CBSE Class 10 Maths Question Paper 2025 PDF (10 March), Exam Analysis, Answer Key

The Central Board of Secondary Education (CBSE) is conducting the CBSE board exam from February 15 to March 18, 2025. According to the CBSE class 10 datasheet 2025, CBSE 10th basic math and slandered math were scheduled on March 10, 2025, from 10:30 am to 1:30 pm which is concluded.CBSE 10th mathematics paper was of moderate difficulty this year and slightly tougher than the last year. Students who have gone through the NCERT and practised the CBSE class 10 maths sample paper can solve the paper easily.

Students can download the CBSE 10th math question paper PDF 2025 from this page along with the solution and answer key of all the questions. CBSE class 10 math answer key will help students calculate the final marks and grades by calculating the total number.

CBSE Class 10 Mathematics Paper: Highlights

CBSE Class 10 Mathematics Question Paper 2025: Download PDF

Students can download the class 10th maths question paper PDF from here and practice to become familiar with the exam trends, types of questions, and marking schemes. The CBSE mathematics question paper PDF link will be activated soon here.

Students who have appeared in the CBSE 10th maths exam must match their answer with the CBSE class 10 maths answer key 2025 which is given below. Check all sets of answer keys and calculate the final score and grade based on the final marks.

CBSE 10th mathematics standard set 2 answer key 2025: QP Code- 30/4/2

CBSE 10th Mathematics Standard Set 2 Solutions 2025: QP Code- 30/4/2

Check the detailed CBSE 10th mathematics standard set 2 solutions of QP code 30/4/2, starting from section B.

SECTION - B

This section consists of 5 questions of 2 marks each.

21. A) Find the value of $x$ for which

$
(\sin A+\csc A)^2+(\cos A+\sec A)^2=x+\tan ^2 A+\cot ^2 A
$

Solution:

$
(\sin A+\csc A)^2+(\cos A+\sec A)^2=x+\tan ^2 A+\cot ^2 A
$

Expanding each term,

$
(\sin A+\csc A)^2=\sin ^2 A+2 \sin A \csc A+\csc ^2 A
$

Since $\sin A \csc A=1$,

$
(\sin A+\csc A)^2=\sin ^2 A+2+\csc ^2 A
$

Using the identity $\csc ^2 A=1+\cot ^2 A$.

$
(\sin A+\csc A)^2=\sin ^2 A+\cot ^2 A+3 \quad -----(1)
$

Similarly,

$
(\cos A+\sec A)^2=\cos ^2 A+2 \cos A \sec A+\sec ^2 A
$

Since $\cos A \sec A=1$,

$
(\cos A+\sec A)^2=\cos ^2 A+2+\sec ^2 A
$

Using the identity $\sec ^2 A=1+\tan ^2 A$,

$
(\cos A+\sec A)^2=\cos ^2 A+\tan ^2 A+3 \quad -----(2)
$

Adding both terms (1) and (2),

$
(\sin A+\csc A)^2+(\cos A+\sec A)^2 = \left(\sin ^2 A+\cot ^2 A+3\right)+\left(\cos ^2 A+\tan ^2 A+3\right)
$

Using $\sin ^2 A+\cos ^2 A=1$,

$
(\sin A+\csc A)^2+(\cos A+\sec A)^2 = 1+\cot ^2 A+3+\tan ^2 A+3=7+\tan ^2 A+\cot ^2 A
$

Equating to the given expression.

$
x=7
$

Hence, the value of x is 7.

21. (B) Evaluate the following:

$
\frac{3 \sin 30^{\circ}-4 \sin ^3 30^{\circ}}{2 \sin ^2 50^{\circ}+2 \cos ^2 50^{\circ}}
$

Solution:

$\sin 30^{\circ}=\frac{1}{2}$, so

$
3 \sin 30^{\circ}-4 \sin ^3 30^{\circ}=3 \times \frac{1}{2}-4 \times\left(\frac{1}{2}\right)^3=\frac{3}{2}-\frac{4}{8}=\frac{3}{2}-\frac{1}{2}=1
$

Using $\sin ^2 A+\cos ^2 A=1$,

$
2 \sin ^2 50^{\circ}+2 \cos ^2 50^{\circ}=2\left(\sin ^2 50^{\circ}+\cos ^2 50^{\circ}\right)=2 \times 1=2
$

Hence, $
\frac{3 \sin 30^{\circ}-4 \sin ^3 30^{\circ}}{2 \sin ^2 50^{\circ}+2 \cos ^2 50^{\circ}} = \frac{1}{2}
$

22. Saima and Aryaa were born in the month of June in the year 2012. Find the probability that :
(i) they have different dates of birth.
(ii) they have the same date of birth.

Solution:

June has 30 days, so each child can be born on any of these 30 days.
The total number of ways to assign birthdates to both Saima and Aryaa is $30 \times 30=900$.
The number of ways they can have the same birthdate is 30 (choosing one day for both).
The probability that they have the same birthdate is

$
\frac{30}{900}=\frac{1}{30}
$

The probability that they have different birthdates is

$
1-\frac{1}{30}=\frac{29}{30}
$

Thus,
(i) Probability that they have different birthdates $=\frac{29}{30}$
(ii) Probability that they have the same birthdate $=\frac{1}{30}$

23. Solve the following system of equations algebraically :

$
\begin{aligned}
& 37 x+63 y=137 \\
& 63 x+37 y=163
\end{aligned}
$

Solution:

Adding both equations:

$
\begin{gathered}
(37 x+63 y)+(63 x+37 y)=137+163 \\
100 x+100 y=300 \\
x+y=3
\end{gathered}
$

Subtracting the second equation from the first:

$
\begin{gathered}
(37 x+63 y)-(63 x+37 y)=137-163 \\
-26 x+26 y=-26 \\
-x+y=1
\end{gathered}
$

Solving the two equations:

$
\begin{gathered}
x+y=3 \\
-x+y=1
\end{gathered}
$

Adding both equations:

$
2 y=4 \Rightarrow y=2
$

Substituting $y=2$ in $x+y=3$ :

$
x+2=3 \Rightarrow x=1
$

Thus, the solution is $x=1, y=2$.

24. (A) A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at the speed of $2.5 \mathrm{~m} / \mathrm{sec}$. Find the length of his shadow after 3 seconds.

Solution:

Let the height of the lamp post be $h_1=12 \mathrm{~m}$ and the height of the boy be $h_2=1.5 \mathrm{~m}$.

Let the distance of the boy from the base of the lamp post after 3 seconds be $d$, and let the length of his shadow be $s$.

Since the boy walks at $2.5 \mathrm{~m} / \mathrm{sec}$, after 3 seconds:

$
d=3 \times 2.5=7.5 \mathrm{~m}
$

Using the property of similar triangles,

$
\frac{h_1}{d+s}=\frac{h_2}{s}
$

Substituting values:

$
\frac{12}{7.5+s}=\frac{1.5}{s}
$

Cross multiplying:

$
\begin{gathered}
12 s=1.5(7.5+s) \\
12 s=11.25+1.5 s \\
12 s-1.5 s=11.25 \\
10.5 s=11.25 \\
s=\frac{11.25}{10.5}=1.07 \mathrm{~m}
\end{gathered}
$

Thus, the length of the shadow after 3 seconds is 1.07 m.

24. (B) In parallelogram $A B C D$, side $A D$ is produced to a point $E$ and $B E$ intersects $C D$ at $F$. Prove that $\triangle A B E \sim \triangle C F B$.

Solution:

In parallelogram $A B C D$, opposite sides are equal:

$
A B=C D \quad \text { and } \quad A D=B C
$

Since $A D$ is extended to $E$, and $B E$ intersects $C D$ at $F$, we need to show similarity between $\triangle A B E$ and $\triangle C F B$.

In $\triangle A B E$ and $\triangle C F B$ :
1. Angle $A B E=C F B$ (Vertically opposite angles)
2. Angle $A E B=B F C$ (Corresponding angles, as $A B \| C D$ and transversal $B E$ creates equal angles)

By AA similarity criterion,

$
\triangle A B E \sim \triangle C F B
$

Thus, it is proved that $\triangle A B E \sim \triangle C F B$.

25. Find the coordinates of the point $C$ which lies on the line $AB$ produced such that $\mathrm{AC}=2 \mathrm{BC}$, where coordinates of points A and B are $(-1,7)$ and $(4,-3)$ respectively.

Solution:

Using the section formula, the coordinates of a point dividing a line segment $A B$ in the ratio $m: n$ are given by:

$
\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)
$

Here, point $C$ lies on the line $A B$ produced such that $A C=2 B C$, meaning $C$ divides $A B$ externally in the ratio $2: 1$.

Given:

$A(-1,7), B(4,-3)$, and ratio $2: 1$.
Applying the external section formula:

$
\begin{gathered}
x=\frac{2(4)-1(-1)}{2-1}=\frac{8+1}{1}=9 \\
y=\frac{2(-3)-1(7)}{2-1}=\frac{-6-7}{1}=-13
\end{gathered}
$

Thus, the coordinates of $C$ are $(9,-13)$.

SECTION - C

This section consists of 6 questions of 3 marks each.

26. $\alpha$ and $\beta$ are zeroes of a quadratic polynomial $x^2-a x-b$. Obtain a quadratic polynomial whose zeroes are $3 \alpha+1$ and $3 \beta+1$.

Solution:

Given that $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2-a x-b$, we have:

$
\begin{gathered}
\alpha+\beta=a \\
\alpha \beta=-b
\end{gathered}
$

We need to find a quadratic polynomial whose zeroes are $3 \alpha+1$ and $3 \beta+1$.
Sum of new zeroes:

$
(3 \alpha+1)+(3 \beta+1)=3(\alpha+\beta)+2=3 a+2
$

Product of new zeroes:

$
\begin{gathered}
(3 \alpha+1)(3 \beta+1)=9 \alpha \beta+3 \alpha+3 \beta+1 \\
=9(-b)+3(\alpha+\beta)+1 \\
=-9 b+3 a+1
\end{gathered}
$

The required quadratic polynomial is:

$
\begin{aligned}
& x^2-(3 a+2) x+(-9 b+3 a+1) \\
& x^2-(3 a+2) x+(3 a-9 b+1)
\end{aligned}
$

27. Rectangle $A B C D$ circumscribes a circle of radius 10 cm. Prove that $A B C D$ is a square. Hence, find the perimeter of $A B C D$.

Solution:

Since the rectangle $A B C D$ circumscribes a circle, it means the circle is inscribed in the rectangle.

A necessary condition for a rectangle to have an inscribed circle is that the sum of its opposite sides must be equal, i.e.,

$
A B+C D=A D+B C
$

Since in a rectangle, opposite sides are equal, this simplifies to:

$
\begin{aligned}
2 A B & =2 A D \\
A B & =A D
\end{aligned}
$

This means the length and breadth of the rectangle are equal, which implies that $A B C D$ is a square.
Now, let the side length of the square be $s$.
Since the circle is inscribed, its diameter is equal to the side length of the square. The given radius of the circle is 10 cm, so the diameter is:

$
2 \times 10=20 \mathrm{~cm}
$

Thus, the side length of the square is:

$
s=20 \mathrm{~cm}
$

The perimeter of square $A B C D$ is:

$
4 \times s=4 \times 20=80 \mathrm{~cm}
$

Thus, $A B C D$ is a square, and its perimeter is 80 cm.

28. (A) Prove that $\sqrt{ } 2$ is an irrational number.

Solution:

Let us assume, for contradiction, that $\sqrt{2}$ is a rational number. This means it can be expressed in the form:

$
\sqrt{2}=\frac{p}{q}
$

where $p$ and $q$ are integers with no common factors other than 1 (i.e., $\frac{p}{q}$ is in its simplest form) and $q \neq 0$.

Squaring both sides:

$
2=\frac{p^2}{q^2}
$

Multiplying both sides by $q^2$ :

$
p^2=2 q^2
$

This shows that $p^2$ is divisible by 2, which implies that $p$ itself is divisible by 2 (since the square of an odd number is always odd).

Let $p=2 k$ for some integer $k$. Substituting this in:

$
\begin{gathered}
(2 k)^2=2 q^2 \\
4 k^2=2 q^2
\end{gathered}
$

Dividing both sides by 2 :

$
2 k^2=q^2
$

This shows that $q^2$ is also divisible by 2, which implies that $q$ is also divisible by 2.
Since both $p$ and $q$ are divisible by 2, they have a common factor of 2, which contradicts our assumption that $\frac{p}{q}$ is in its simplest form.
Thus, our initial assumption that $\sqrt{2}$ is rational is incorrect.
Therefore, $\sqrt{2}$ is an irrational number.

28. (B) Let $x$ and $y$ be two distinct prime numbers and $p=x^2 y^3, q=xy^4, r=x^5 y^2$. Find the HCF $(p, q, r)$ and $\operatorname{LCM}(p, q, r)$. Further check if $H C F(p, q, r) \times L C M(p, q, r)=p \times q \times r$ or not.

Solution:

The HCF is obtained by taking the lowest power of each prime factor in $p, q, r$. The prime factor powers are:
- $p=x^2 y^3$
- $q=x y^4$
- $r=x^5 y^2$

The lowest powers of $x$ and $y$ are $\min (2,1,5)=1$ and $\min (3,4,2)=2$.

$
\operatorname{HCF}(p, q, r)=x^1 y^2=x y^2
$

The LCM is obtained by taking the highest power of each prime factor. The highest powers are $\max (2,1,5)=5$ and $\max (3,4,2)=4$.

$
\operatorname{LCM}(p, q, r)=x^5 y^4
$

To verify the condition:

$
\begin{aligned}
& \operatorname{HCF}(p, q, r) \times \operatorname{LCM}(p, q, r)=x y^2 \times x^5 y^4=x^{1+5} y^{2+4}=x^6 y^6 \\
& p \times q \times r=\left(x^2 y^3\right) \times\left(x y^4\right) \times\left(x^5 y^2\right)=x^{2+1+5} y^{3+4+2}=x^8 y^9
\end{aligned}
$

Since $x^6 y^6 \neq x^8 y^9$, the given condition is not satisfied.

29. The two angles of a given triangle are in the ratio $2: 3$. The third angle is greater than the largest angle by $10^{\circ}$. Find all the angles of the triangle algebraically as a system of linear equations.

Solution:

Let the two angles of the triangle be $2 x$ and $3 x$. The third angle is given as $10^{\circ}$ more than the largest angle, so it is $3 x+10^{\circ}$.

Since the sum of the angles in a triangle is always $180^{\circ}$, we get the equation:

$
\begin{gathered}
2 x+3 x+(3 x+10)=180 \\
8 x+10=180 \\
8 x=170 \\
x=21.25
\end{gathered}
$

Substituting $x$ into the expressions for the angles:

$
\begin{gathered}
\qquad 2 x=2(21.25)=42.5^{\circ} \\
\qquad 3 x=3(21.25)=63.75^{\circ} \\
\text { Third angle }=3 x+10=63.75+10=73.75^{\circ}
\end{gathered}
$

Thus, the three angles of the triangle are $42.5^{\circ}, 63.75^{\circ}, 73.75^{\circ}$.

30. $\mathrm{P}(x, y), \mathrm{Q}(-2,-3)$ and $\mathrm{R}(2,3)$ are the vertices of a right triangle $P Q R$ right angled at P . Find the relationship between $x$ and $y$. Hence, find all possible values of $x$ for which $y=2$.

Solution:

Since $\triangle P Q R$ is right-angled at $P$, the product of the slopes of $P Q$ and $P R$ should be -1.
The slope of $P Q$ is

$
\frac{y+3}{x+2}
$

The slope of $P R$ is

$
\frac{y-3}{x-2}
$

By the perpendicularity condition,

$
\left(\frac{y+3}{x+2}\right) \times\left(\frac{y-3}{x-2}\right)=-1
$

Using the identity $(a+b)(a-b)=a^2-b^2$.

$
\frac{y^2-9}{x^2-4}=-1
$

Multiplying both sides by $x^2-4$,

$
y^2-9=-x^2+4
$

Rearranging,

$
x^2+y^2=13
$

This is the required relationship between $x$ and $y$.
For $y=2$,

$
\begin{gathered}
x^2+2^2=13 \\
x^2+4=13 \\
x^2=9 \\
x= \pm 3
\end{gathered}
$

Thus, the possible values of $x$ for $y=2$ are $x=3$ and $x=-3$.

31. (A) Prove that $\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\operatorname{cosec} A-\cot A$

Solution:

We need to prove the identity:

$
\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\csc A-\cot A
$

Rewriting the right-hand side:

$
\csc A-\cot A=\frac{1}{\sin A}-\frac{\cos A}{\sin A}=\frac{1-\cos A}{\sin A}
$

This means we need to prove:

$
\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\frac{1-\cos A}{\sin A}
$

Cross multiplying:

$
(\cos A+\sin A-1) \sin A=(1-\cos A)(\cos A-\sin A+1)
$

Expanding both sides:

$
\cos A \sin A+\sin ^2 A-\sin A=(1-\cos A)(\cos A-\sin A+1)
$

Expanding $(1-\cos A)(\cos A-\sin A+1)$ :

$
\begin{gathered}
1(\cos A-\sin A+1)-\cos A(\cos A-\sin A+1) \\
=(\cos A-\sin A+1)-\left(\cos ^2 A-\cos A \sin A+\cos A\right) \\
=\cos A-\sin A+1-\cos ^2 A+\cos A \sin A-\cos A \\
=1-\cos ^2 A-\sin A+\cos A \sin A
\end{gathered}
$

Since $1-\cos ^2 A=\sin ^2 A$, we get:

$
\sin ^2 A-\sin A+\cos A \sin A
$

Both sides simplify to:

$
\cos A \sin A+\sin ^2 A-\sin A=\sin ^2 A-\sin A+\cos A \sin A
$

Since both sides are equal, the given equation is proven:

$
\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\csc A-\cot A
$

Hence Proved

31. (B) If $\cot \theta+\cos \theta=p$ and $\cot \theta-\cos \theta=q$. prove that $p^2-q^2=4 \sqrt{p q}$

Solution:

We are given the following equations:

$
\begin{aligned}
& \cot \theta+\cos \theta=p \\
& \cot \theta-\cos \theta=q
\end{aligned}
$

We are tasked with proving that:

$
p^2-q^2=4 \sqrt{p q}
$

Subtract the second equation from the first:

$
(\cot \theta+\cos \theta)-(\cot \theta-\cos \theta)=p-q
$

Simplifying the left-hand side:

$
2 \cos \theta=p-q
$

This gives:

$
\cos \theta=\frac{p-q}{2}
$

Now, add the first and second equations:

$
(\cot \theta+\cos \theta)+(\cot \theta-\cos \theta)=p+q
$

Simplifying the left-hand side:

$
2 \cot \theta=p+q
$

This gives:

$
\cot \theta=\frac{p+q}{2}
$

Next, recall the identity $\cot \theta=\frac{\cos \theta}{\sin \theta}$. Substituting the expressions for $\cot \theta$ and $\cos \theta$ into this equation:

$
\frac{p+q}{2}=\frac{\frac{p-q}{2}}{\sin \theta}
$

Simplifying:

$
\frac{p+q}{2}=\frac{p-q}{2 \sin \theta}
$

Multiplying both sides by 2 :

$
p+q=\frac{p-q}{\sin \theta}
$

Solving for $\sin \theta$ :

$
\sin \theta=\frac{p-q}{p+q}
$

Now, using the identity $\cot ^2 \theta+1=\csc ^2 \theta$, we get

$
\left(\frac{p+q}{2}\right)^2+1=\frac{1}{\sin ^2 \theta}
$

Substitute the expression for $\sin \theta$ :

$
\left(\frac{p+q}{2}\right)^2+1=\frac{(p+q)^2}{(p-q)^2}
$

Now, compute $p^2-q^2$ :

$
p^2-q^2=(p+q)(p-q)
$

Thus, the required identity is proven:

$
p^2-q^2=4 \sqrt{p q}
$

Hence Proved.

How to Calculate Marks Based on CBSE Class 10 Math Answer Key

To calculate the total marks based on the answer key, follow the steps given below:

• First, match the question paper set and QP code.
• If your marked option matches with the given correct answer then add +1, if not match means incorrect and count zero.
• There is no negative marking for wrong answers. so simply add the total number of questions which are correct.

Suggested Articles:

• CBSE Class 12 English Answer Key 2025
• CBSE Class 10 Maths Answer Key 2025
• CBSE Class 10 Maths Question Paper 2025
• CBSE Class 12 Hindi Answer key 2025
• CBSE Class 10 Maths Paper Analysis 2025