods each of volume 1540 cu. cm and length 10 cm.

Based on the above, answer the following questions :
(A) What is the diameter of the base of a cylindrical rod?
OR
What is the diameter of the sphere? [2]
(B) What is the curved surface area of a cylindrical rod? [1]
(C) How much curved surface area of the sphere is covered by 11 cylindrical rods ? [1]
Answer:
(A) Volume of cylinderical shape = 1540 cu. Cm
Height of cylinderical shape h = 10 cm
Let r be the radius of the cylinderical shape
∴ πr²h = V
⇒ \(\frac{22}{7}\) × r² × 10 = 1540
⇒ r² = \(\frac{1540 \times 7}{10 \times 22}\)
⇒ r = 7 cm
∴ Diameter = 2r = 14 cm
OR
Volume of sphere = 38808 cm³
Let R be the radius of the sphere
∴ \(\frac{4}{3}\)πR³ = 38808
⇒ R³ = \(\frac{38808 \times 3}{4 \pi}\)
= \(\frac{38808 \times 3}{4 \times 22}\) × 7
= 9261
⇒ R = 21cm
∴ Diameter = 2R = 42 cm

(B) For cylindericol rod,
CSA = 2πrh
= 72 × \(\frac{22}{7}\) × 7 × 7
= 440 sq cm

(C) Base Area of each cylindricaL rod = it,2
= \(\frac{22}{7}\) × 7 × 7
= 154 sq. cm
Total area covered = 11 × 154
= 1694 sq.cm

Question 38.
Prime Minister’s National Relief Fund (PMNRF) was established to help the families of earthquake affected village. The allotment officer is trying to come up with a method to calculate fair division of funds across various affected families so that the fund amount and amount received per family can be easily adjusted based on daily revised numbers.

The total fund allotted is formulated by the officer as:
x³ – 5x² – 2x – 6
The officer has also divided the fund equally among families of the village and each family receives an amount of x² + 2x + 1. After distribution, an amount of 11x + 1 should be left to have some buffer for future disbursements.
On the basis of the above information, answer the following questions:
(A) If an amount of ? 540 is left after distribution, what is value of x? [1]
(B) How much amount (In rupees) does each family receive? [1]
(C) What is the amount of fund (In rupees) allocated ?
OR
Find the value of k, if x – 1 is a factor of p(x) = 2x² + kx + √2. [2]
Answer:
(A) Amount lift = 11x + 1
∴ 11x + 1 = 540
x = \(\frac{539}{11}\) = 49

(B) Since, x = 49
∴ Amount received by each family is
x² + 2x + 1 = (49)² + 2(49) + 1
= 2401 + 98 + 1
= 2500

Let g(x) = x – 1
g(x) = 0 = x – 1 ⇒ x = 1
so, p(x) = 2x² + kx + √2
P( 1) = 2(1)² + k(1) + S.
= 2 + k + √2 ………. (i)
As (x – 1) is a factor of p(x)
∴ P(x) = 0
2 + k+ √2 = 0
k = -[2 + √2]