CBSE Sample Papers for Class 10 Maths Basic Set 9 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 9 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.
Two concentric circles are of radii 5 cm and 3 cm. The length of the chord of the largest circle which touches the smaller circle is: [1]
(a) 4 cm
(b) 7 cm
(c) 6 cm
(d) 8 cm
Answer:
(d) 8 cm

Explanation:
Here, AB is a chord of Larger circle which touches the smaller circle at P.
So, AB is tangent to the smaller circle at P.
So, AB is perpendicular to P and perpendicular from the centre to the chord, bisects the chord.
Now,
OP = 3 cm; OA = 5 cm
So, AP = \(\sqrt{5^2-3^2}\) = \(\sqrt{25-9}\)
= √16 = 4 cm

⇒ AB = 2 × AP = 2 × 4 cm = 8 cm.

Question 2.
In the figure, P (2, 3) is the mid-point of the line segment AB. The coordinates of A and B are, respectively: [1]

(a) (0, 6), (4, 0)
(b) (0, 5), (0, 5)
(c) (6, 0), (0, 4)
(d) (3, 0), (0,2)
Answer:
(a) (0, 6), (4, 0)

Explanation:
Let A (0, y) and B (x, 0). Then, P is the mid-point of \(\overline{A B}\)
So, P(2, 3) = P\(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\)
2 = \(\frac{x}{2}\) and 3 = \(\frac{y}{2}\)
x = 4 and y = 6
So, A (0, 6) and B (4, 0).

Question 3.
The smallest 4-digit number, which can be divided exactly by 24 and 36 is: [1]
(a) 1008
(b) 1004
(c) 1009
(d) 996
Answer:
(a) 1008

Explanation:
The smallest number, exactly divisible by 24 and 36, is the LCM (24, 36).
24 = 2 × 2 × 2 × 3, or 2³ × 3¹
36 = 2 × 2 × 3 × 3, or 2² × 3²
So, LCM (24, 36) = 2³ × 3², i.e. 72
The smallest 4-digit number which is the multiple of 72 is the required number.
So, the required number is 1008.

Question 4.
If P is the point (- cos θ, sin θ), then the length of OP, where O is the origin, is: [1]
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) 1
Answer:
(d) 1

Explanation:
OP = \(\sqrt{(-\cos \theta-0)^2+(\sin \theta-0)^2}\)
= \(\sqrt{\cos ^2 \theta+\sin ^2 \theta}\)
= 1.

Question 5.
If a metalic cube of edge 1 cm is drawn into a wire of diameter 3.5 mm, then the length of the wire is: [1]
(a) 10 cm
(b) 9.5 cm
(c) 10.4 cm
(d) 5 cm
Answer:
(c) 10.4 cm

Explanation:
Length of wire
= \(\frac{\text { Volume of cube }}{\pi\left(\frac{3.5}{2}\right)^2}\)
= \(\frac{4000}{\pi(3.5)^2}\) = \(\frac{4000 \times 7}{22(3.5)^2}\)
= 104 mm or 10.4 cm.

Question 6.
For the following frequency distribution, [1]

1 – 10	13
10 – 20	16
20 – 30	28
30 – 40	23
40 – 50	20

(a) 23.2
(b) 27.5
(c) 28.4
(d) 29.3
Answer:
(b) 27.5

Explanation:

x_i	f_i
35	5
38	9
40	10
42	7
44	2

Here \(\frac{N}{2}=\frac{100}{2}\) = 50
Cumulative frequency just greater than 50 is 57, which belongs to class 20-30.
∴ Median class = 20-30
∴ l = 20, f = 28, c.f = 29 and h = 10
we know
Median = l + \(\left(\frac{\frac{N}{2}-c . f}{f}\right)\) × 10
= 20 + \(\left(\frac{50-29}{28}\right)\) × 10
= 20 + 7.5
= 27.5

Question 7.
If a cubical block of side 7 cm is surmounted by a hemisphere, then the greatest diameter that a hemisphere can have is: [1]
(a) 7 cm
(b) 3.5 cm
(c) 10 cm
(d) 5 cm
Answer:
(a) 7 cm

Explanation:
The greatest diameter of the hemisphere is equal to the edges Length of the cubical box.
Hence, diameter = 7 cm.

Question 8.
If the n^th term of an A.P. is \(\frac{3+n}{4}\), then the common difference of A.P is: [1]
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{3}\)
(c) 1
(d) \(\frac{1}{4}\)
Answer:
(d) \(\frac{1}{4}\)

Explanation:
Given,

Question 9.
In the given figure, if \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{CD}}\), then the ∠ABD is: [1]

(a) 20°
(b) 30 °
(c) 40°
(d) 60°
Answer:
(c) 40°

Explanation:
Since \(\frac{A B}{A C}=\frac{B D}{C D}\), so by angle-bisector theorem, AD is the bisector of A.
So, A = 60°
⇒ ABD = 180° – (60° + 80°) = 40°

Question 10.
The mode of the following data is: [1]

Class	Frequency	Cumulative frequency
0-10	13	13
10-20	16	29
20-30	28	57
30-40	23	80
40-50	20	100
	N = 100

(a) 38
(b) 40
(c) 42
(d) 44
Answer:
(b) 40

Explanation:
Since the observation 40 has maximum frequency of 10.
∴ Mode of the data = 40

Question 11.
The perimeter of a rectangle is 82 m and its area is 400 sq m. The breadth of the rectangle is: [1]
(a) 16 or 25
(b) 16 or 19
(c) 23
(d) 25 or 13
Answer:
(a) 16 or 25

Explanation:
Let, the breadth of rectangle be ‘b’ cm.
Then, length, l = 41 – b
And, Area = l × b
400 = (41 – b) b
⇒ 400 = 41b – b²
⇒ b<sup2 – 41b + 400 = 0
⇒ b² – 25b – 16b + 400 = 0
⇒ b(b – 25) – 16 (b – 25) = 0
⇒ (b – 16) (b – 25) = 0
⇒ b = 16 cm or b = 25 cm
Then, l = 41 – 16 or l = 41 – 25
= 25 cm = 16 cm
Hence, breadth of the rectangle is either 16 cm or 25 cm.

Question 12.
The value of ‘k’ for which the pair of equations x + 2y = 3, 5x + ky = – 7 have no solution is: [1]
(a) 20
(b) 10
(c) 15
(d) 9
Answer:
(b) 10

Explanation:
Given, equations are,
x + 2y = 3
5x + ky = – 7
For the given pair of equations to have no solution, we have
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
⇒ \(\frac{1}{5}=\frac{2}{k} \neq \frac{3}{-7}\)
⇒ k = 10.

Question 13.
The value of x : y√2 x² + 7x + 5√2 = 0 is: [1]
(a) – 4, 3
(b) 2, \(\frac{5}{\sqrt{2}}\)
(c) – 2,
(d) – 2, \(\frac{-5}{\sqrt{2}}\)
Answer:
(d) – 2, \(\frac{-5}{\sqrt{2}}\)

Explanation:
Given,
√2x² + 7x + 5√2 = 0
⇒ √2x² + 5x + 2x + 5√2 = 0
⇒ x(√2 + 5) + 2 (√2x + 5) = 0
⇒ (x + 2) (√2x + 5) = 0
x = – 2 or x = \(\frac{-5}{\sqrt{2}}\)

Question 14.
The coordinates of point A, where AB is a diameter of a circle whose centre is (2, -3) and B is the point (1, 4) is: [1]
(a) (0, 1)
(b) (5, 7)
(c) (2, 8)
(d) (3, -10)
Answer:
(d) (3, -10)

Explanation:
Let, the coordinates c (x, y)
Since, O is the centre of the circle,
Then, O is the mid-point of AB.

∴ \(\frac{x+1}{2}\) = 2, \(\frac{y+4}{2}\) = – 3
⇒ x = 3, y = – 10

Question 15.
The distance between the points (0, 5) and (- 5, 0) is: [1]
(a) 5
(b) 3√2
(c) 5√2
(d) √2
Answer:
(c) 5√2

Explanation:
By distance formula

Question 16.
What is the distance between two parallel tangents to a circle of radius 5 cm? [1]
(a) 8 cm
(b) 15 cm
(c) 6 cm
(d) 10 cm
Answer:
(d) 10 cm

Explanation:
Distance between two parallel tangents = Diameter of circle
= 2 × 5 = 10 cm

Question 17.
Which term of the A.P. 21, 42, 63, 84,… is 210? [1]
(a) 15^th
(b) 10^th
(c) 9^th
(d) 3^rd
Answer:
(b) 10^th

Explanation:
Given, A.P is 21, 42, 63, 84
Here, a = 21
d = 42 – 21 = 63 – 42 = 21
Let a_n = 210
∴ a_n = a + (n – 1) d
⇒ 210 = 21 + (n – 1) × 21
⇒ 210 – 21 = (n – 1) × 21
⇒ n – 1 = \(\frac{189}{21}\)
⇒ n – 1 = 9
⇒ n = 10
Hence, 10^th term of given A.P. is 210.

Question 18.
The values x and y; x – y = 3 and x + 2y = 6 is: [1]
(a) 2, 3
(b) 4, 1
(c) 4, 2
(d) 3, 5
Answer:
(b) 4, 1

Explanation:
Given equations are:
x – y = 3 …….. (i)
and x + 2y = 6 …….. (ii)
Subtract equation (ii) from (i), we get

⇒ y = 1
∴ x = 3 + y = 4.
Thus, x = 4, y = 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 but reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (IQ is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Assertion (A): In the graph shown below the number of zeros of the polynomial 2. [1]

Reason (R) : The number of zeroes of a polynomial f(x) is the number of points at which f(x) intersects the x-axis.
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
Here, group intersects x-axis at 3 points so the number of zeroes 3.

Question 20.
Assertion (A): A wire is in the shape of a circle of radius 21 cm. It is bent to form a square. The side of the square is 30 cm.
Reason (R): Circumference of circle = perimeter of the square. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
Circumference of circle = Perimeter of the square.
So, 2πr = 4a
4a = 2 × \(\frac{22}{7}\) × 21
4a = 132
a = 33 cm

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

Question 21.
In the adjoining factor tree, find the numbers m and n. [2]

OR
In the figure, if AB = AC, prove that: BE = ECT

Answer:

Thus, m = 160 and n = 40.

We know that the Lengths of tangents drawn from an external point to a circle are equal.
So, AD = AF, BE = BD, CE = CF.
Now,
⇒ AB = AC (Given)
⇒ AD + DB = AF + CF
⇒ DB = CF (∵ AD = AF)
⇒ BE = EC (∵ BD = BE and CF = EC)
Hence Proved

Question 22.
If sin θ = \(\frac{12}{13}\), find the value of : [2]
\(\frac{\sin ^2 \theta-\cos ^2 \theta}{2 \sin \theta \cos \theta}-\frac{1}{\tan ^2 \theta}\)
Answer:
sin θ = \(\frac{12}{13}\) gives tan θ = \(\frac{12}{5}\) and cos θ = \(\frac{5}{13}\)

Question 23.
Find the area and perimeter of a sheet of a paper which is a sector of a circle of radius 21 cm central angle 45°. [2]
Answer:

Question 24.
If the perimeter of a circle is equal to area of a square, then find the ratio of the area of the circle to the area of the square. [2]
Answer:
Let, the radius of circle be ‘r’ and side of the square be ‘a’.
Given: 2πr = a² …… (i)
\(\frac{\text { Area of circle }}{\text { Area of square }}=\frac{\pi r^2}{a^2}\)
= \(\frac{\pi r^2}{2 \pi r}\)
= \(\frac{r}{2}\), which is required ratio

Question 25.
The two opposite verticles of a square (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.
OR
Find a relationship between x and y such that the point (x, y) is equidistant from the points (2, 5) and (-1, 4). [2]
Answer:
Let ABCD be a square and Let A(-1, 2) and C(3, 2) be the given two vertices.
Also, Let B(X, y) be the unknown vertex.

Then, AB = BC
⇒ (x + 1)² + (y – 2)² = (3 – x)² + (2 – y)²
⇒ x² + 2x + 1 + y² – 4y + 4 = 9 – 6x + x² + 4 + y² – 4y
⇒ 2x + 1 – 4y + 4 = – 6x + 9 – 4y + 4
⇒ 8x = 8
⇒ x = 1
Further, in ▢ABC,
⇒ AC² = AB² + BC²
⇒ (3 + 1)² + (2 – 2)² = (x + 1)² + ( y – 2)² + (3 – x)² + (2 – y)²
⇒ 16 = 2x² + 2y² + 2x + 1 – 4y + 4 – 6x + 9 – 4y + 4
Using x = 1, we get:
16 = 2 + 2y² + 2 + 1 – 4y + 4 – 6 + 9 – 4y + 4
⇒ 2y² – 8y = 0
⇒ 2y(y – 4) = 0
⇒ y = 0 or y = 4
Thus, the other two vertices are (1, 0) and (1, 4).

Given, points P (2, 5) and Q (- 1, 4) are equidistant from point O (x, y)
PO = OQ or
∴ PO² = OQ²
⇒ (2 – x)² + (5 – y)² = (- 1 – x)² + (4 – y)²
⇒ 4 + x² – 4x + 25 + y² – 10y = 1 + x² + 2x + 16 + y² – 8y
⇒ 29 – 4x – 10y = 2x – 8y + 17
⇒ – 6x – 2y + 12 = 0
⇒ 3x + y – 6 = 0, is the required relation.

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

Question 26.
Find the greatest number which when divides 245 and 1029 leaves remainder 5 in each case. [3]
Answer:
The required number is H.C.F. (245 – 5, 1029 – 5), i.e. HCF (240, 1024)
Now, 240 = 2 × 2 × 2 × 2 × 3 × 5,
i.e. 2⁴ × 3¹ × 5¹
and 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2, i.e. 210
So, HCF (240, 1024) = 2⁴, i.e. 16.
Thus, the required greatest number is 16.

Question 27.
Solve for x: x⁴ – 20x² + 64 = 0. [3]
Answer:
Put x² = y.
Then given equation becomes:
y² – 20y + 64 = 0
⇒ y² – 16y – 4y + 64 = 0
⇒ y(y – 16) – 4( y – 16) = 0
⇒ (y – 16)( y – 4) = 0
⇒ y = 16 or y = 4
For y = 16, x² = 16 ⇒ x = ± 4
For y = 4, x² = 4 ⇒ x = ± 2
Thus, x = ± 4, ± 2.

Question 28.
Prove that the area of an equilateral triangle described on a side of a right- angle isosceles triangle is half the area of the equilateral triangle, described on the hypotenuse.
OR
In the figure, two tangents PT and PS are drawn to a circle with centre O and radius r, from an external point P.
(A) If OP = 2r, show that ∠OTS = ∠OST = 30°
(B) if ∠TPS = 60°, find the length of OP. [3]

Answer:

Given: ∆ABC, in which ∠ABC = 90° and
AB = BC. Also, ∆ABD and ACE are equilateral.
To Prove: ar (∆ABC = \(\frac{1}{2}\) ar (∆ CAE)
Proof: Let AB = BC = x units
∴ CA = x√2 units
Now, ∆ABD and ∆CAE being equilateral has each angle equal to 60°
∴ ∆ABD ∆DAE
But ratio of of similar As is equal to the ratio of their corresponding sides

Hence, ar (∆ABD) = \(\frac{1}{2}\) × (∆CAE)
Hence Proved.

(A) ∆ OTS, OT = OS = radii of the circle
∴ OTS = TPQ
In right ∆ OTP
∴ \(\frac{O T}{O P}\) = sin TPO
⇒ \(\frac{r}{2 r}\) = sin TPO
⇒ TPO = 30°
⇒ TPS = 2 × TPO = 2 × 30° = 60°
And POT = 90° – 30° = 60°
Now, In triangle PTS,
PT = PS (tangents from same external point P)
∴ PST = PTS = x (say)
⇒ PTS + PST + SPT = 180°
⇒ x + x + 60° = 180°
⇒ 2x = 180° – 60° = 120°
⇒ x = 60°
Also, OTP = 90°. so, OTS = 90° – 60°
= 30°
Thus, OTS = OST = 30°

(B) Here TPO = ∆∆ TPS = 30°
In right ∆ OTP,
\(\frac{O T}{O P}\) = sin 30°
⇒ \(\frac{r}{Q P}=\frac{1}{2}\)
⇒ OP = 2r

Question 29.
If α and β are zeros of a quadratic polynomial 4x² + 4x + 1, then find the quadratic polynomial whose zeros are α² + β² and 2αβ. [3]
Answer:
Given, are the zeroes of polynomial 4x² + 4x + 1.
α + β = – \(\frac{b}{a}\) = – \(\frac{4}{4}\) = – 1
αβ = \(\frac{c}{a}\) = \(\frac{1}{4}\)
Now, (α + β)² = α² + β² + 2αβ
(- 1)² = α² + β² + 2 × \(\frac{1}{4}\)
⇒ α² + β² = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
And 2αβ = \(\frac{1}{2}\)
∴ Sum of roots = \(\frac{1}{2}+\frac{1}{2}\) = 1
Product of roots = \(\frac{1}{2} \times \frac{1}{2}\) = \(\frac{1}{4}\)
Quadratic Equation is
x² – (1)x + \(\frac{1}{4}\) = 0
⇒ 4x² – 4x + 1 = 0, is the required equation.

Question 30.
A farmer connects a pipe of internal diameter 20 cm from a canal into cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 3 km per hour, in how much time will the tank be filled completely ? [3]
OR
Find the mean and modal marks of students from the following frequency distribution:

Answer:
Water speed = 3 km/h = 50 m
Diameter of pipe = 20 cm = \(\frac{1}{5}\) m
Water tank has 2m depth with radius 5 m
Let the time required to fill the tank be ‘n’ minutes.
Then, water flowing through the pipe in ‘n’ minutes = volume of water in the tank
⇒ π\(\left(\frac{1}{10}\right)^2\) × n × 50
= π × (5)² × 2
⇒ n = 100
Hence, the required time is 100 minutes.

Here, the modal class is 50-60 [class with max. frequency]
For this class,
l = 50, h = 10, f₁ = 15, f0 = 12, f₂ = 12

Calculation of Mean:

Question 31.
How many terms of the A.P.: 9, 17, 25, ……….. must be taken to give a sum of 636? [3]
Answer:
Let ‘n terms of A.P. gives sum of 636.
In the given A.P., a = 9 and d = 8
So, 636 = \(\frac{n}{2}\) [2 × 9 + (n – 1) (8)]
⇒ n(18 + 8n – 8) = 1272
⇒ 8n² + 10n – 1272 = 0
⇒ 4n² + 5n – 636 = 0
⇒ 4n² + 53n – 48n – 636 = 0
⇒ n(4n + 53) – 12 (4n + 53) = 0
⇒ (n – 12) (4n – 53) = 0
⇒ n – 12 = 0
⇒ n = 12
⇒ [∵ (4n + 53) cannot be zero]
Thus, 12 terms of A.P. gives sum of 636.

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

Question 32.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that: [5]

(A) TP = TQ
(B) ∠PTQ = 2∠OPQ
Answer:
(A) Join OT.
Consider ∆s OPT and OQT.
Here, P = Q = 90° (each)
OP = OQ (radii)
OT = OT (common)

So, by RHS congruence criterion,
∆OPT ≅ AOQT
⇒ TP = TQ

(B) In ∆OPQ,
OQP = OPQ (∵ OP = OQ)
= \(\frac{1}{2}\) [180° – POQ]
= \(\frac{1}{2}\) [PTQ]
[∵ In quad. OPTQ, P = Q = 90°
So, POQ + PTQ = 180°]
⇒ PTQ = 2 OPQ

Question 33.
5 books and 7 pens together cost ? 434, whereas 7 books and 5 pens together cost ₹ 550. Find the total cost of 1 book and 2 pens.
OR
The length of a rectangular plot is greater than thrice its breadth by 2 m. If the area of the plot is 120 m², find the dimensions of the plot. [5]
Answer:
Let, the cost of 1 book be ₹ x and cost of 1 pen be ₹ y.
∴ 5x + 7y = 434 …(i)
7x + 5y = 550 …(ii)
Add equations (i) and (ii) we get 12x + 12y = 984
x + y = 82 …(iii)
Subtract equation (ii) from (i), we get 2x – 2y = 116
x – y = 58 …(iv)
Now, add equations (iii) and (iv), we get
2x = 140
x = 70
∴ y = 70 – 58 = 12
Hence, the cost of 1 book is ₹ 70 and the cost of 2 pens is ₹ 24.

Let ‘l’ m and ‘b’ m be the Length and breadth of the plot respectively.
As per the question,
l = 3b + 2 and l × b = 120
Solving the two equations, we get
(3b + 2) b = 120
⇒ 3b² + 2b – 120 = 0
3b² + 20b – 18b – 120 = 0
b (3b + 20) – 6 (3b + 20) = 0
(3b + 20) (b – 6) = 0
3b + 20 = 0 or b – 6 = 0 20
b = – \(\frac{20}{3}\) or b = 6
Thus, breadth (b) = 6m
From the equation l = 3b + 2,
we have l = (3 × 6 + 2) m
i.e. l = 20m
Hence, 20m × 6m are the dimensions of the plot.

Question 34.
A tower is 50 m high. Its shadow is x metres shorter, when the sun’s altitude is 45° than when it is 30°. Find x correct to the nearest cm. [5]
Answer:
PA and QA represent shadows Lengths at sun’s altitude of 30° and 45° respectively.
It is given that:
PA – QA = PQ = x metres
We need to find the value of x.

In right angle triangle PAB,
\(\frac{A B}{P A}\) = tan30° = \(\frac{1}{\sqrt{3}}\)
PA = 5o√3 metres
In right angle triangle QAB,
\(\frac{A B}{P A}\) = tan 45° = 1
⇒ QA = 50 metres
So, x = PQ = PA – QA
x = (50√3 – 50) m
= 50(√3 – 1) m
= [50 × (1.732 – 1)] m
= (50 × 0.732) m
= (50 × 0.732 × 100) cm
x = 3660 cm

Question 35.
If sin A = m sin B and tan A = n tan B then show that (n² – 1) cos² A = m² – 1.
OR
The length and breadth of a rectangle are (3x + 1) cm and (2x – 1) cm respectively. If the area of the rectangle is 144 sq. cm, then find the value of x. [5]
Answer:
Given: tan A = n tan B
and sin A = m sin B
To Prove:(n² – 1) cos2A = m² – 1
Proof: sin A = m sin B (given) …….. (i)
tan A = n tan B
\(\frac{\sin A}{\cos A}=n \frac{\sin B}{\cos B}\) ………. (ii)
on substituting sin B from eq. (i),
we get
⇒ cos B = \(\frac{n}{m}\) cos A …….. (iii)
and sin²A = m² sin²B
⇒ (1 – cos²A) = m² (1 – cos²B)
substituting eq. (iii), we get

1 – cos²A = m² – n² cos²A
n² cos² A – cos²A = m² – 1
cos²A (n² – 1) = m² – 1
Hence, Proved

Given, Length of a rectangle = (3x + 1) cm
Breadth of a rectangle = (2x – 1) cm
Area of a rectangle = 144 sq. cm
ATQ.,
Area of rectangle = l × b
⇒ (3x + 1) (2x – 1) = 144
⇒ 6x² + 2x – 3x – 1 = 144
⇒ 6x² – x – 1 = 144
6x² – x – 145 = 0
Solving by using quadratic formula,
x = \(\frac{-b \pm \sqrt{(-1)^2-4 \times 6 \times(-145)}}{2 \times 6}\)
Here, a = 6, b = -1, c = -145

Hence the value of x is 5.

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

Question 36.
NITI aayog has tasked their statistical officer to create a model for farmers to be able to predict their produce output based on various factors.
To test the model out, the officer picked a local farmer who sells apples to check various factors like weight, bad apples, half-cooked, green vs red etc.
A box containing 250 apples was opened and each apple was weighed.

The distribution of the masses of the apples is given in the following table:

On the basis of the above information, answer the following questions:
(A) Find the value of p. [1]
(B) Find the lower limit of the modal class. [1]
(C) Find the mean mass of the apples.
OR
Find the upper limit of the median. [2]
Answer:
(A) Since, total apples are 250.
Then
20 + 60 + 70 + p + 60 = 250
p = 250 – 210 = 40

(B) Here, modal class is 120-140, with maximum frequency 70. Then, lower limit of modal class is 120.

Marks	Frequency	c.f.
80-100	20	20
100-120	60	80
120-140	70	150
140-160	40	190
160-180	60	250
	N = 250

Then, \(\frac{N}{2}=\frac{250}{2}\) = 125
So, cumulative frequency just greater than 125 is 150, which belong to class 120-140.
So. median class is 120-140.
Hence, upper limit of median class is 140.

Question 37.
A ticket machine in a car park takes ₹ 1 coin and ₹ 2 coin. A ticket cost ₹ 3. The probability that the machine will accept a particular ₹ 1 coin is 0.9 and that it will accept a particular ₹ 2 coin is 0.8.

On the basis of the above information, answer the following questions:
(A) Find the probability that the machine will not accept a particular ₹ 1 coin. [1]
(B) Urmila put one ₹ 1 coin and one ₹ 2 coin into the machine. Find The probability that the machine will not accept either of these coins.
OR
Jayant only has three ₹ 1 coins. Find the probability that the machine accept all these coins, and find the probability that Jayant will not get a ticket. [2]
(C) Pritam has one ₹ 1 coins and two ₹ 2 coins. Find the probability that pritam will get a ticket. [1]
Answer:
(A) P(will not accept ₹ 1 coin)
= 1 – P( will accept ₹ 1 coin)
= 1 – 0.9
= 0.1

(B) P(it will not accept either coin)
= P( will not accept ₹ 1 coin) × P( will not accept ₹ 2 coin)
= [1 – P( will accept ₹ 1 coin)] × [1 – P( will accept ₹ 2 coin)]
= (1 – 0.9) × (1 – 0.8)
= 0.1 × 0.2 = 0.02

P(All three coins will be accepted)
= 0.9 × 0.9 × 0.9
= 0.729
P(He will not get ticket)
= 1 – 0.729
= 0.271

Question 38.
On either side of a mountain are two hotels that are situated at street level. Two cottages are located as shown in the figure after travelling a certain distance towards the mountain’s peak. The ratio of the distances from Hotel B to Hut 2 to the mountain peak is 3:7.

On the basis of the above information, answer the following questions:
(A) What is the ratio of the perimeters of the triangle formed by both hotels and mountain top to the triangle formed by both huts and mountain top?
OR
Find the distance between the hotel A and hut-1 and if the horizontal distance between the hut-1 and hut- 2 is 8 miles, then find the distance between the two hotels. [2]
(B) If the distance from mountain top to hut-1 is 5 miles more than that of distance from hotel B to mountain top, then what is the distance between hut-2 and mountain top? [1]
(C) What is the ratio of areas of two parts formed in the complete figure? [1]
Answer:
(A) Let ∆ABC is the triangle formed by both hotels and mountain top. ∆CDE is the triangle formed by both huts and mountain top.
Clearly, DE || AB and so
∆ABC ~ ∆DEC
[By AA-similarity criterion]

Now, required ratio = Ratio of their
corresponding sides = \(\frac{B C}{E C}=\frac{10}{7}\) i.e., 10 : 7

Since, DE || AB, therefore

= 4.29 miles
since ∆ABC ~ ∆DEC
∴ \(\frac{B C}{E C}=\frac{A B}{D E}\)
[∵ Corresponding sides of similar triangles are proportional]
⇒ \(\frac{10}{7}=\frac{A B}{8}\)
⇒ AB = \(\frac{80}{7}\) = 11.43 miles

(B) Given, DC = 5 + BC
Clearly, BC = 10 – 5 = 5 miles
Now, CE = \(\frac{7}{10}\) × BC = \(\frac{7}{10}\) × 5 = 3.5 miles