CBSE Maths › Class 10 PYQs › 2025

CBSE Class 10 Maths Basic 2025 Question Paper 430/3/3 with Solutions

All 44 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/3/3 (2025), with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.

Set 430/1/1 Set 430/1/2 Set 430/1/3 Set 430/2/1 Set 430/2/2 Set 430/2/3 Set 430/3/1 Set 430/3/2 Set 430/3/3 Set 430/4/1 Set 430/4/2 Set 430/4/3 Set 430/5/1 Set 430/5/2 Set 430/5/3 Set 430/6/1 Set 430/6/2 Set 430/6/3

Q11 markMCQAreas Related to Circles

The perimeter of a quadrant of a circle of circumference 22 cm is :

(A)29 cm
(B)22 cm
(C)12.5 cm
(D)5.5 cm

Show answer & solution

Answer: (C) 12.5 cm

$2 π r = 22 \Rightarrow r = \frac{22 \times 7}{2 \times 22} = 3.5$ cm.
Perimeter of quadrant $= \frac{22}{4} + 2 r = 5.5 + 7 = 12.5$ cm.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q21 markMCQSurface Areas and Volumes

A cone and cylinder have same height and same radius. The volume of the cone and the volume of the cylinder are in the ratio :

(A)1 : 1
(B)1 : 3
(C)3 : 1
(D)1 : 2

Show answer & solution

Answer: (B) 1 : 3

Volume of cone : volume of cylinder $= \frac{1}{3} π r^{2} h : π r^{2} h = 1 : 3$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q31 markMCQStatistics

The following table shows the marks scored by 23 students of a class.
Marks: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50
Number of Students: 5, 3, 4, 8, 3
The lower limit of the modal class is :

(A)10
(B)20
(C)30
(D)40

Show answer & solution

Answer: (C) 30

The highest frequency is 8, for the class 30 – 40.
Modal class = 30 – 40, so its lower limit is 30.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q41 markMCQStatistics

For a distribution, if mean = 15 and mode = 12, then its median is :

(A)12
(B)13
(C)14
(D)15

Show answer & solution

Answer: (C) 14

Empirical relation: $3 Median = Mode + 2 Mean$ .
$3 Median = 12 + 30 = 42$ , so Median $= 14$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q51 markMCQProbability

A pair of dice is thrown simultaneously. Let E denote the event that “The sum of numbers obtained on both dice is at least 9.” The number of outcomes in favour of event E is :

(A)4
(B)6
(C)10
(D)26

Show answer & solution

Answer: (C) 10

Sum 9: (3, 6), (4, 5), (5, 4), (6, 3), which is 4 outcomes.
Sum 10: (4, 6), (5, 5), (6, 4), which is 3 outcomes.
Sum 11: (5, 6), (6, 5), which is 2 outcomes. Sum 12: (6, 6), which is 1 outcome.
Total $= 4 + 3 + 2 + 1 = 10$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q61 markMCQReal Numbers

If $p = 2^{3} \times 3^{2} \times 5$ and $q = 2^{2} \times 3^{3}$ , then the LCM of $p$ and $q$ is :

(A) $2^{3} \times 3^{3}$
(B) $2^{2} \times 3^{2}$
(C) $2^{2} \times 3^{2} \times 5$
(D) $2^{3} \times 3^{3} \times 5$

Show answer & solution

Answer: (D)

2^{3} \times 3^{3} \times 5

LCM = product of the highest powers of all prime factors.
Highest powers: $2^{3}$ , $3^{3}$ and $5$ .
LCM $= 2^{3} \times 3^{3} \times 5$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q71 markMCQReal Numbers

$3^{n}$ , where $n$ is a natural number, cannot end with the digit :

(A)3
(B)5
(C)7
(D)9

Show answer & solution

Answer: (B) 5

The only prime factor of $3^{n}$ is 3.
A number ending in 5 must be divisible by 5, so $3^{n}$ can never end with 5.
(The last digits of $3^{n}$ cycle through 3, 9, 7, 1.)

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q81 markMCQReal Numbers

A prime number has :

(A)exactly two prime factors
(B)exactly one prime factor
(C)at least one prime factor
(D)at least two prime factors

Show answer & solution

Answer: (B) exactly one prime factor

A prime number $p$ has only the factors 1 and $p$ .
1 is not prime, so its only prime factor is $p$ itself.
Hence a prime number has exactly one prime factor.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q91 markMCQPair of Linear Equations in Two Variables

The system of linear equations given by $x = a$ and $y = b$ is :

(A)Consistent with a unique solution.
(B)Consistent with infinitely many solutions.
(C)Consistent with two solutions.
(D)Inconsistent.

Show answer & solution

Answer: (A) Consistent with a unique solution.

$x = a$ is a line parallel to the y-axis and $y = b$ is a line parallel to the x-axis.
These lines are perpendicular and meet at exactly one point $(a, b)$ .
So the system is consistent with a unique solution.

Q101 markMCQQuadratic Equations

If $(x + 1)^{3} = x^{3} + 1$ is expressed as a quadratic equation in the form $p x^{2} + q x + r = 0$ , then the value of $p - q + r$ is :

(A)0
(B)1
(C)3
(D)6

Show answer & solution

Answer: (A) 0

$(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1$ .
So $x^{3} + 3 x^{2} + 3 x + 1 = x^{3} + 1$ , i.e. $3 x^{2} + 3 x = 0$ .
$p = 3$ , $q = 3$ , $r = 0$ , so $p - q + r = 3 - 3 + 0 = 0$ .

Q111 markMCQCoordinate Geometry

If point $(a, 2 b)$ is the mid-point of the line segment joining the points $(3, 5)$ and $(- 1, - 1)$ , then $(a, b)$ is equal to :

(A) $(1, 2)$
(B) $(2, 2)$
(C) $(2, 1)$
(D) $(1, 1)$

Show answer & solution

Answer: (D)

(1, 1)

Mid-point $= (\frac{3 + ( - 1 )}{2}, \frac{5 + ( - 1 )}{2}) = (1, 2)$ .
So $a = 1$ and $2 b = 2 \Rightarrow b = 1$ .
$(a, b) = (1, 1)$ .

Q121 markMCQCoordinate Geometry

The distance of point $(- 3, 4)$ from y-axis is :

(A) $- 3$
(B)3
(C)4
(D)5

Show answer & solution

Answer: (B) 3

Distance of a point $(x, y)$ from the y-axis is $∣ x ∣$ .
$∣ - 3∣ = 3$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q131 markMCQIntroduction to Trigonometry

The value of $θ$ for which $sin 2 θ = tan 4 5^{\circ}$ is :

(A) $22. 5^{\circ}$
(B) $3 0^{\circ}$
(C) $4 5^{\circ}$
(D) $9 0^{\circ}$

Show answer & solution

Answer: (C)

4 5^{\circ}

$tan 4 5^{\circ} = 1$ , so $sin 2 θ = 1 = sin 9 0^{\circ}$ .
$2 θ = 9 0^{\circ} \Rightarrow θ = 4 5^{\circ}$ .

Q141 markMCQTriangles

Which types of triangles are always similar ?

(A)Right-angled triangles
(B)Acute-angled triangles
(C)Isosceles triangles
(D)Equilateral triangles

Show answer & solution

Answer: (D) Equilateral triangles

All angles of every equilateral triangle are $6 0^{\circ}$ .
So any two equilateral triangles are similar (AAA). The other types need not have equal angles.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q151 markMCQTriangles

What values of $x$ and $y$ will make $△ A B C$ similar to $△ QR P$ in the figures given below ?

Diagram for CBSE 2025 Class 10 Maths question 15

(A) $x = 6$ , $y = 5$
(B) $x = 5$ , $y = 6$
(C) $x = 6$ , $y = 6$
(D) $x = 12$ , $y = 3.2$

Show answer & solution

Answer: (B)

x = 5

y = 6

$△ A B C \sim △ QR P$ means A ↔ Q, B ↔ R, C ↔ P.
So $\frac{A B}{QR} = \frac{B C}{R P} = \frac{C A}{P Q}$ , i.e. $\frac{2}{4} = \frac{2.5}{x} = \frac{3}{y}$ .
$x = 5$ cm and $y = 6$ cm.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q161 markMCQCircles

How many tangents can be drawn from the point P on the outer circle to the inner circle in the given figure ?

Diagram for CBSE 2025 Class 10 Maths question 16

(A)0
(B)1
(C)2
(D)Infinite

Show answer & solution

Answer: (C) 2

P lies on the outer circle, so it lies outside the inner (smaller concentric) circle.
Exactly two tangents can be drawn to a circle from a point outside it.

Q171 markMCQSome Applications of Trigonometry

In the given figure, the angle of elevation of point A from point C is :

Diagram for CBSE 2025 Class 10 Maths question 17

(A) $3 0^{\circ}$
(B) $4 5^{\circ}$
(C) $6 0^{\circ}$
(D)Cannot be determined

Show answer & solution

Answer: (A)

3 0^{\circ}

In right $△ A B C$ , $tan C = \frac{A B}{B C} = \frac{2}{2 3} = \frac{1}{3}$ .
So $∠ C = 3 0^{\circ}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q181 markMCQAreas Related to Circles

The angle of the sector of a circle whose area is one-eighth of the area of the circle is :

(A) $22 \frac{1}{2}^{\circ}$
(B) $4 5^{\circ}$
(C) $6 0^{\circ}$
(D) $9 0^{\circ}$

Show answer & solution

Answer: (B)

4 5^{\circ}

$\frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{1}{8} π r^{2}$ .
$θ = \frac{36 0 ^{\circ}}{8} = 4 5^{\circ}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q191 markAssertion–ReasonIntroduction to Trigonometry

Assertion (A) : For an angle $θ$ , $sec θ = 1 \Rightarrow tan θ = 0$ .
Reason (R) : $sec^{2} θ + tan^{2} θ = 1$ .

(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.

Show answer & solution

Answer: (C) Assertion (A) is true, but Reason (R) is false.

A: $tan^{2} θ = sec^{2} θ - 1 = 1 - 1 = 0$ , so $tan θ = 0$ . A is true.
R: the identity is $sec^{2} θ - tan^{2} θ = 1$ ; $sec^{2} θ + tan^{2} θ = 1$ is false in general (e.g. $θ = 4 5^{\circ}$ gives $2 + 1 = 3$ ). R is false.

Q201 markAssertion–ReasonQuadratic Equations

Assertion (A) : Every quadratic equation has two real roots.
Reason (R) : A quadratic polynomial can have at most two zeroes.

(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.

Show answer & solution

Answer: (D) Assertion (A) is false, but Reason (R) is true.

A: $x^{2} + 1 = 0$ has no real roots, so A is false.
R: a polynomial of degree 2 has at most 2 zeroes, so R is true.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q212 marksVery Short AnswerAreas Related to Circles

From a circular sheet of radius 10 cm, a quadrant is cut. Find the perimeter of the remaining sheet.

Show answer & solution

Answer:

\frac{470}{7}

\approx 67.14

The remaining sheet is bounded by three-quarters of the circle and two radii.
Arc length $= \frac{3}{4} \times 2 π r = \frac{3}{4} \times 2 \times \frac{22}{7} \times 10 = \frac{330}{7}$ cm.
Perimeter $= \frac{330}{7} + 2 \times 10 = \frac{330}{7} + 20 = \frac{470}{7} \approx 67.14$ cm.

Q222 marksVery Short AnswerPair of Linear Equations in Two Variables

Solve the following system of linear equations graphically :
$x + y = 5$ and $x - y = 3$

Show answer & solution

Answer:

x = 4

y = 1

$x + y = 5$ : points $(0, 5)$ , $(5, 0)$ , $(4, 1)$ . Plot and join them.
$x - y = 3$ : points $(3, 0)$ , $(0, - 3)$ , $(4, 1)$ . Plot and join them.
The two lines meet at $(4, 1)$ .
Hence $x = 4$ , $y = 1$ .

Q232 marksVery Short AnswerTriangles

A vertical pole of height 10 m casts a shadow of 15 m on the ground and at the same time, a tower casts a shadow of 45 m on the ground. Find the height of the tower.

Show answer & solution

Answer: 30 m

At the same time the sun's rays make the same angle with the ground, so the pole–shadow triangle and the tower–shadow triangle are similar (AA).
$\frac{h}{45} = \frac{10}{15}$ .
$h = 30$ m.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q23 (OR) (OR)2 marksVery Short AnswerTriangles

In the given figure, $\frac{QR}{QS} = \frac{QT}{P R}$ and $∠1 = ∠2$ . Prove that $△ P QS \sim △ T QR$ .

Diagram for CBSE 2025 Class 10 Maths question 23 (OR)

Show answer & solution

Answer: Proved.

In $△ P QR$ , $∠ P R Q = ∠ P QR$ ( $∠1 = ∠2$ ), so $P Q = P R$ .
Given $\frac{QR}{QS} = \frac{QT}{P R}$ , so $\frac{QR}{QS} = \frac{QT}{P Q}$ , i.e. $\frac{QS}{QR} = \frac{P Q}{QT}$ .
In $△ P QS$ and $△ T QR$ : $\frac{P Q}{T Q} = \frac{QS}{QR}$ and $∠ Q$ is common.
Hence $△ P QS \sim △ T QR$ (SAS similarity).

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2023 Basic 430/6/1, 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q242 marksVery Short AnswerCircles

In the given figure, TP and TQ are two tangents. If $∠ P T Q = 5 0^{\circ}$ , then find the measure of $∠ O P Q$ .

Diagram for CBSE 2025 Class 10 Maths question 24

Show answer & solution

Answer:

∠ O P Q = 2 5^{\circ}

TP = TQ (tangents from an external point), so $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - 5 0 ^{\circ}}{2} = 6 5^{\circ}$ .
$∠ O P T = 9 0^{\circ}$ (radius $⊥$ tangent).
$∠ O P Q = 9 0^{\circ} - 6 5^{\circ} = 2 5^{\circ}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q252 marksVery Short AnswerIntroduction to Trigonometry

If $sin 3 A = 1$ , then find the value of $cos 2 A - tan^{2} 4 5^{\circ}$ .

Show answer & solution

Answer:

- \frac{1}{2}

$sin 3 A = 1 = sin 9 0^{\circ}$ , so $3 A = 9 0^{\circ}$ and $A = 3 0^{\circ}$ .
$cos 2 A - tan^{2} 4 5^{\circ} = cos 6 0^{\circ} - 1 = \frac{1}{2} - 1 = - \frac{1}{2}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q25 (OR) (OR)2 marksVery Short AnswerIntroduction to Trigonometry

If $(sec A + tan A) (1 - sin A) = k cos A$ , then find the value of $k$ .

Show answer & solution

Answer:

k = 1

$(sec A + tan A) (1 - sin A) = \frac{1 + s i n A}{c o s A} \times (1 - sin A)$ .
$= \frac{1 - s i n ^{2} A}{c o s A} = \frac{c o s ^{2} A}{c o s A} = cos A$ .
So $k cos A = cos A$ , giving $k = 1$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q263 marksShort AnswerProbability

A box contains 6 blue, 4 white and 8 red marbles. A marble is drawn at random from this box. Find the probability that the marble so drawn is :
(i) white
(ii) white or red
(iii) not red

Show answer & solution

Answer: (i)

\frac{2}{9}

(ii)

\frac{2}{3}

(iii)

\frac{5}{9}

Total marbles $= 6 + 4 + 8 = 18$ .
(i) P(white) $= \frac{4}{18} = \frac{2}{9}$ .
(ii) P(white or red) $= \frac{4 + 8}{18} = \frac{12}{18} = \frac{2}{3}$ .
(iii) P(not red) $= \frac{6 + 4}{18} = \frac{10}{18} = \frac{5}{9}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q273 marksShort AnswerCircles

In the given figure, a circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively. Prove that $∠ A O B + ∠ C O D = 18 0^{\circ}$ .

Diagram for CBSE 2025 Class 10 Maths question 27

Show answer & solution

Answer: Proved.

Join OP, OQ, OR, OS. In $△ O A P$ and $△ O A S$ : $A P = A S$ (tangents from A), $O P = O S$ (radii), OA common, so they are congruent (SSS) and $∠ O A P = ∠ O A S$ .
Similarly OB, OC and OD bisect $∠ B$ , $∠ C$ and $∠ D$ . Let $∠ O A B = a$ , $∠ O B A = b$ , $∠ O C D = c$ , $∠ O D C = d$ .
Then $2 a + 2 b + 2 c + 2 d = ∠ A + ∠ B + ∠ C + ∠ D = 36 0^{\circ}$ , so $a + b + c + d = 18 0^{\circ}$ .
In $△ A O B$ : $∠ A O B = 18 0^{\circ} - (a + b)$ . In $△ C O D$ : $∠ C O D = 18 0^{\circ} - (c + d)$ .
$∠ A O B + ∠ C O D = 36 0^{\circ} - (a + b + c + d) = 36 0^{\circ} - 18 0^{\circ} = 18 0^{\circ}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q283 marksShort AnswerReal Numbers

Prove that $2$ is an irrational number.

Show answer & solution

Answer: Proved.

Suppose $2$ is rational. Then $2 = \frac{a}{b}$ , where $a, b$ are integers, $b \neq = 0$ , and $a, b$ have no common factor other than 1.
Squaring: $a^{2} = 2 b^{2}$ , so 2 divides $a^{2}$ , hence 2 divides $a$ . Let $a = 2 c$ .
Then $4 c^{2} = 2 b^{2}$ , i.e. $b^{2} = 2 c^{2}$ , so 2 divides $b^{2}$ and hence 2 divides $b$ .
So 2 is a common factor of $a$ and $b$ , a contradiction.
Hence $2$ is irrational.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/3/3, 2026 Basic 430/5/1, 2026 Basic 430/5/2, 2026 Basic 430/5/3, 2025 Standard 30/4/1, 2025 Standard 30/4/2, 2025 Standard 30/4/3, 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/4/3, 2023 Standard 30/6/1, 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q28 (OR) (OR)3 marksShort AnswerReal Numbers

Find which among the following numbers a, b and c is/are composite numbers.
$a = 7 \times 11 \times 13 + 13$
$b = 6 \times 5 \times 4 + 4$
$c = 7 \times 13 + 6$

Show answer & solution

Answer: a and b are composite numbers (c = 97 is prime).

$a = 13 (7 \times 11 + 1) = 13 \times 78 = 1014$ , which has factors other than 1 and itself, so $a$ is composite.
$b = 4 (6 \times 5 + 1) = 4 \times 31 = 124$ , so $b$ is composite.
$c = 91 + 6 = 97$ , which is not divisible by 2, 3, 5 or 7 ( $7 \times 14 = 98 > 97$ ), so $c$ is prime.
Hence $a$ and $b$ are composite.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q293 marksShort AnswerPolynomials

Find the zeroes of the polynomial $25 a^{2} - 10 a + 1$ and verify the relationship between the zeroes and coefficients of the given polynomial.

Show answer & solution

Answer: Zeroes:

\frac{1}{5}

\frac{1}{5}

$25 a^{2} - 10 a + 1 = (5 a - 1)^{2}$ , so the zeroes are $\frac{1}{5}$ and $\frac{1}{5}$ .
Sum of zeroes $= \frac{2}{5}$ and $- \frac{b}{a} = - \frac{- 10}{25} = \frac{2}{5}$ . Verified.
Product of zeroes $= \frac{1}{25}$ and $\frac{c}{a} = \frac{1}{25}$ . Verified.

Q303 marksShort AnswerPair of Linear Equations in Two Variables

A fraction becomes $\frac{1}{3}$ , when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ , when 8 is added to its denominator. Find the fraction.

Show answer & solution

Answer:

\frac{5}{12}

Let the fraction be $\frac{x}{y}$ .
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow y = 3 x - 3$ ... (1)
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4 x = y + 8$ ... (2)
Substituting (1) in (2): $4 x = 3 x - 3 + 8$ , so $x = 5$ and $y = 12$ .
The fraction is $\frac{5}{12}$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q30 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables

Find the value of $k$ for which the following pair of linear equations will have infinitely many solutions :
$k x + 3 y - (k - 3) = 0$ and $12 x + k y - k = 0$
Hence, find any two solutions of the given pair of equations.

Show answer & solution

Answer:

k = 6

; two solutions, e.g.

(0, 1)

and

(1, - 1)

For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ .
$\frac{k}{12} = \frac{3}{k} \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$ .
$\frac{3}{k} = \frac{k - 3}{k} \Rightarrow k - 3 = 3 \Rightarrow k = 6$ . So $k = 6$ .
The equations become $6 x + 3 y - 3 = 0$ and $12 x + 6 y - 6 = 0$ , both equivalent to $2 x + y = 1$ .
Two solutions: $x = 0, y = 1$ and $x = 1, y = - 1$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q313 marksShort AnswerIntroduction to Trigonometry

Prove the following trigonometric identity :
$\frac{t a n θ}{1 + c o t θ} + \frac{c o t θ}{1 + t a n θ} = tan θ + cot θ - 1$

Show answer & solution

Answer: Proved.

Let $t = tan θ$ , so $cot θ = \frac{1}{t}$ .
$\frac{t}{1 + \frac{1}{t}} = \frac{t ^{2}}{t + 1}$ and $\frac{\frac{1}{t}}{1 + t} = \frac{1}{t ( t + 1 )}$ .
LHS $= \frac{t ^{3} + 1}{t ( t + 1 )} = \frac{( t + 1 ) ( t ^{2} - t + 1 )}{t ( t + 1 )} = \frac{t ^{2} - t + 1}{t}$ .
$= t - 1 + \frac{1}{t} = tan θ + cot θ - 1 =$ RHS. Hence proved.

Q325 marksLong AnswerStatistics

The following table shows the daily expenditure of 25 households of a locality.
Daily Expenditure (in ₹): 500 – 750, 750 – 1000, 1000 – 1250, 1250 – 1500, 1500 – 1750
Number of Households: 4, 2x + 1, 12, x, 2
Find the value of x. Hence find the mean daily expenditure.

Show answer & solution

Answer:

x = 2

; mean daily expenditure = ₹ 1055

$4 + (2 x + 1) + 12 + x + 2 = 25 \Rightarrow 3 x + 19 = 25 \Rightarrow x = 2$ .
Frequencies: 4, 5, 12, 2, 2. Class marks: 625, 875, 1125, 1375, 1625.
$\sum f_{i} x_{i} = 2500 + 4375 + 13500 + 2750 + 3250 = 26375$ .
Mean $= \frac{26375}{25} = 1055$ .
Mean daily expenditure = ₹ 1055.

Q335 marksLong AnswerQuadratic Equations

Find two consecutive odd integers, sum of whose squares is 290.

Show answer & solution

Answer: 11 and 13, or −13 and −11

Let the integers be $x$ and $x + 2$ .
$x^{2} + (x + 2)^{2} = 290 \Rightarrow 2 x^{2} + 4 x + 4 = 290 \Rightarrow x^{2} + 2 x - 143 = 0$ .
$(x + 13) (x - 11) = 0$ , so $x = 11$ or $x = - 13$ .
The integers are 11 and 13, or $- 13$ and $- 11$ .
Check: $121 + 169 = 290$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q33 (OR) (OR)5 marksLong AnswerQuadratic Equations

A charity trust decides to build a rectangular hall having an area of 300 $m^{2}$ . The length of the hall is one metre more than twice its width. Find the length and breadth of the hall.

Show answer & solution

Answer: Length = 25 m, breadth = 12 m

Let the breadth be $x$ m; then the length is $(2 x + 1)$ m.
$x (2 x + 1) = 300 \Rightarrow 2 x^{2} + x - 300 = 0$ .
$2 x^{2} + 25 x - 24 x - 300 = 0 \Rightarrow (2 x + 25) (x - 12) = 0$ .
$x = 12$ (breadth cannot be negative).
Breadth = 12 m, length = $2 \times 12 + 1 = 25$ m.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q345 marksLong AnswerTriangles

State and Prove “Basic Proportionality Theorem”.

Show answer & solution

Answer: Proved.

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .
$\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} A D \times E N}{\frac{1}{2} D B \times E N} = \frac{A D}{D B}$ .
$\frac{a r ( △ A D E )}{a r ( △ D E C )} = \frac{\frac{1}{2} A E \times D M}{\frac{1}{2} E C \times D M} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so $a r (△ B D E) = a r (△ D E C)$ .
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2026 Standard 30/1/1, 2026 Standard 30/1/2, 2026 Standard 30/1/3, 2025 Basic 430/2/3, 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2024 Standard 30/3/1

Q34 (OR) (OR)5 marksLong AnswerTriangles

In the given figure, CM and RN are respectively, the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , prove that :
(i) $△ A M C \sim △ P N R$
(ii) $∠ B C M = ∠ QR N$
(iii) $△ B M C \sim △ QN R$

Diagram for CBSE 2025 Class 10 Maths question 34 (OR)

Show answer & solution

Answer: Proved.

Since $△ A B C \sim △ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ , $∠ A = ∠ P$ , $∠ B = ∠ Q$ .
(i) M and N are mid-points of AB and PQ, so $\frac{A M}{P N} = \frac{\frac{1}{2} A B}{\frac{1}{2} P Q} = \frac{A B}{P Q} = \frac{C A}{R P}$ .
With $∠ A = ∠ P$ , $△ A M C \sim △ P N R$ (SAS similarity).
(iii) Similarly $\frac{B M}{QN} = \frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ , so $△ B M C \sim △ QN R$ (SAS similarity).
(ii) From (iii), corresponding angles are equal, so $∠ B C M = ∠ QR N$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q355 marksLong AnswerSurface Areas and Volumes

A necklace is made up of wooden beads. Each bead is in the form of a sphere of diameter 4.2 mm. A cylinder is hollowed out from each bead. If the radius of the cylinder is 1 mm, find the volume of wood left in each bead.

Diagram for CBSE 2025 Class 10 Maths question 35

Show answer & solution

Answer: 25.608

mm^{3}

Radius of sphere $R = 2.1$ mm; the cylinder has radius $r = 1$ mm and, as in the figure, height $h = 4.2$ mm.
Volume of sphere $= \frac{4}{3} π R^{3} = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 = 38.808$ $mm^{3}$ .
Volume of cylinder $= π r^{2} h = \frac{22}{7} \times 1 \times 4.2 = 13.2$ $mm^{3}$ .
Wood left $= 38.808 - 13.2 = 25.608$ $mm^{3}$ .

Q364 marksCase StudyCoordinate Geometry

There is a semicircular park in Aman’s society. He wishes to plant saplings along the boundary of the park. There is a borewell at the centre O of the park along the diameter AB as shown in the figure below.
Based on the above information, answer the following questions :
(i) Find the coordinates of point O. (1)
(ii) Find the radius of the semicircular park. (1)
(iii) (a) One sapling is kept at point C(12, y). Find the coordinates of C. (2)
OR
(b) One sapling is kept at point P along AB so that PA = $\frac{1}{3}$ PB. Find the coordinates of P. (2)

Diagram for CBSE 2025 Class 10 Maths question 36

Show answer & solution

Answer: (i) O(12, 3) (ii) 10 units (iii) (a) C(12, −7) OR (b) P(7, 3)

A(2, 3), B(22, 3).
(i) O is the mid-point of AB: $(\frac{2 + 22}{2}, \frac{3 + 3}{2}) = (12, 3)$ .
(ii) $A B = 22 - 2 = 20$ , so radius $= 10$ units.
(iii) (a) C lies on the semicircle below AB with $O C = 10$ and the same x-coordinate as O, so $y = 3 - 10 = - 7$ . C(12, $- 7$ ).
(iii) (b) $P A = \frac{1}{3} P B$ , so $A P : P B = 1 : 3$ . $P = (\frac{1 \times 22 + 3 \times 2}{4}, \frac{1 \times 3 + 3 \times 3}{4}) = (7, 3)$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q374 marksCase StudyArithmetic Progressions

In a society, a yoga instructor was hired to train the people of the society to live a healthy lifestyle. Yoga sessions were held daily from 5 p.m. to 7 p.m. in the society park. On day one, 5 people joined the yoga session, on day two, 3 more people joined, on day three, another 3 people joined and in this manner every next day, 3 more people kept on joining.
Based on the given information, answer the following questions :
(i) On which day did 59 people join the yoga session ? (1)
(ii) How many people joined the yoga session on the 31st day ? (1)
(iii) (a) The yoga instructor was paid ₹100 for each person attending the yoga session. On which day would he earn ₹5,000 ? (2)
OR
(b) What was the total amount earned by the yoga instructor in 16 days ? (2)

Show answer & solution

Answer: (i) 19th day (ii) 95 (iii) (a) 16th day OR (b) ₹44,000

Number of people on day $n$ forms an AP: 5, 8, 11, ... with $a = 5$ , $d = 3$ , so $a_{n} = 3 n + 2$ .
(i) $5 + (n - 1) \times 3 = 59 \Rightarrow n - 1 = 18 \Rightarrow n = 19$ . On the 19th day.
(ii) $a_{31} = 5 + 30 \times 3 = 95$ people.
(iii) (a) ₹5,000 ÷ ₹100 = 50 people. $5 + (n - 1) \times 3 = 50 \Rightarrow n = 16$ . On the 16th day.
(iii) (b) Total attendance in 16 days $= \frac{16}{2} (2 \times 5 + 15 \times 3) = 8 \times 55 = 440$ . Amount $= 440 \times ₹ 100 = ₹ 44, 000$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

Q384 marksCase StudySome Applications of Trigonometry

Two motorboats A and B are waiting at the opposite banks of a river in order to reach the opposite side. From a point P on the bridge, 20 m above the river, the angles of depression of the boats are $3 0^{\circ}$ and $4 5^{\circ}$ respectively, as shown in the figure given below. Both the boats leave at the same time at the speed of 10 m/s and 5 m/s, respectively
Based on the above information, answer the following questions :
(i) Find the distance travelled by boat A to reach point D in the river, vertically below the point P. (Use $3 = 1.73$ ) (1)
(ii) What is the width of the river ? (1)
(iii) (a) Which boat will reach point D first, and how much earlier, than the other boat ? (2)
OR
(b) What is the distance between the two boats after 3 seconds ? (2)

Diagram for CBSE 2025 Class 10 Maths question 38

Show answer & solution

Answer: (i) 34.6 m (ii) 54.6 m (iii) (a) Boat A, 0.54 s earlier OR (b) 9.6 m

PD = 20 m; angle of depression of A is $3 0^{\circ}$ , of B is $4 5^{\circ}$ , so $∠ P A D = 3 0^{\circ}$ and $∠ P B D = 4 5^{\circ}$ .
(i) $tan 3 0^{\circ} = \frac{P D}{A D} \Rightarrow A D = 203 = 20 \times 1.73 = 34.6$ m.
(ii) $tan 4 5^{\circ} = \frac{P D}{D B} \Rightarrow D B = 20$ m. Width $= A D + D B = 34.6 + 20 = 54.6$ m.
(iii) (a) Time for A $= \frac{34.6}{10} = 3.46$ s; time for B $= \frac{20}{5} = 4$ s. Boat A reaches D first, $4 - 3.46 = 0.54$ s earlier.
(iii) (b) In 3 s, A travels 30 m and B travels 15 m towards each other. Distance between them $= 54.6 - 30 - 15 = 9.6$ m.

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2

← 2025 Basic 430/3/2 2025 Basic 430/4/1 →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →