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

CBSE Class 10 Mathematics Paper 2025: Highlights

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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+tan2⁡A+cot2⁡A

Solution:

(sin⁡A+csc⁡A)2+(cos⁡A+sec⁡A)2=x+tan2⁡A+cot2⁡A

Expanding each term,

(sin⁡A+csc⁡A)2=sin2⁡A+2sin⁡Acsc⁡A+csc2⁡A

Since sin⁡Acsc⁡A=1,

(sin⁡A+csc⁡A)2=sin2⁡A+2+csc2⁡A

Using the identity csc2⁡A=1+cot2⁡A.

(sin⁡A+csc⁡A)2=sin2⁡A+cot2⁡A+3−−−−−(1)

Similarly,

(cos⁡A+sec⁡A)2=cos2⁡A+2cos⁡Asec⁡A+sec2⁡A

Since cos⁡Asec⁡A=1,

(cos⁡A+sec⁡A)2=cos2⁡A+2+sec2⁡A

Using the identity sec2⁡A=1+tan2⁡A,

(cos⁡A+sec⁡A)2=cos2⁡A+tan2⁡A+3−−−−−(2)

Adding both terms (1) and (2),

(sin⁡A+csc⁡A)2+(cos⁡A+sec⁡A)2=(sin2⁡A+cot2⁡A+3)+(cos2⁡A+tan2⁡A+3)

Using sin2⁡A+cos2⁡A=1,

(sin⁡A+csc⁡A)2+(cos⁡A+sec⁡A)2=1+cot2⁡A+3+tan2⁡A+3=7+tan2⁡A+cot2⁡A

Equating to the given expression.

x=7

Hence, the value of x is 7.

21. (B) Evaluate the following:

3sin⁡30∘−4sin3⁡30∘2sin2⁡50∘+2cos2⁡50∘

Solution:

sin⁡30∘=12, so

3sin⁡30∘−4sin3⁡30∘=3×12−4×(12)3=32−48=32−12=1

Using sin2⁡A+cos2⁡A=1,

2sin2⁡50∘+2cos2⁡50∘=2(sin2⁡50∘+cos2⁡50∘)=2×1=2

Hence, 3sin⁡30∘−4sin3⁡30∘2sin2⁡50∘+2cos2⁡50∘=12

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 30×30=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

30900=130

The probability that they have different birthdates is

1−130=2930

Thus,
(i) Probability that they have different birthdates =2930
(ii) Probability that they have the same birthdate =130

23. Solve the following system of equations algebraically :

37x+63y=13763x+37y=163

Solution:

Adding both equations:

(37x+63y)+(63x+37y)=137+163100x+100y=300x+y=3

Subtracting the second equation from the first:

(37x+63y)−(63x+37y)=137−163−26x+26y=−26−x+y=1

Solving the two equations:

x+y=3−x+y=1

Adding both equations:

2y=4⇒y=2

Substituting y=2 in x+y=3 :

x+2=3⇒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 m/sec. Find the length of his shadow after 3 seconds.

Solution:

Let the height of the lamp post be h1=12 m and the height of the boy be h2=1.5 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 m/sec, after 3 seconds:

d=3×2.5=7.5 m

Using the property of similar triangles,

h1d+s=h2s

Substituting values:

127.5+s=1.5s

Cross multiplying:

12s=1.5(7.5+s)12s=11.25+1.5s12s−1.5s=11.2510.5s=11.25s=11.2510.5=1.07 m

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

24. (B) In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that △ABE∼△CFB.

Solution:

In parallelogram ABCD, opposite sides are equal:

AB=CD and AD=BC

Since AD is extended to E, and BE intersects CD at F, we need to show similarity between △ABE and △CFB.

In △ABE and △CFB :
1. Angle ABE=CFB (Vertically opposite angles)
2. Angle AEB=BFC (Corresponding angles, as AB‖CD and transversal BE creates equal angles)

By AA similarity criterion,

△ABE∼△CFB

Thus, it is proved that △ABE∼△CFB.

25. Find the coordinates of the point C which lies on the line AB produced such that AC=2BC, 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 AB in the ratio m:n are given by:

(mx2+nx1m+n,my2+ny1m+n)

Here, point C lies on the line AB produced such that AC=2BC, meaning C divides AB externally in the ratio 2:1.

Given:

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

x=2(4)−1(−1)2−1=8+11=9y=2(−3)−1(7)2−1=−6−71=−13

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

SECTION - C

This section consists of 6 questions of 3 marks each.

26. α and β are zeroes of a quadratic polynomial x2−ax−b. Obtain a quadratic polynomial whose zeroes are 3α+1 and 3β+1.

Solution:

Given that α and β are the zeroes of the polynomial x2−ax−b, we have:

α+β=aαβ=−b

We need to find a quadratic polynomial whose zeroes are 3α+1 and 3β+1.
Sum of new zeroes:

(3α+1)+(3β+1)=3(α+β)+2=3a+2

Product of new zeroes:

(3α+1)(3β+1)=9αβ+3α+3β+1=9(−b)+3(α+β)+1=−9b+3a+1

The required quadratic polynomial is:

x2−(3a+2)x+(−9b+3a+1)x2−(3a+2)x+(3a−9b+1)

27. Rectangle ABCD circumscribes a circle of radius 10 cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.

Solution:

Since the rectangle ABCD 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.,

AB+CD=AD+BC

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

2AB=2ADAB=AD

This means the length and breadth of the rectangle are equal, which implies that ABCD 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:

2×10=20 cm

Thus, the side length of the square is:

s=20 cm

The perimeter of square ABCD is:

4×s=4×20=80 cm

Thus, ABCD is a square, and its perimeter is 80 cm.

28. (A) Prove that 2 is an irrational number.

Solution:

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

2=pq

where p and q are integers with no common factors other than 1 (i.e., pq is in its simplest form) and q≠0.

Squaring both sides:

2=p2q2

Multiplying both sides by q2 :

p2=2q2

This shows that p2 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=2k for some integer k. Substituting this in:

(2k)2=2q24k2=2q2

Dividing both sides by 2 :

2k2=q2

This shows that q2 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 pq is in its simplest form.
Thus, our initial assumption that 2 is rational is incorrect.
Therefore, 2 is an irrational number.

28. (B) Let x and y be two distinct prime numbers and p=x2y3,q=xy4,r=x5y2. Find the HCF (p,q,r) and LCM(p,q,r). Further check if HCF(p,q,r)×LCM(p,q,r)=p×q×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=x2y3
- q=xy4
- r=x5y2

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

HCF(p,q,r)=x1y2=xy2

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.

LCM(p,q,r)=x5y4

To verify the condition:

HCF(p,q,r)×LCM(p,q,r)=xy2×x5y4=x1+5y2+4=x6y6p×q×r=(x2y3)×(xy4)×(x5y2)=x2+1+5y3+4+2=x8y9

Since x6y6≠x8y9, 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∘. Find all the angles of the triangle algebraically as a system of linear equations.

Solution:

Let the two angles of the triangle be 2x and 3x. The third angle is given as 10∘ more than the largest angle, so it is 3x+10∘.

Since the sum of the angles in a triangle is always 180∘, we get the equation:

2x+3x+(3x+10)=1808x+10=1808x=170x=21.25

Substituting x into the expressions for the angles:

2x=2(21.25)=42.5∘3x=3(21.25)=63.75∘ Third angle =3x+10=63.75+10=73.75∘

Thus, the three angles of the triangle are 42.5∘,63.75∘,73.75∘.

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

Solution:

Since △PQR is right-angled at P, the product of the slopes of PQ and PR should be -1.
The slope of PQ is

y+3x+2

The slope of PR is

y−3x−2

By the perpendicularity condition,

(y+3x+2)×(y−3x−2)=−1

Using the identity (a+b)(a−b)=a2−b2.

y2−9x2−4=−1

Multiplying both sides by x2−4,

y2−9=−x2+4

Rearranging,

x2+y2=13

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

x2+22=13x2+4=13x2=9x=±3

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

31. (A) Prove that cos⁡A+sin⁡A−1cos⁡A−sin⁡A+1=cosecA−cot⁡A

Solution:

We need to prove the identity:

cos⁡A+sin⁡A−1cos⁡A−sin⁡A+1=csc⁡A−cot⁡A

Rewriting the right-hand side:

csc⁡A−cot⁡A=1sin⁡A−cos⁡Asin⁡A=1−cos⁡Asin⁡A

This means we need to prove:

cos⁡A+sin⁡A−1cos⁡A−sin⁡A+1=1−cos⁡Asin⁡A

Cross multiplying:

(cos⁡A+sin⁡A−1)sin⁡A=(1−cos⁡A)(cos⁡A−sin⁡A+1)

Expanding both sides:

cos⁡Asin⁡A+sin2⁡A−sin⁡A=(1−cos⁡A)(cos⁡A−sin⁡A+1)

Expanding (1−cos⁡A)(cos⁡A−sin⁡A+1) :

1(cos⁡A−sin⁡A+1)−cos⁡A(cos⁡A−sin⁡A+1)=(cos⁡A−sin⁡A+1)−(cos2⁡A−cos⁡Asin⁡A+cos⁡A)=cos⁡A−sin⁡A+1−cos2⁡A+cos⁡Asin⁡A−cos⁡A=1−cos2⁡A−sin⁡A+cos⁡Asin⁡A

Since 1−cos2⁡A=sin2⁡A, we get:

sin2⁡A−sin⁡A+cos⁡Asin⁡A

Both sides simplify to:

cos⁡Asin⁡A+sin2⁡A−sin⁡A=sin2⁡A−sin⁡A+cos⁡Asin⁡A

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

cos⁡A+sin⁡A−1cos⁡A−sin⁡A+1=csc⁡A−cot⁡A

Hence Proved

31. (B) If cot⁡θ+cos⁡θ=p and cot⁡θ−cos⁡θ=q. prove that p2−q2=4pq

Solution:

We are given the following equations:

cot⁡θ+cos⁡θ=pcot⁡θ−cos⁡θ=q

We are tasked with proving that:

p2−q2=4pq

Subtract the second equation from the first:

(cot⁡θ+cos⁡θ)−(cot⁡θ−cos⁡θ)=p−q

Simplifying the left-hand side:

2cos⁡θ=p−q

This gives:

cos⁡θ=p−q2

Now, add the first and second equations:

(cot⁡θ+cos⁡θ)+(cot⁡θ−cos⁡θ)=p+q

Simplifying the left-hand side:

2cot⁡θ=p+q

This gives:

cot⁡θ=p+q2

Next, recall the identity cot⁡θ=cos⁡θsin⁡θ. Substituting the expressions for cot⁡θ and cos⁡θ into this equation:

p+q2=p−q2sin⁡θ

Simplifying:

p+q2=p−q2sin⁡θ

Multiplying both sides by 2 :

p+q=p−qsin⁡θ

Solving for sin⁡θ :

sin⁡θ=p−qp+q

Now, using the identity cot2⁡θ+1=csc2⁡θ, we get

(p+q2)2+1=1sin2⁡θ

Substitute the expression for sin⁡θ :

(p+q2)2+1=(p+q)2(p−q)2

Now, compute p2−q2 :

p2−q2=(p+q)(p−q)

Thus, the required identity is proven:

p2−q2=4pq

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.

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