TS 10th Class Maths Question Paper June 2025

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TS 10th Class Maths Question Paper June 2025

Time: 3 Hours
Maximum Marks: 80

General 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 : log₁₀\(\left(\frac{405}{32}\right)\).
Solution:

Question 2.
Find the distance of the point (log₁₀ 1000, log₂16) from the origin.
Solution:

Question 3.
“The angle of elevation of the top of a tower from a point on the ground at a distance of 70 m from the foot of the tower is 60°. Draw a suitable diagram for this situation.
Solution:
Tower – BC
Top of tower is ‘C’
Point on the ground is ‘A’.
AB = 70 m,
∠CÄB = ∠BAC = angle of elevation of top = 60°

Question 4.
Is (x + 1)² = 2 (x – 3) a quadratic equation ? Justify.
Solution:
LHS = (x + 1)² = x² + 2x + 1
RHS = 2(x – 3) = 2x – 6
Given is x² + 2x + 1 = 2x – 6
= x² + 2x + 1 – 2x + 6 = 0
x² + 7 = 0 is a quadratic equation
So, given is a quadratic equation.

Question 5.
A = {1, 2, 3, 4}, B= {2, 4, 6, 8, 10}. Find A – B and B – A.
Solution:
Given A = {1, 2, 3, 4},
B = {2, 4, 6, 8, 10}
Then A – B = {x/x ∈ A and x ∉ B}
So A – B = {1, 3}
and B – A = {x/x ∈ B ∧ x ∉ A}
So B – A = {2, 4, 6, 8, 10} – {1, 2, 3, 4}
= {6, 8, 10}

Question 6.
The length of the minute hand of a clock is 3.5 cm. Find the area swept by the minute hand in 30 minutes (π = \(\frac{22}{7}\))
Solution:
Length of a minute hand of a clock 7
OA = r = 3.5 cm \(\frac{7}{2}\) cm
Now, the area swept by minute hand in 30 minutes is the area of sector with OA = OB = \(\frac{7}{2}\) cm and θ = 15°
(∵ angle made by minute hand in one hour is 30°)

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 median of a grouped data and explain each term of it.
Solution:
Formula for median of a grouped data is median = l + \(\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\) × h
Where.
‘l‘ = lower boundary of median class
n = number of observations
cf = cumulative frequency of class preceding median class
f = frequency of median class
h = size of median class.

Question 8.
ABC is a triangle right – angled at C. P and Q are A points on sides CB and CA respectively. Show that AP² + BQ² = AB² + PQ².

Solution:
Give, in ΔABC, ∠C = 90°
P, Q are two points on CR and CA respectively.
RTP: AP² + BQ² = AB² + PQ²
Construction: Join PQ

Proof: In ΔBCQ, ∠C = 90°
∴ QC² + BC² = QB² …… (1)
(From Pythagorous theorem)
In ΔACP, ∠C = 90°
∴ CP² + AC² = AP² …… (2)
Now adding (1) and (2) we get

We get QC² + BC² + CP² + AC² = QB² + AP² ……. (3)
But from ΔABC, AC² + BC² = AB² …….. (4)
and QCP, QC² + CP² = QP² = PQ² …… (5)
Now adding (4) and (5), we get
AC² + BC² + QC² + CP² = AB² + PQ² ……. (6)
So, from (3) and (6), we get
AB² + PQ² = BQ² + AP²
Hence proved.

Question 9.
A strip of width 5 cm is attached to one side of square to form a rectangle. The area of the rectangle formed is 204 cm², then find the length of the side of the square.

Solution:
Let ABCD is a square with a side of ‘x cm.
And ‘BCEF’ is a rectangle strip of width 5 cm.

From given, area of rectangle
AFED = 204 cm²
Now let the side of square AB = BC = x cm
Then length of rectangle (l)
= AF = AB + BF
= (x + 5) cm
and breadth of rectangle (B) = EF = x cm
Then area of rectangle
= l × b
= (x + 5) x = x² + 5x = 204
∴ x² + 5x – 204 = 0
⇒ x² + 17x – 12x – 204 = 0
⇒ x(x + 17) -12(x + 17) = 0
⇒ (x – 12)(x + 17) = 0
So, x – 12 = 0 ⇒ x = 12 or
x + 17 = 0 ⇒ x = -17
But ‘x’ cannot be negative.
Hence side of square = x = 12 cm

Question 10.
Solve following pair of equations :
3x + 4y = 10,
4x – 3y = 5
Solution:
Given equations are
3x + 4y = 10 ……. (1)
4x – 3y = 5 …….. (2)
We solve this by elimination method (3x + 4y = 10) multiply both sides by ’4’
(3x + 4y = 10) multiply both sides by ‘4’
(4x – 3y = 5) multiply both sides by 3, we get, then

Question 11.
Two dice are rolled simultaneously and the sum of the numbers appearing on them is noted. Find the probability of getting the sum.
i) 11
ii) a perfect square.
Solution:
When two dice are rolled simultaneously the possible outcomes are as shown below

Now favourable outcomes for the sum their numbers is 11 are (5, 6) and (6, 5)
So no. of favourable outcomes = 2
no. of total outcomes = 6 × 6 = 36
So, the probability of getting the sum of 11 is \(\frac{2}{36}\) = \(\frac{1}{18}\)
and now, favourable outcomes for sum as perfect square are(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3).
So the number of favourable outcomes for getting a square number = 7
Hence probability of getting a sum as a square = \(\frac{7}{36}\).

Question 12.
A girl says “The points A(6, 4), B(4, -6) and C(-2, 4) are collinear”. Do you agree with her ? Justify your answer.
Solution:
Let us assume the above given points are not collinear. Then they form a triangle. Then let us find the area of possible triangle.
A(6, 4) so x₁ = 6, y₁ = 4
B(4, -6) so x₂ = 4, y₂ = -6
C(-2, 4) so x₃ = -2, y₃ = y
then formula for area of triangle

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.
Draw the graph of the quadratic polynomial P(x) = x² + x – 6 and find the zeroes of the polynomial from the graph.
Solution:
Given polynomial is
p(x) = y = x² + x – 6
Let us find the co – ordinates of points through which the polynomial passes.

Question 14.
Construct an equilateral triangle ABC with side BC = 6 cm. The construct another triangle similar to ΔABC whose sides are \(\frac{3}{2}\) times of the corresponding sides of ΔABC.
Solution:
Steps of construction :
1) Draw an equilateral triangle ΔABC with sides AB = BC = C A = 6 cm as shown below.
2) Draw a ray \(\overline{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A

3) Mark 3 points B₁, B₂, B₃ such that BB₁ = B₁B₂ = B₂B₃
4) Now join B₂C

5) Now draw a line parallel to B₂C through B₃ meeting BC produced C’.

6)Draw a line parallel to AC through C’ meeting BA produced A’.
7) Now A’BC’ is the required triangle which is similar and sides are \(\frac{3}{2}\) times of ΔABC

Here AB = BC = AC = 6 cm. and A’B’ = BC’ = A’C’ = 9 cm.

Question 15.
Find the Mode for the following data.

Solution:

Since the maximum number (frequency) is 31. Hence the modal class is 30 – 40.
So, the lower boundary of modal class l = 30
Model class size (h) = 10
Frequency of modal class = f₁ = 31
Frequency of class preceding modal class = f₀ = 26
Frequency of class succeeding modal class = f₂ = 16
Now, formula for mode is

Question 16.
The volume of a solid cuboid is 210 cm³ and its lateral surface area is 130 cm². If its height is 5 cm, then find its length and breadth.
Solution:
Let length, breadth and height of given cuboid are l, b, h respectively.
Given that its height (h) = 5 cm
and it volume (y) = lbh = 210 cm³
∴ v = l × b × 5 = 210 cm³
⇒ fb = \(\frac{210}{5}\) = 42 cm²
So lb = 42cm² …….. (1)
and now formula for its lateral surface area = 130 cm²
formula for lateral surface area of cuboid = A = 2h (l + b)
So 2h (l + b) = 130 cm² (given)
⇒ 2(5)(l + b) = 130 cm² (∵ h = 5 cm)
∴ (l + b) = \(\frac{130}{10}\) = 13 cm
∴ l + b = 13 cm ……… (2)
Then b = 13 – l ………. (3)
Now put b = (13 – l) in equation (1)
lb = 42, we get
l(13 – l) = 42
⇒ 13l – l² = 42
⇒ l² – 13l + 42 = 0
⇒ l² – 6l – 7l + 42 = 0
⇒ l(l – 6) -7(l – 6) = 0
⇒ (l – 6)(l – 7) = 0
So, l = 6 or l = 7
If l = 6 then b = 13 – l = 13 – 6 = 7
So, l = 6, b = 7 (or)
If l = 7 then b = 13 – l = 13 – 7 = 6
So, l = 7, b = 6
So, length, breadth of given cuboid or 6, 7 or 7, 6 cm.

Question 17.
Show that \(\frac{\sin \theta}{1-\cos \theta}+\frac{1-\cos \theta}{\sin \theta}\), where ‘θ’ is an acute angle.
Solution:

Question 18.
If the sum of first 8 terms of an arithmetic progression is 64 and the sum of first 17 terms is 289, then find the sum of first 30 terms.
Solution:
Let a, a + d, a + 2d ……. is the given AP.
Then formula for sum of first ‘n’ terms =
S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
So, for n = 8, sum of first ‘8’ terms =
S₈ = \(\frac{8}{2}\)[2a + (8 – 1)d] = 64
⇒ 4(2a + 7d) = 64
⇒ (2a + 7d) = \(\frac{64}{4}\) = 16
∴ 2a + 7d = 6 ….. (1)
and for n = 17, sum of first 17 terms =

So d = 2, then a + 8d = 17.
⇒ a + 8(2) = 17
a + 16 = 17
⇒ a = 1
So, in the given AP, a = 1,d = 2
Now sum of first 30 terms =
S₃₀ = \(\frac{30}{2}\)[2(1) + (30 – 1)2]
(∵ a = 1, d = 2, n = 30)
= 15[2 + (29 × 2)]
= 15(2 + 58) = 15 × 60
= 900
So, sum of first 30 terms of given AP = S₃₀ = 900

Part – B (20 Marks)

Note :

1. ALL question are to be answered.
2. Each questions 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 (A, B, C, D) of the correct answer in the brackets provided against each question, 20 × 1 = 20

Question 1.
If quadratic equation x² – 5x + 6 = 0 and x² – 7x + k = 0 have a common root, then the possible value of k is : [ ]
A) 8, 9
B) 2, 3
C) 10, 12
D) 5, 3
Solution:
C) 10, 12

Question 2.
Which term of the arithmetic progression 24, 21, 18, 15, …….. is zero ? [ ]
A) 10^th term
B) 8^th term
C) 7^th term
D) 9^th term
Solution:
D) 9^th term

Question 3.
If sin A = \(\frac{7}{25}\), then the value of cos A (A is an acute angle) is :
A) \(\frac{7}{24}\)
B) \(\frac{25}{24}\)
C) \(\frac{25}{7}\)
D) \(\frac{24}{25}\)
Solution:
D) \(\frac{24}{25}\)

Question 4.
If x = 3, y = -1 is a solution of the linear equation x + ky = 1 and x – y = 4, then the value of k is : [ ]
A) 2
B) -3
c) 3
D) -1
Solution:
A) 2

Question 5.
If the slope of the line segment joining the points (-1, -1) and (1, x) is 2, then the value of x is : [ ]
A) -2
B) -3
C) 2
D) 3
Solution:
D) 3

Question 6.
If the ratio of surface areas of two spheres is 1 : 4, then the ratio of their volumes is :
A) 1 : 64
B) 2 : 8
C) 1 : 8
D) 1 : 16
Solution:
C) 1 : 8

Question 7.
In ΔABC, DE || BC. \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{4}{3}\) and EC = 4.8 cm, then AE is equal to: [ ]

A) 6.4 cm
B) 4.5 cm
C) 3.6 cm
D) 4.2 cm
Solution:
A) 6.4 cm

Question 8.
At a particular instance, if the length of the shadow of a tower of height 100m is \(\frac{100}{\sqrt{3}}\)m, then the angle made by Sun’s rays with the ground at that time is :
A) 30°
B) 60°
C) 45°
D) 90°
Solution:
B) 60°

Question 9.
If sin θ. cos θ = \(\frac{1}{2}\), then the value of (sin θ + cos θ)² is :
A) 0
B) 4
C) 2
D) 1
Solution:
C) 2

Question 10.
Which of the following is not a rational number ?
A) log₁₀3
B) \(5 . \overline{23}\)
C) 123.123
D) \(\frac{10}{19}\)
Solution:
A) log₁₀3

Question 11.
LCM of least prime number and least composite number is
A) 2
B) 4
C) 1
D) 3
Solution:
A) 2

Question 12.
From the given Venn diagram A ∩ B is :

A) {5, 6}
B) {7, 8}
C) {5, 6, 7, 8}
D) φ
Solution:
D) φ

Question 13.
Set A = {F, W, L, O}, Which of the following is not a set builder form for set A ? [
A) {x : x is a letter of the word FOLLOW}
B) {x : x is a letter of the word FLOW}
C) {x : x is a letter of the word WOLF}
D) {x : x is a letter of the word SLOW}
Solution:
D) {x : x is a letter of the word SLOW}

Question 14.
If P(x) = 3x² – x – 4, then the value of P(-1) is
A) 2
B) 0
C) -2
D) 1
Solution:
B) 0

Question 15.
The \(\frac{P}{Q}\) (Q ≠ 0) form of the decimal 0.125 is
A) \(\frac{2}{5}\)
B) \(\frac{3}{4}\)
C) \(\frac{1}{8}\)
D) \(\frac{1}{6}\)
Solution:
C) \(\frac{1}{8}\)

Question 16.
The formula for the n^th term of a Geometric progression is a_n = a.r^n-1. In this formula ‘r’ represents : [ ]
A) Common ratio
B) Radius
C) First term
D) Common difference
Solution:
A) Common ratio

Question 17.
If (6, k) is one of the solutions of 3x + 2y – 26 = 0, then the value of log₂k is : [ ]
A) 1
B) 2
C) 3
D) 4
Solution:
B) 2

Question 18.
The value of log₁₀(sec θ + tan θ) + log₁₀(sec θ – tan θ) is : [ ]
A) 0
B) 1
C) -1
D) 2
Solution:
A) 0

Question 19.
The exponential form logg₁₀ 0.001 = -3 is : [ ]
A) (0.01)¹⁰ = -3
B) (-3)¹⁰ = 0.001
C) 10³ = -0.001
D) 10^-3 = 0.001
Solution:
D) 10^-3 = 0.001

Question 20.
In the given figure AP and AQ are two tangents to a circle with centre O such that ∠PAQ = 60°, then ∠POQ is :

A) 30°
B) 60°
C) 120°
D) 130°
Solution:
C) 120°