The CBSE 2025 Class 10 Mathematics exam will be held on 10th March, from 10:30 AM to 1:30 PM. Approximately 26.60 lakh students are expected to appear for the exam across 7,800 centers in India and 28 other countries.
The Mathematics theory paper is of 80 marks, while 20 marks are allocated for the internal assessment. The paper covers topics such as Algebra, Geometry, Trigonometry, Mensuration, Statistics & Probability, and Coordinate Geometry. It includes formula-based, conceptual, and application-based problems.
CBSE Class 10 Mathematics Question Paper 2025 (Set-2) with Answer Key
CBSE Class 10 2025 Mathematics Question Paper with Answer Key | Download PDF | Check Solutions |

Section - A
Question 1:
The system of equations \(x+5=0\) and \(2x-1=0\) has
View Solution
Solving both: \[ x+5=0 \Rightarrow x=-5 \] \[ 2x-1=0 \Rightarrow x=\frac{1}{2} \]
Since the values are different, the system has no solution. Quick Tip: If two linear equations in one variable yield different values, the system is inconsistent.
In a right-angled triangle ABC at A, if \(\sin B = \frac{1}{4}\), then the value of \(\sec B\) is:
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\(\sqrt{0.4}\) is a/an
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Which of the following cannot be the unit digit of \(8^n\), where \(n\) is a natural number?
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Which of the following quadratic equations has real and distinct roots?
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If the zeroes of the polynomial \(ax^2+bx+\frac{2a}{b}\) are reciprocal of each other, then the value of \(b\) is:
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The distance of point \((a, -b)\) from the \(x\)-axis is
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In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ cm, PX=1.5\ cm, BX=4\ cm\). If \(QY=0.75\ cm\), then \(AQ+CY =\)
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Given \(\triangle ABC \sim \triangle PQR\), \(\angle A=30^\circ\), \(\angle Q=90^\circ\). The value of \((\angle R + \angle B)\) is:
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Two coins are tossed simultaneously. The probability of getting at least one head is
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In the adjoining figure, \(PA\) and \(PB\) are tangents to a circle with centre \(O\) such that \(\angle P = 90^\circ\). If \(AB = 3\sqrt{2}\ cm\), then the diameter of the circle is:
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If \(x = \cos 30^\circ - \sin 30^\circ\) and \(y = \tan 60^\circ - \cot 60^\circ\), then
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For a circle with centre O and radius 5 cm, which of the following statements is true?
P : Distance between every pair of parallel tangents is 5 cm.
Q : Distance between every pair of parallel tangents is 10 cm.
R : Distance between every pair of parallel tangents must be between 5 cm and 10 cm.
S : There does not exist a point outside the circle from where length of tangent is 5 cm.
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In the adjoining figure, TS is a tangent to a circle with centre O. The value of \(2x^\circ\) is
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A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is \(10\sqrt{3}\) m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is
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If a cone of greatest possible volume is hollowed out from a solid wooden cylinder, then the ratio of the volume of remaining wood to the volume of cone hollowed out is
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If the mode of some observations is 10 and sum of mean and median is 25, then the mean and median respectively are
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If the maximum number of students has obtained 52 marks out of 80, then
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Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option from the following:
(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.
Question 19:
Assertion (A) : For two prime numbers \(x\) and \(y\) (\(x < y\)), HCF\((x, y) = x\) and LCM\((x, y) = y\).
Reason (R): HCF\((x, y) \leq \) LCM\((x, y)\), where \(x, y\) are any two natural numbers.
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In an experiment of throwing a die,
Assertion (A): Event \(E_1\): getting a number less than 3 and Event \(E_2\): getting a number greater than 3 are complementary events.
Reason (R): If two events E and F are complementary events, then \(P(E) + P(F) = 1\).
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Section - B
This section has 5 very short answer type questions of 2 marks each
Question 21:
In the adjoining figure, if \(\dfrac{AD}{BD} = \dfrac{AE}{EC}\) and \(\angle BDE = \angle CED\), prove that \(\triangle ABC\) is an isosceles triangle.
View Solution
Given: \[ \frac{AD}{BD} = \frac{AE}{EC} \]
and \(\angle BDE = \angle CED\)
By applying Basic Proportionality Theorem (Thales' theorem) and congruence criteria, we can prove \(\triangle ABD \cong \triangle CBE\)
Therefore: \[ AB = AC \]
Hence, \(\triangle ABC\) is isosceles. Quick Tip: Use the Basic Proportionality Theorem and congruence rules for triangle equality.
A bag contains cards numbered from 5 to 100 such that each card bears a different number. A card is drawn at random. Find the probability that the number on the card is:
[(i)] a perfect square
[(ii)] a 2-digit number
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Solve the following pair of equations algebraically: \[ 101x + 102y = 304 \] \[ 102x + 101y = 305 \]
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In a pair of supplementary angles, the greater angle exceeds the smaller by 50°. Express the given situation as a system of linear equations in two variables and hence obtain the measure of each angle.
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If \(a \sec \theta + b \tan \theta = m\) and \(b \sec \theta + a \tan \theta = n\), prove that: \[ a^2 + n^2 = b^2 + m^2 \]
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Use the identity: \[ \sin^2 A + \cos^2 A = 1 \]
to prove that: \[ \tan^2 A + 1 = \sec^2 A \]
Then, find the value of \(\tan A\) when \(\sec A = \frac{5}{3}\), where A is an acute angle.
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Prove that abscissa of a point P which is equidistant from points with coordinates A(7, 1) and B(3, 5) is 2 more than its ordinate.
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Section - C
This section has 6 short answer type questions of 3 marks each.
Question 26(a):
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]
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Factor numerator and denominator: \[ = \frac{\cos \theta (1 - 2 \cos^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} + \cot \theta \]
Use identity: \[ 1 - 2 \cos^2 \theta = - (1 - 2 \sin^2 \theta) \]
Simplify, and sum terms to prove zero. Quick Tip: Factor cubic terms and use \(\sin^2 \theta + \cos^2 \theta = 1\) identity to simplify expressions.
Given that \(\sin \theta + \cos \theta = x\), prove that: \[ \sin^4 \theta + \cos^4 \theta = \frac{2 - (x^2 - 1)^2}{2} \]
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In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If \(\angle OPQ = 15^\circ\) and \(\angle PTQ = \theta\), then find the value of \(\sin 2 \theta\)
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Prove that \(\sqrt{5}\) is an irrational number.
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Let \(p, q, r\) be three distinct prime numbers. Check whether \(p \cdot q \cdot r + q\) is a composite number or not.
Further, give an example for 3 distinct primes \(p, q, r\) such that:
[(i)] \(p \cdot q \cdot r + 1\) is a composite number.
[(ii)] \(p \cdot q \cdot r + 1\) is a prime number.
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Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \]
Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)
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Check whether the following system of equations is consistent or not.
If consistent, solve graphically: \[ x - 2y + 4 = 0, \quad 2x - y - 4 = 0 \]
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If the points \(A(6, 1)\), \(B(p, 2)\), \(C(9, 4)\), and \(D(7, q)\) are the vertices of a parallelogram \(ABCD\), then find the values of \(p\) and \(q\). Hence, check whether \(ABCD\) is a rectangle or not.
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Section - D
This section has 4 long answer questions of 5 marks each
The following data shows the number of family members living in different bungalows of a locality:
If the median number of members is found to be 5, find the values of \(p\) and \(q\).
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There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter.
An entry gate is to be constructed at a point \(P\) on the boundary of the park such that distance of \(P\) from \(A\) is 35 m more than the distance of \(P\) from \(B\).
Find the distance of point \(P\) from \(A\) and \(B\) respectively.
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Find the smallest value of \(p\) for which the quadratic equation \[ x^2 - 2(p+1)x + p^2 = 0 \]
has real roots. Hence, find the roots of the equation so obtained.
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On the day of her examination, Riya sharpened her pencil from both ends as shown below.
The diameter of the cylindrical and conical part of the pencil is 4.2 mm.
If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.
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From one face of a solid cube of side 14 cm, the largest possible cone is carved out.
Find the volume and surface area of the remaining solid. \[ (Use \pi = \frac{22}{7}, \, \sqrt{5} = 2.2) \]
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Section - E
This section has 3 case study based questions of 4 marks each.
Question 36:
In order to organise Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below:
The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane.
Based on given information, answer the following questions, using concept of Arithmetic Progression.
[(i)] What is the length of the 6th lane?
[(ii)] How much longer is the 8th lane than the 4th lane?
[(iii)]
[(a)] While practicing for a race, a student took one round each in the first six lanes. Find the total distance covered by the student.
OR
[(b)] A student took one round each in lanes 4 to 8. Find the total distance covered by the student.
View Solution
Given:
First term of A.P. (length of innermost lane) = \(a = 400\) m
Common difference = \(d = 7.6\) m
(i) Length of 6th lane:
\[ a_6 = a + (6 - 1)d = 400 + 5 \cdot 7.6 = 400 + 38 = \boxed{438 m} \]
(ii) Difference between 8th and 4th lanes:
\[ a_8 - a_4 = \left[a + (8 - 1)d\right] - \left[a + (4 - 1)d\right] = (a + 7d) - (a + 3d) = 4d = 4 \cdot 7.6 = \boxed{30.4 m} \]
(iii) (a) Total distance in 1st to 6th lane:
This forms an A.P. of 6 terms: \[ S_6 = \frac{n}{2} \left[2a + (n - 1)d\right] = \frac{6}{2} \left[2 \cdot 400 + 5 \cdot 7.6\right] = 3 \cdot (800 + 38) = 3 \cdot 838 = \boxed{2514 m} \]
OR
(iii) (b) Total distance in lanes 4 to 8:
This is a sum of 5 terms starting from 4th lane: \[ a_4 = a + 3d = 400 + 22.8 = 422.8 \]
Using \(n = 5\), \(a' = a_4 = 422.8\), \(d = 7.6\): \[ S_5 = \frac{5}{2} \left[2 \cdot 422.8 + (5 - 1) \cdot 7.6\right] = \frac{5}{2} \left[845.6 + 30.4\right] = \frac{5}{2} \cdot 876 = \frac{4380}{2} = \boxed{2190 m} \] Quick Tip: In A.P. problems, use \(a_n = a + (n - 1)d\) for specific terms, and \(S_n = \frac{n}{2}[2a + (n - 1)d]\) for sums.
The Statue of Unity situated in Gujarat is the world’s largest Statue which stands over a 58 m high base.
As part of the project, a student constructed an inclinometer and wishes to find the height of the Statue of Unity using it.
He noted the following observations from two places:
Situation – I:
The angle of elevation of the top of the Statue from Place A which is \(80\sqrt{3}\) m away from the base of the Statue is found to be \(60^\circ\).
Situation – II:
The angle of elevation of the top of the Statue from a Place B which is 40 m above the ground is found to be \(30^\circ\) and the entire height of the Statue including the base is found to be 240 m.
Based on given information, answer the following questions:
[(i)] Represent the Situation – I with the help of a diagram.
[(ii)] Represent the Situation – II with the help of a diagram.
[(iii)] Calculate the height of the Statue excluding the base and also find the height including the base with the help of Situation – I.
OR
[(iv)] Find the horizontal distance of point B (Situation – II) from the Statue and the value of \(\tan \alpha\), where \(\alpha\) is the angle of elevation of the top of base of the Statue from point B.
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Anurag purchased a farmhouse which is in the form of a semicircle of diameter \(70\, m\). He divides it into three parts by taking a point \(P\) on the semicircle in such a way that \(\angle PAB = 30^\circ\) as shown in the following figure, where \(O\) is the centre of the semicircle.
In part I, he planted saplings of Mango tree; in part II, he grew tomatoes; and in part III, he grew oranges. Based on the given information, answer the following questions:
[(i)] What is the measure of \(\angle POA\)?
[(ii)] Find the length of wire needed to fence the entire piece of land.
[(iii)]
[(a)] Find the area of the region in which saplings of Mango tree are planted.
OR
[(b)] Find the length of wire needed to fence the region III.
View Solution
[(i)] Since \(\angle PAB = 30^\circ\), and triangle \(OAB\) is isosceles with \(OA = OB\), then \(\angle POA = \angle POB = 60^\circ\)
[(ii)] Diameter = 70 m \(\Rightarrow\) Radius \(r = 35\, m\)
Length of semicircle = \(\frac{1}{2} \times 2\pi r = \pi r = \frac{22}{7} \times 35 = 110\, m\)
Add lengths of straight sides \(AB\), \(PA\), and \(PB\): use geometry to calculate.
Total fencing length \(\approx AB + PA + PB + semicircle arc\)
[(iii)(a)] Area of Sector \(POA\) =
\[ \frac{\theta}{360^\circ} \cdot \pi r^2 = \frac{60}{360} \cdot \frac{22}{7} \cdot 35^2 = \frac{1}{6} \cdot \frac{22}{7} \cdot 1225 \approx 642.86 \, m^2 \]
[(iii)(b)] To fence Region III (Sector POB), repeat similar arc + straight sides calculation. Quick Tip: Use formulas for arc length: \(L = \frac{\theta}{360^\circ} \cdot 2\pi r\) and area of sector: \(A = \frac{\theta}{360^\circ} \cdot \pi r^2\)
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