TS 10th Class Maths Question Paper March 2024

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TS 10th Class Maths Question Paper March 2024

Time: 3 Hours
Maximum Marks: 80

Instructions:

1. Answer all the questions under Part – A on a separate answer book.
2. Write the answers to the questions under Part B on the question paper itself and attach it to the answer book of Part – A.

Part – A (60 Marks)
Section – I (6 × 2 = 12 Marks)

Note :

1. Answer ALL the following questions.
2. Each question carries 2 marks.

Question 1.
Express 360 as a product of prime factors.
Solution:

Question 2.
Is the pair of linear equations 3x – 5y = 7 and 6x – 10y = 13 are inconsistent? Justify your answer.
Solution:

Question 3.
A flag pole stands vertically on the ground. From a point which is 15 metres away from the foot of the tower, the angle of elevation of the top of the tower is 45°. Draw a suitable diagram for the given data.
Solution:

Where AB = Flag pole height
BC = Distance between foot of tower and foot that pole
θ = Angle of elevation

Question 4.
AOB is the diameter of a circle with centre ‘O’ and AC is a tangent to the circle at A. If ∠BOC = 130°, then find ∠ACO.

Solution:

Given that
AOB is a diameter and centre is O
AC is a tangent
∠BOC = 130° and point of contact at A.
We know that
The angle at point of contact is 90°
The angle of straight line is 180°
∠AOC + ∠BOC = 180°
∠AOC + 130° = 180°
∠AOC = 180° – 130°
∠AOC = 50°
In a triangle
The sum of angles = 180°
∠AOC + ∠OAC + ∠ACO = 180°
∠ACO + 90° + 50° = 180°
∠ACO = 180° – 140°
∠ACO = 40°
So, the angle ∠ACO is 40°

Question 5.
Express ‘sin θ’ in terms of ‘tan θ’.
Solution:

Question 6.
Construct a Quadratic equation having the roots log₂8 and log₁₀100 .
Solution:
log₂ 8 = log₂2³ = 3log₂2 = 3 × 1 = 3
l0g₁₀ 100 = log₁₀ 10²
= 2log₁₀ 10 = 2 × 1 =2
So, the roots are 3 and 2.
x² – x (sum of the roots) + (product of the roots) = 0
x² – x(3 + 2) + (3 × 2) = 0
⇒ x² – 5x + 6 = 0
∴ The required quadratic equation
⇒ x² – 5x + 6 = 0

Section – II (6 × 4 = 24 Marks)

Note :

1. Answer ALL the following questions.
2. Each question carries 4 marks.

Question 7.
Write the formula for Mode of a grouped data and explain each term.
Solution:
Mode of a grouped data:
Mode = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right]\) × h
Where,
l – lower boundary of the modal class.
h – size of the modal class interval.
f₁ – frequency of modal class.
f₀ – frequency of the class preceding the modal class.
f₂ – frequency of the class succeeding the modal class.

Question 8.
Prove that \(\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\) = 2 cosec θ.
Solution:

Question 9.
From the Venn diagram, find the following sets,

i) X∪Y
ii) X∩Y
iii) X – Y
iv) Y – X
Solution:

Question 10.
In an arithmetic progression, if 4 times of fourth term is equal to 8 times of the eighth term, then prove that twelth term of the progression is zero.
Solution:
Given that
In an AP it 4 times of fourth term is equal to 8 times of the eighth term.
4(4^th term) = 8(8^th term)
4(a + 3d) = 8(a + 7d)
4a + 12d = 8a + 56d
8a – 4a + 56d – 12d = 0
4a + 44 d = 0
4a = – 44d
a = \(\frac{-44}{4}\)
a = -11d
So, 12^th term ⇒ a₁₂ = a + 11d
We know that a = -11d
So, a₁₂ = -11d + 11d
a₁₂ = 0
∴ Hence proved.

Question 11.
In a bag, there are 5 Red balls, 2 Black balls and 3 White balls. If one ball is selected randomly from the bag, then find the probability of
i) getting a Red ball.
ii) getting not a Red ball.
Solution:
Red balls = 5
Blackballs = 2
White balls = 3
Total balls = 5 + 2 + 3 = 10
∴ P(E) \(=\frac{\text { (Number of favourable outcomes }}{\text { Total number of outcomes }}\)

i) P(getting a Red ball) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

ii) P(Not getting a Red ball)
\(=\frac{\text { (Number of favourable outcomes }}{\text { Total number of outcomes }}\)
= \(\frac{2+3}{10}\) – \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 12.
In a rectangle ABCD, AB = 2x – y, BC = 15, CD = 2 and DA = x + 3y, then find the values of x and y.
Solution:
Given that
In a rectangle ABCD

AB = 2x – y
BC = 15; CD = 2; OA = x + 3y
So,
In a rectangle opposite sides are equal

Section – III (4 × 6 = 24 Marks)

Note :

1. Answer any 4 questions from the given 6 questions.
2. Each question carries 6 marks.

Question 13.
Prove that 3 \(\sqrt{5}\) + \(\sqrt{7}\) is an irrational number.
Solution:
We have to prove that 3\(\sqrt{5}\) + \(\sqrt{7}\) is an irrational number. Let us assume that 3\(\sqrt{5}\) + \(\sqrt{7}\) is a rational number.

\(\frac{p^2-38 q^2}{2 p q}\) is a rational number but \(\sqrt{7}\) is an irrational number. This is a contradiction to the fact that \(\sqrt{7}\) is an irrational number. So our assumption is wrong.
So, \(\sqrt{7}\) is an irrational number.

Question 14.
Draw the graph of the Quadratic polynomial p(x) = x² + x -12 and find the zeroes of the polynomial from the graph.
Solution:
Given that
p(x) = x² + x – 12

Question 15.
Find the Arithmetic mean of the following data.

Solution:

Question 16.
Construct a triangle ABC with AB = 5.6 cm, BC = 7.2 cm and CA = 4.8 cm. Construct another triangle similar to ΔABC, whose sides are \(\frac{3}{5}\) times of the corresponding sides of ΔABC.
Solution:

Step of Construction:

1. Construct ΔABC with given measurements.
2. Draw a Ray AX in the downward direction at point A such that ∠BAX is an acute angle.
3. Draw 5 arcs on AX such that AP₁ = P₁P₂ = P₃P₄ = P₄P₅.
4. Joint P₅ and B.
5. Draw a line parallel to P₅B through point P₃ which cuts AB at B¹.
6. Draw a line parallel to BC through point B¹, which cuts AC at C¹.
7. ΔB¹C¹ is required similar triangle.

Question 17.
The three vertices of a parallelogram ABCD are A(- 1, – 2), B(4, – 1) and C(6, 3). Find the coordinates of vertex D and find the area of parallelogram ABCD.
Solution:

Question 18.
Due to heavy floods in the state thousands were rendered homeless. The State Government decided to provide canvas for 1500 tents. The lower part of each tent is cylindrical of base radius 2.8 meters and height 3.5 meters with conical upper part of same base radius but of height 2.1 meters. If the canvas used to make the tent costs Rs. 100 per square meter, find the total cost of canvas to construct the tents.
Solution:

Curved surface area of the cylinder = 2πrh
= 2 × \(\sqrt{7}\) × 2.8 × 3.5 = 61.6 sq.m
= (2.8)² + (2.1)² = 7.84 + 4.41
l = \(\sqrt{12.25}\) = 3.5 m
Curved surface area of the cone = πrl
= \(\frac{22}{7}\) × 2.8 × 3.5 = 30.8m².
The area of the cloth required for one tent = 61.6 + 30.8 = 92.4 m²
Cost of the 1m² cloth = Rs. 1oo
∴ Total cost of 1500 tents
= ₹ 92.4 × 100 × 100
= ₹ 9240 × 1500
= ₹ 1,38,60,000

Part – B (20 Marks)

Note :

1. All questions are to be answered.
2. Each question carries 1 mark.
3. Answers are to be written in the Question paper only.
4. Marks will not be given for over-writing, rewriting or erased answers.

I. Write the CAPITAL LETTERS (A, B, C, D) of the correct answer in the brackets provided against each question, 20 × 1 = 20 M

Question 1.
The perimeter of a triangle, whose vertices are (0, 5), (0, 0) and (12, 0) is …….. ( )
A) 15
B) 13
C) 30
D) 10
Solution:
C) 30

Question 2.
The ten’s digit of the product 6²⁰×5² is …… ( )
A) 0
B) 6
C) 4
D) 5
Solution:
A) 0

Question 3.
If p(x) = 5x⁴ + 6x² + 3x – 7. then p(0) is …… ( )
A) 5
B) 3
C) 6
D) -7
Solution:
D) -7

Question 4.
Which of the following cannot be probability of an event? ( )
A) 0.35
B) 40%
C) 1\(\frac{1}{2}\)
D) \(\frac{2}{3}\)
Solution:
C) 1\(\frac{1}{2}\)

Question 5.
Set A = {x : x is a letter in the word “CHANDRAYAN”}, then n(A) is …….. ( )
A) 10
B) 7
C) 8
D) 6
Solution:
B) 7

Question 6.
The first degree equation in two variables ax – by + c = 0 represents ……..
A) Straight line
B) Parabola
C) Line segment
D) Ray
Solution:
A) Straight line

Question 7.
The Quadratic equation, whose sum of the roots is -3 and product of the roots is 2. ( )
A) x² + 6x + 5 = 0
B) x² – x – 6 = 0
C) x² – 3x + 2 = 0
D) x² + 3x + 2 = 0
Solution:
D) x² + 3x + 2 = 0

Question 8.
The number of terms in the AP; 2, 5, 8,…….., 32 is …….. ( )
A) 9
B) 10
C) 11
D) 32
Solution:
C) 11

Question 9.
In ΔABC, DE || BC, where AD = 2cm, DB = 3 cm and AE = 4 cm, then AC is …….. ( )

A) 5 cm
B) 10 cm
C) 6 cm
D) 9 cm
Solution:
B) 10 cm

Question 10.
In a circle, ‘O’ is the centre, P is the external point and AP is the tangent drawn to the circle from P, OA is the radius. If ∠APO = 30°, then ∠POA = ? ( )
A) 120°
B) 90°
C) 30°
D) 60°
Solution:
D) 60°

Question 11.
The base diameter and height of a Right circular cone are 12 cm and 8 cm, then the slant height is ……. ( )
A) 10 cm
B) 9 cm
C) 20 cm
D) 4 cm
Solution:
A) 10 cm

Question 12.
In ΔABC, ∠H = 90°, then cos(A + C) ?
A) 1
B) 0
C) \(\frac{1}{2}\)
D) \(\frac{\sqrt{3}}{2}\)
Solution:
B) 0

Question 13.
If the angle of elevation of Sun increases from 0° to 90°, then the length of the shadow of
the tower is …… ( )
A) No change
B) Increases
C) Decreases
D) Can’t be decided
Solution:
C) Decreases

Question 14.
The probability of an event P(E) is always
A) 0 ≤ P (E) ≤ 1
B) P(E) < 1
C) P(E) ≥ 1
D) P(E) ≤ 0
Solution:
A) 0 ≤ P (E) ≤ 1

Question 15.
The roots of a Quadratic equation 2x² + x + 4 = 0 are …….. ( )
A) One is positive and other is negative.
B) Both are positive.
C) Both are negative.
D) No real roots.
Solution:
D) No real roots.

Question 16.
Which of the following equation has ‘2’ as a root? ( )
A) x² – 4x + 5 = 0
B) x² + 3x – 12 = 0
C) 2x² – 7x + 6 = 0
D) 3x² – 6x – 2 = 0
Solution:
C) 2x² – 7x + 6 = 0

Question 17.
The number of subsets of a set is 16, then the number of elements of the set is …… ( )
A) 8
B) 16
C) 10
D) 4
Solution:
D) 4

Question 18.
‘O’ is the centre. PA and PB are tangents drawn to the circle from point P. If angle between the tangents is 60° and length of the tangent is \(\sqrt{3}\) cm, then the radius of the circle is ……….

A) 2 cm
B) 1 cm
C) 2 \(\sqrt{3}\) cm
D) \(\frac{\sqrt{3}}{2}\)
Solution:
B) 1 cm

Question 19.
The number of secants can be drawn to a circle from an external point is ……. ( )
A) Infinite
B) 1
C) 2
D) 0
Solution:
A) Infinite

Question 20.
If the ratio of the height of a pole and the length of it’s shadow is \(\sqrt{3}\) : 1, then the angle of elevation of Sun is …… ( )
A) 30°
B) 60°
C) 45°
D) 90°
Solution:
B) 60°