TS 10th Class Maths Question Paper June 2024

The strategic use of TS 10th Class Maths Model Papers and TS 10th Class Maths Question Paper June 2024 can significantly enhance a student’s problem-solving skills.

TS 10th Class Maths Question Paper June 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.
Expand log \(\frac{243}{32}\).
Solution:
We know that log_a\(\frac{x}{y}\) = log_a x – log_a y
So, log \(\frac{243}{32}\) = log 243 – log 32
= log3⁵ – log2⁵ = 5log 3 – 5 log 2
(Since log_a x^m = nlog_a x)
∴ log \(\frac{243}{32}\) = 5(log3 – log2)

Question 2.
Whether the following pair of linear equations represents parallel lines ? Justify your answer.
2x + 3y = 10 and 6x + 9y = 15.
Solution:
2x + 3y – 10 = 0
6x + 9y – 15 = 0
\(\frac{a_1}{a_2}\) = \(\frac{2}{6}\) = \(\frac{1}{6}\), \(\frac{b_1}{b_2}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\), \(\frac{c_1}{c_2}\) = \(\frac{-10}{-15}\) = \(\frac{2}{3}\)
Since \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\), the pair of linear equations are inconsistent and parallel.

Question 3.
“The top of the tower is observed at an angle of elevation 45° and the food of the tower is at the distance of 20 meters from the observer”. Draw a suitable diagram for this data.
Solution:
In the above figure, 8 represents the top of the tower and A is its foot. AC represents the distance between the foot of the tower and the observer (C).

Question 4.
Express ‘tanθ’ in terms of ‘sinθ’.
Solution:
We know that, tanθ = \(\frac{\sin \theta}{\cos \theta}\) and sin²θ + cos²θ = 1
sin²θ + cos²θ = 1
∴ tan θ = \(\frac{\sin \theta}{\cos \theta}\) = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
(Since, cos²θ = 1 – sin²θ)
∴ cosθ = \(\sqrt{1-\sin ^2 \theta}\)

Question 5.
Construct a Quadratic equation having the roots log₂ 8 and cosec 30°.
Solution:
Given that the roots of the quadratic equation are \(\log _2^8\) and cosec 30°.
\(\log _2^8\) can be written as \(\log _2 2^3\) = 3 log₂ 2
(∵ 8 = 2 × 2 × 2 = 2³)
We know that log₂ 2 = 1
∴ log₂ 8=3log₂ 2 = 3 × 1 = 3
Since sin30° = \(\frac{1}{2}\),
cosec 30° = \(\frac{1}{\sin 30^{\circ}}\) = \(\frac{1}{\frac{1}{2}}\) = 2.
So, the roots of the quadratic equation are 3 and 2.
∴ The required Q.E
= x² – x (Sum of the roots) + Products of the roots = 0
i.e., x² – x(3 + 2) + (3 × 2) = 0
x² – 5x + 6 = 0
The Q.E having roots log₂ 8 and cosec 30° is x² – 5x + 6 = 0.

Question 6.
In the given figure, PQ and PR are tangents to a circle with centre ‘O’. If ∠QOR = 120°, then find the ∠RPO.

Solution:
From the figure, it is clear that PQ and PR are the tangents to the circle with centre ‘O’ from an external point P.
PQ is the tangent and QO’ is the radius drawn through the point of contact Q.

∴ ∠PQO = 90°
Similarly, ∠PRO= 90°.
PQOR is a quadrilateral.
∠PQO + ∠QOR + ∠PRO + ∠PRQ = 360°
(∵ Sum of the interior angles of the quadrilateral is 360°)
90° + 120° + 90° + ∠RPQ = 360°
300° + ∠RPQ = 360°
∴ ∠RPQ = 360° – 300° = 60°
∴ ∠RPQ = \(\frac{60^{\circ}}{2}\) = 30°
Since, ΔPQO and ΔPRO are congruent,
SAS (Property)
OQ = OR (radii)
PQ = PR (tangents from p to the given circle
∠PQO = ∠PRO = 90°
∴ ∠RPO = ∠QPO
i.e., OP bisects ∠RPQ.

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 finding the sum of first n terms of an arithmetic progression and explain each term.
Solution:
The sum of the first ‘n’ terms of an AP is given by
S_n = [2a + (n – 1) d] (or) S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
[∵ a_n = a + (n – 1)d]
Here, S_n denotes the sum of the first n terms of the AP.
‘a’ denotes the first term of the AP.
‘d’ denotes the common difference.
If the first and last terms of an AP are given and the common difference is not given, then S_n = \(\frac{\mathrm{n}}{2}\) (a + l) where ‘a’ is the first term and ‘l’ is the last term.

Question 8.
Prove that \(\frac{1+\sin \theta}{1-\sin \theta}\) = (sec θ + tan θ)².
Solution:
Given, \(\frac{1+\sin \theta}{1-\sin \theta}\)
Divide the numerator and denominator by Cos θ. We get

Question 9.
Cards numbered 1 to 30 are put in a bag. If a card is drawn at randomly, find the probability that the drawn card is …….
Solution:
Number of cards (numbered 1 to 30, put in the bag is 30.

i) Let E be the event of the getting a prime number on the card, then the outcomes favourable to E are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
∴ The number of outcomes favourable to E is 10.

ii) Let E be the event of getting a perfect square on the card. Then the outcomes favourable to E are 1, 4, 9, 16, 25.
∴ The number of outcomes favourable to E is 5.
So, P(E) = P (a perfect square below 30)

Question 10.
If A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}, then show that n(A∪B) = n(A) + n(B) – n(A∩B).
Solution:
A = {1, 2, 3, 4, 5, 6}
B = {2, 4, 6, 8}
Then A∪B
= {1, 2, 3, 4, 5, 6} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 5, 6, 8}
∴ n(A∪B) = 7 ………. (1)
n(A ∩ B) = 3, n(A) = 6, n(B) = 4
n(A) + n(B) – n(A∩B)
= 6 + 4 – 3 = 7 …….. (2)
From (1) and (2)
n(A∪B) = n(A) + n(B) – n(A ∩ B)

Question 11.
Find the dimensions of a rectangle, whose perimeter is 36 cm and whose area is 65 square centimeters.
Solution:
Let the length of the rectangle be x cm,
Let the breadth of the rectangle be y cm.
∴ The perimeter of the rectangle = 2 (x + y) cm.
Area of the rectangle = x × y = xy sq.cm
By problem,
2(x + y) = 36
⇒ x + y = \(\frac{36}{2}\) = 18 …… (1)
xy = 65 ……… (2)
We know that
(x – y)² = (x + y)² – 4xy ……. (3)
Substituting the above values in (3),
we have (x – y)² = (18)² – (4 × 65)
= 324 – 260 = 64
∴ x – y = \(\sqrt{64}\) = 8 …….. (4)
Solving (1) and (4)

Substituting the value of ‘x’ in (1),
13 + y = 18
i.e., y = 18 – 13 = 5
∴ The dimensions of the rectangle are 13 cm and 5 cm.

Question 12.
Solve the pair of linear equations : 3x + 2y = 11, 2x + 3y = 4.
Solution:
Given : 3x + 2y = 11 …….. (1)
2x + 3y = 4 ……. (2)
Adding (1) and (2),
5x + 5y = 15
⇒ x + y = 3 …….. (3)
Subtracting (2) from (1),
x – y = 7 ………… (4)
Adding (3) and(4), we get
= 2x = 10 ⇒ x = \(\frac{10}{2}\) = 5
Substituting the value of ‘x’ in (1),
we get 5 + y = 3
⇒ y = 3 – 5 = -2
∴ Solution : x = 5 and y = -2

Section – III (4 × 6 = 24 Marks)

Note :

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

Question 13.
If x² + y² = 34xy, then prove that 2 log (x + y) = 2 log 6 + log x + log y.
Solution:
x² + y² = 34xy (Given)
Adding 2xy on both sides, we get
x² + y² + 2xy = 34xy + 2xy
(x + y)² = 36xy
Taking log both sides,
log(x+y)² = log36xy
2 log (x + y) = log 36 + log x + log y
2 log (x + y) = log6² + log x + log y
∴ 2 log (x + y) = 2 log 6 + log x + log y

Question 14.
Draw the graph of the Quadratic polynomial p(x) = x² + x – 6 and find the zeroes of the polynomial from the graph.
Solution:
The given quadratic polynomial p(x) = x² + x – 6

Locate the points listed above on a graph paper and join the points with a smooth curve. We get a U shaped curve. It intersects the X-axis at two points (2, 0) and (-3, 0).
∴ The zeroes of the polynomial from the graph are 2 and -3.

Question 15.
Find the mode of the following data.

Solution:
The given data is :

Here, the maximum frequency is 20, and the class corresponding to this frequency is 20 – 30. So, the modal class is 20 – 30.
Now, boundary limit (l) of modal class = 20
class size (h) = 10
frequency of the modal class (f₁) = 20 frequency of the class preceding the modal class (f₀) = 8 frequency of the class succeeding the modal class (f₂) = 12 Substituting these values in the formula,

Question 16.
Construct a triangle ABC with AB = 6 cm, AC = 6 cm and BC = 8 cm. Construct another triangle similar to ΔABC, whose sides are \(\frac{2}{3}\) times of the corresponding sides of ΔABC.
Solution:

Steps of construction:

1. Draw a triangle ABC in which BC = 8cm, CA = 6 cm and AB = 6 cm.
2. Draw a ray BX, making an acute angle with BC on the side opposite to vertex A.
3. Locate 3 points B₁, B₂ and B₃ on BX such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃C and draw a line from B₂ to C’ which is parallel to B₃C and intersection BC at C’.
5. Draw a line through C parallel to CA to intersect AB at A’.
6. Now ΔA’BC’ is the required triangle.

Question 17.
Find the coordinates of the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4).
Solution:

\(\overline{\mathrm{AB}}\) is divided into three equal parts
(i.e.,) \(\overline{\mathrm{AP}}\) = \(\overline{\mathrm{PQ}}\) = \(\overline{\mathrm{QB}}\).
P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 2 internally.
Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.
∴ The co-ordinates of P are

Question 18.
A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2 cm and the height of the cylindrical and conical portion are 12 cm and 7 cm respectively. Find the volume of the solid toy. (Take π = \(\frac{22}{7}\)).
Solution:

Volume of the toy = Volume of the hemisphere + Volume of the cylinder + Volume of the cone
Diameter of the solid toy (d) =4.2 cm
∴ radius (r) = \(\frac{d}{2}\) = \(\frac{4.2}{2}\) = 2.1cm = \(\frac{21}{10}\)

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 H.C.F. of 12 and 21 is ………..
A) 4
B) 7
C) 3
D) 6
Solution:
C) 3

Question 2.
If A and B are disjoint sets such that n(A) = 5 and n(A∪B) = 8, then n(B) = ………
A) 3
B) 4
C) 5
D) 7
Solution:
A) 3

Question 3.
The degree of the polynomial, whose graph is given in adjacent figure ………
A) 1
B) 2
C) 0
D) 3

Solution:
B) 2

Question 4.
The value of k for which the pair of linear equations 2x + 3y = 7 and 4x + ky = 11 has no solution is ……… ( )
A) 8
B) -6
C) 18
D) 6
Solution:
D) 6

Question 5.
If x + \(\frac{1}{x}\) = 2 then x² + \(\frac{1}{x^2}\) = …….. ( )
A) 4
B) 6
C) 2
D) 1
Solution:
C) 2

Question 6.
The 15^th term of the AP; 3, 6, 9, …… ( )
A) 15
B) 45
C) 30
D) 360
Solution:
B) 45

Question 7.
The distance between two points (0, 0) and (sinθ, cosθ), where 0° < θ < 90° is ( )
A) 1/2
B) \(\sqrt{2}\)
C) tanθ
D) 1
Solution:
D) 1

Question 8.
In ΔABC, BC² = AB² + AC², then ……. is the right angle. ( )
A) ∠A
B) ∠B
C) ∠C
D) None
Solution:
A) ∠A

Question 9.
PA and PB are tangents to a circle drawn from the external point P to the circle. If PA = 7 cm, then PB = ……… ( )
A) 7/2 cm
B) 7 cm
C) 14 cm
D) 49 cm
Solution:
B) 7 cm

Question 10.
If the ratio of surface areas of two spheres is 4 : 9, then the ratio of their volumes is ( )
A) 27 : 8
B) 9 : 4
C) 16 : 81
D) 8 : 27
Solution:
D) 8 : 27

Question 11.
If sinθ = \(\frac{4}{5}\), then the value of secθ + tanθ is …….. ( )
A) 3
B) 4
C) 5
D) 9
Solution:
A) 3

Question 12.
In the given figure BC = 20 m, AC = 10 m, then θ = ( )

A) 60°
B) 90°
C) 45°
D) 30°
Solution:
D) 30°

Question 13.
E and \(\overline{\mathrm{E}}\) are complementary events in an experiment. If P(E) = 0.7, then P(\(\overline{\mathrm{E}}\)) = …….. ( )
A) 0.7
B) 1
C) 0.3
D) 0
Solution:
C) 0.3

Question 14.
The mean of first primes is ……….
A) 5.6
B) 8.1
C) 7.3
D) 5
Solution:
A) 5.6

Question 15.
The curved surface area of a right circular cone is πrl, where l is …….. ( )
A) Diameter
B) Radius
C) Height
D) Slant Height
Solution:
D) Slant Height

Question 16.
If A ⊂ B, then A∩B = ( )
A) B
B) A
C) µ
D) φ
Solution:
B) A

Question 17.
Which terms of the AP : 50, 40, 30, ………. is zero ? ( )
A) 6
B) 5
C) 10
D) 7
Solution:
A) 6

Question 18.

A) 0
B) tan²θ
C) cot²θ
D) 1
Solution:
D) 1

Question 19.
The probability of getting 8 as sum of numbers on two dice, when they are rolled is……
A) \(\frac{8}{36}\)
B) \(\frac{1}{12}\)
C) \(\frac{1}{9}\)
D) \(\frac{11}{36}\)
Solution:
Correct answer : \(\frac{5}{36}\)

Question 20.
ΔABC ΔPQR , ∠Q + ∠R = 130°, then ∠A = …….. ( )
A) 60°
B) 150°
C) 50°
D) 130°
Solution:
C) 50°