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 – 1 (30/6/1) with Answer Key
| CBSE Class 10 2025 Mathematics Question Paper with Answer Key | Download Now | Check Solution |

Question 1:
\(\sqrt{0.4}\) is a/an
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\(\sqrt{0.4} = \sqrt{\dfrac{2}{5}} = \dfrac{\sqrt{2}}{\sqrt{5}}\), which is an irrational number because it cannot be expressed as a ratio of two integers.
Quick Tip: The square root of a non-perfect square rational number is always irrational.
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 equal roots?
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If the zeroes of the polynomial \(ax^2 + bx + \dfrac{2a}{b}\) are reciprocal of each other, then the value of \(b\) is
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The distance of the point A(\(-3\), \(-4\)) from x-axis is
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Given \(\triangle ABC \sim \triangle PQR\), \(\angle A = 30^\circ\) and \(\angle Q = 90^\circ\). 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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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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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 10 cm.
Q: Distance between every pair of parallel tangents must be between 5 cm and 10 cm.
R: Distance between every pair of parallel tangents is 5 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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If \(\dfrac{2\tan30^\circ}{1+\tan^2 30^\circ} = \dfrac{2\tan30^\circ}{\sqrt{1 - \tan^2 30^\circ}}\), then \(x : y =\)
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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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The system of equations \(2x + 1 = 0\) and \(3y - 5 = 0\) has
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In a right triangle ABC, right-angled at A, if \(\sin B = \dfrac{1}{4}\), then the value of \(\sec B\) 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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Using the empirical relationship: \[ Mode = 3 \cdot Median - 2 \cdot Mean \] \[ 10 = 3m - 2M,\quad and also: M + m = 25 \]
Solve equations:
1. \(3m - 2M = 10\)
2. \(M + m = 25\)
Substitute \(M = 25 - m\) into (1): \[ 3m - 2(25 - m) = 10 \Rightarrow 3m - 50 + 2m = 10 \Rightarrow 5m = 60 \Rightarrow m = 12,\ M = 13 \] Quick Tip: Use the empirical formula: Mode = \(3 \times\) Median – \(2 \times\) Mean
If the maximum number of students has obtained 52 marks out of 80, then
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Assertion (A): For any two prime numbers \(p\) and \(q\), their HCF is 1 and LCM is \(p + q\).
Reason (R): For any two natural numbers, HCF × LCM = product of 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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(a) Solve the following pair of equations algebraically:
\[ 101x + 102y = 304 \quad (i)
102x + 101y = 305 \quad (ii) \]
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Question 21:
(b) In a pair of supplementary angles, the greater angle exceeds the smaller by \(50^\circ\). 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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Question 22:
(b) Use the identity: \(\sin^2A + \cos^2A = 1\) to prove that \(\tan^2A + 1 = \sec^2A\). Hence, find the value of \(\tan A\), when \(\sec A = \dfrac{5}{3}\) where A is an acute angle.
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Prove that the 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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P is a point on the side BC of \(\triangle ABC\) such that \(\angle APC = \angle BAC\). Prove that \(AC^2 = BC \cdot CP\).
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The number of red balls in a bag is three more than the number of black balls. If the probability of drawing a red ball at random from the given bag is \(\dfrac{12}{23}\), find the total number of balls in the given bag.
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(a) Prove that \(\sqrt{5}\) is an irrational number.
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Question 26:
(b) Let \(p\), \(q\) and \(r\) be three distinct prime numbers. Check whether \(pqr + q\) is a composite number or not. Further, give an example for three distinct primes \(p\), \(q\), \(r\) such that
(i) \(pqr + 1\) is a composite number
(ii) \(pqr + 1\) is a prime number
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Find the zeroes of the polynomial \(p(x) = 3x^2 - 4x - 4\). Hence, write a polynomial whose each of the zeroes is 2 more than the zeroes of \(p(x)\).
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Check whether the following pair of equations is consistent or not. If consistent, solve graphically:
\[ x + 3y = 6
3y - 2x = -12 \]
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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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(a) Prove that: \[ \frac{\cos\theta - 2\cos^3\theta}{\sin\theta - 2\sin^3\theta} + \cot\theta = 0 \]
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Question 30:
(b) Given that \(\sin \theta + \cos \theta = x\), prove that \(\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}\).
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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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Question 32:
(a) 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 distance of point P from A and B respectively.
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Question 32:
(b) 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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(a) If a line drawn parallel to one side of triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to third side. State and prove the converse of the above statement.
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Question 33:
(b) In the adjoining figure, \(\triangle CAB\) is a right triangle, right angled at A and \(AD \perp BC\). Prove that \(\triangle ADB \sim \triangle CDA\). Further, if \(BC = 10\) cm and \(CD = 2\) cm, find the length of AD.
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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 \(\pi = \dfrac{22{7}, \sqrt{5} = 2.2\))
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The following distribution shows the marks of 230 students in a particular subject. If the median marks are 46, then find the values of \(x\) and \(y\).
Marks Number of Students
10 -- 20 12
20 -- 30 30
30 -- 40 \(x\)
40 -- 50 65
50 -- 60 \(y\)
60 -- 70 25
70 -- 80 18
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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: \(A = \frac{\theta}{360^\circ} \cdot \pi r^2\)
Question 37:
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.
View Solution
(i) and (ii) — Draw right triangles representing the situations with height and distance as sides and angle of elevation at the observer’s location.
(iii-a) From Situation – I: \[ \tan(60^\circ) = \frac{h}{80\sqrt{3}} \quad \Rightarrow \quad \sqrt{3} = \frac{h}{80\sqrt{3}} \Rightarrow h = 240 m \]
So, height of statue = \(240 - 58 = \boxed{182 m}\)
(iii-b) From Situation – II:
Let \(x\) be the horizontal distance from point B to the base.
\[ \tan(30^\circ) = \frac{240 - 40}{x} = \frac{200}{x} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{200}{x} \Rightarrow x = 200\sqrt{3} \approx \boxed{346.4 m} \]
Now for the base only: \[ \tan(\alpha) = \frac{58 - 40}{346.4} = \frac{18}{346.4} \Rightarrow \boxed{\tan \alpha \approx 0.052} \] Quick Tip: Use trigonometric ratios like \(\tan \theta = \frac{opposite}{adjacent}\) when height and distances are involved with angles of elevation.



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