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

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

Edited By Vishal kumar | Updated on Mar 18, 2025 10:45 AM IST | #CBSE Class 10th

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.

This Story also Contains
  1. CBSE Class 10 Mathematics Paper: Highlights
  2. CBSE Class 10 Mathematics Question Paper 2025: Download PDF
  3. CBSE Class 10 Mathematics Standard Answer Key 2025: Check Solutions
  4. CBSE 10th Mathematics Standard Set 2 Solutions 2025: QP Code- 30/4/2
  5. How to Calculate Marks Based on CBSE Class 10 Math Answer Key
CBSE Class 10  Maths Question Paper 2025 PDF (10 March), Exam Analysis, Answer Key
CBSE Class 10 Maths Question Paper 2025 PDF (10 March), Exam Analysis, Answer Key

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

Particulars

Details

Conducting Body

CBSE

Subject

Mathematics (Basic and Standerd)

Mode

Offline

Date

March 10, 2025

Duration

3 Hours

Medium

English / Hindi

Question Types

MCQs, Short & Long Answers

Theory Marks

80

Internal Assessment

20

Total Marks

100

Passing Marks

33% (in aggregate)

Difficulty Level

Moderate

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.

Question paper

Download PDF

CBSE Class 10 Mathematics Basic Question Paper 2025

Available Soon

CBSE Class 10 Mathematics Standard Question Paper and Solution 2025

Click Here

Some of the CBSE class 10 mathematics standard set 2 of QP code: 30/4/2 are given below:


1. Which of the following statements is true for a polynomial p(x) of degree 3?
(a) p(x) has at most two distinct zeroes.
(b) p(x) has at least two distinct zeroes.
(c) p(x) has exactly three distinct zeroes.
(d) p(x) has at most three distinct zeroes.


2. A pair of dice is thrown once. The probability that the sum of numbers appearing on top faces is at least 4 is :
(a) 111
(b) 1011
(c) 56
(d) 1112

5. The value of ' a ' for which ax2+x+a=0 has equal and positive roots is :
(a) 2
(b) -2
(c) 12
(d) 12


6. The distance of which of the following points from the origin is less than 5 units?
(a) (3,4)
(b) (2,6)
(c) (3,4)
(d) (1,4)


7. The number of red balls in a bag is 10 more than the number of black balls. If the probability of drawing a red ball at random from this bag is 35, then the total number of balls in the bag is :
(a) 50
(b) 60
(c) 80
(d) 40


8. The value of ' p ' for which the equations px+3y=p3,12x+py=p has infinitely many solutions is :
(a) -6 only
(b) 6 only
(c) ±6
(d) Any real number except ±6


11. Which of the following statements is true?
(a) sin20>sin70
(d) tan20>tan70
(c) cos20>cos70


12. A 30 m long rope is tightly stretched and tied firm the top of pole to the ground. If the rope makes an angle of 60 with the ,(i) 153 m the pole is:
(b) 303 m
(a) 103 m
(c) 15 m are 10 and 13 respectively.


CBSE Class 10 Mathematics Standard Answer Key 2025: Check Solutions

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

Question numberCorrect OptionQuestion numberCorrect Option
1D11C
2D12D
3C13C
4C14C
5D15B
6D16A
7A17A
8C18B
9C19B
10C20D
Background wave

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

(sinA+cscA)2+(cosA+secA)2=x+tan2A+cot2A

Solution:

(sinA+cscA)2+(cosA+secA)2=x+tan2A+cot2A

Expanding each term,

(sinA+cscA)2=sin2A+2sinAcscA+csc2A

Since sinAcscA=1,

(sinA+cscA)2=sin2A+2+csc2A

Using the identity csc2A=1+cot2A.

(sinA+cscA)2=sin2A+cot2A+3(1)

Similarly,

(cosA+secA)2=cos2A+2cosAsecA+sec2A

Since cosAsecA=1,

(cosA+secA)2=cos2A+2+sec2A

Using the identity sec2A=1+tan2A,

(cosA+secA)2=cos2A+tan2A+3(2)

Adding both terms (1) and (2),

(sinA+cscA)2+(cosA+secA)2=(sin2A+cot2A+3)+(cos2A+tan2A+3)

Using sin2A+cos2A=1,

(sinA+cscA)2+(cosA+secA)2=1+cot2A+3+tan2A+3=7+tan2A+cot2A

Equating to the given expression.

x=7

Hence, the value of x is 7.

21. (B) Evaluate the following:

3sin304sin3302sin250+2cos250

Solution:

sin30=12, so

3sin304sin330=3×124×(12)3=3248=3212=1

Using sin2A+cos2A=1,

2sin250+2cos250=2(sin250+cos250)=2×1=2

Hence, 3sin304sin3302sin250+2cos250=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

1130=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)=13716326x+26y=26x+y=1

Solving the two equations:

x+y=3x+y=1

Adding both equations:

2y=4y=2

Substituting y=2 in x+y=3 :

x+2=3x=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+1.5s12s1.5s=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 ABECFB.

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 ABCD and transversal BE creates equal angles)

By AA similarity criterion,

ABECFB

Thus, it is proved that ABECFB.

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)21=8+11=9y=2(3)1(7)21=671=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 x2axb. Obtain a quadratic polynomial whose zeroes are 3α+1 and 3β+1.

Solution:

Given that α and β are the zeroes of the polynomial x2axb, 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+(3a9b+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 q0.

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 x6y6x8y9, 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)=42.53x=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

y3x2

By the perpendicularity condition,

(y+3x+2)×(y3x2)=1

Using the identity (a+b)(ab)=a2b2.

y29x24=1

Multiplying both sides by x24,

y29=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 cosA+sinA1cosAsinA+1=cosecAcotA

Solution:

We need to prove the identity:

cosA+sinA1cosAsinA+1=cscAcotA

Rewriting the right-hand side:

cscAcotA=1sinAcosAsinA=1cosAsinA

This means we need to prove:

cosA+sinA1cosAsinA+1=1cosAsinA

Cross multiplying:

(cosA+sinA1)sinA=(1cosA)(cosAsinA+1)

Expanding both sides:

cosAsinA+sin2AsinA=(1cosA)(cosAsinA+1)

Expanding (1cosA)(cosAsinA+1) :

1(cosAsinA+1)cosA(cosAsinA+1)=(cosAsinA+1)(cos2AcosAsinA+cosA)=cosAsinA+1cos2A+cosAsinAcosA=1cos2AsinA+cosAsinA

Since 1cos2A=sin2A, we get:

sin2AsinA+cosAsinA

Both sides simplify to:

cosAsinA+sin2AsinA=sin2AsinA+cosAsinA

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

cosA+sinA1cosAsinA+1=cscAcotA

Hence Proved

31. (B) If cotθ+cosθ=p and cotθcosθ=q. prove that p2q2=4pq

Solution:

We are given the following equations:

cotθ+cosθ=pcotθcosθ=q

We are tasked with proving that:

p2q2=4pq

Subtract the second equation from the first:

(cotθ+cosθ)(cotθcosθ)=pq

Simplifying the left-hand side:

2cosθ=pq

This gives:

cosθ=pq2

Now, add the first and second equations:

(cotθ+cosθ)+(cotθcosθ)=p+q

Simplifying the left-hand side:

2cotθ=p+q

This gives:

cotθ=p+q2

Next, recall the identity cotθ=cosθsinθ. Substituting the expressions for cotθ and cosθ into this equation:

p+q2=pq2sinθ

Simplifying:

p+q2=pq2sinθ

Multiplying both sides by 2 :

p+q=pqsinθ

Solving for sinθ :

sinθ=pqp+q

Now, using the identity cot2θ+1=csc2θ, we get

(p+q2)2+1=1sin2θ

Substitute the expression for sinθ :

(p+q2)2+1=(p+q)2(pq)2

Now, compute p2q2 :

p2q2=(p+q)(pq)

Thus, the required identity is proven:

p2q2=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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Frequently Asked Questions (FAQs)

1. When is the CBSE Class 10 Maths Exam 2025?

The CBSE Class 10 Maths Exam was scheduled for March 10, 2025.

2. What is the exam duration?

The exam duration was 3 hours long.

3. What is the passing mark for the exam?

Students need to score at least 33% in aggregate to pass.

4. What types of questions will be asked?

The paper includes MCQs, short answer, and long answer questions.

5. What is the difficulty level of the exam?

The exam is considered moderate and focuses on concept-based questions with application-based problem-solving.

6. Is maths standard paper tough 2025 class 10

The difficulty level of maths standard paper 2025 was moderate.

7. which set is hardest in board exam class 10 2025

Set 3 was overall difficult as compared to other sets of CBSE 10th maths paper 2025.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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