CBSE Class 10 Maths Standard 2025 Question Paper (Set 30/6/1) with Solutions

1. $\sqrt{0.4} = \sqrt{\dfrac{4}{10}} = \dfrac{2}{\sqrt{10}}$.
2. $10$ is not a perfect square, so $\sqrt{10}$ is irrational, and a non-zero rational number divided by an irrational number is irrational.

Answer: (D) irrational number

Common mistake: Thinking $0.4$ is rational so its root must be. A square root of a rational number is rational only when it is a perfect square ratio, like $\sqrt{0.25} = 0.5$.

1. The unit digits of $8^1, 8^2, 8^3, 8^4, \dots$ are $8, 4, 2, 6$ and then repeat.
2. So the unit digit can be $8, 4, 2$ or $6$, but never $0$. ($8^n = 2^{3n}$ has no factor $5$, so it is never divisible by $10$.)

Answer: (C) 0

1. Check the discriminant $D = b^2 - 4ac$ of each.
2. (A) $x^2 + 2x + 1 = 2x + 1$ gives $x^2 = 0$: $D = 0$, equal roots $0, 0$. ✓
3. (B) $D = 1 > 0$ (roots $0, -1$). (C) $D = 16 > 0$ (roots $\pm 2$). (D) $D = 1 - 4 = -3 < 0$ (no real roots).

Answer: (A) $(x + 1)^2 = 2x + 1$

Common mistake: Reading (A) as already in standard form. Expand it first.

1. Zeroes that are reciprocals of each other have product $1$.
2. Product of zeroes $= \dfrac{c}{a} = \dfrac{2a/b}{a} = \dfrac{2}{b}$.
3. $\dfrac{2}{b} = 1$, so $b = 2$.

Answer: (A) 2

1. The distance of a point from the $x$-axis is the absolute value of its $y$-coordinate.
2. $|-4| = 4$.

Answer: (B) 4

Common mistake: Answering $5$ (the distance from the origin). The question is about the $x$-axis only.

1. $PQ \parallel XY$ in $\triangle AXY$: $\dfrac{AP}{PX} = \dfrac{AQ}{QY}$, so $\dfrac{2}{1.5} = \dfrac{AQ}{0.75}$ and $AQ = 1$ cm.
2. $AY = AQ + QY = 1.75$ cm and $AX = AP + PX = 3.5$ cm.
3. $XY \parallel BC$ in $\triangle ABC$: $\dfrac{AX}{XB} = \dfrac{AY}{YC}$, so $\dfrac{3.5}{4} = \dfrac{1.75}{CY}$ and $CY = 2$ cm.
4. $AQ + CY = 1 + 2 = 3$ cm.

Answer: (C) 3 cm

Common mistake: Using $\dfrac{AP}{PB}$ directly. $P$ and $Q$ are parallel to $XY$, so compare with $AX$, not $AB$.

1. $\triangle ABC \sim \triangle PQR$ means $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$. So $\angle P = 30^\circ$, $\angle B = \angle Q = 90^\circ$.
2. In $\triangle PQR$: $\angle R = 180^\circ - 30^\circ - 90^\circ = 60^\circ$.
3. $\angle R + \angle B = 60^\circ + 90^\circ = 150^\circ$.

Answer: (C) $150^\circ$

Common mistake: Taking $\angle A = 30^\circ$ to be in $\triangle PQR$ without matching the letters. The order $ABC \leftrightarrow PQR$ decides which angles are equal.

1. Outcomes: $HH, HT, TH, TT$ — $4$ in all.
2. At least one head: $HH, HT, TH$ — $3$ outcomes (only $TT$ has none).
3. $P = \dfrac{3}{4}$.

Answer: (C) $\dfrac{3}{4}$

1. $OA \perp PA$ and $OB \perp PB$, and $\angle P = 90^\circ$. So $OAPB$ has four right angles, and $OA = OB = r$ gives a square of side $r$.
2. $AB$ is a diagonal of this square: $AB = r\sqrt{2} = 3\sqrt{2}$, so $r = 3$ cm.
3. Diameter $= 2r = 6$ cm.

Answer: (D) 6 cm

Common mistake: Calling AB the diameter. AB is a chord, the diagonal of the square OAPB.

1. A pair of parallel tangents touch the circle at the two ends of a diameter, so the distance between them is the diameter, $2 \times 5 = 10$ cm, for every pair. So Q is true (and P, R are false).
2. S is false: from a point $5\sqrt{2}$ cm from the centre, the tangent length is $\sqrt{(5\sqrt{2})^2 - 5^2} = 5$ cm.

Answer: (B) Q

1. $TS$ is a tangent, so $OS \perp TS$ and $\angle OST = 90^\circ$.
2. In $\triangle OST$: $3x + x + 90^\circ = 180^\circ$, so $4x = 90^\circ$ and $x = 22.5^\circ$.
3. $2x = 45^\circ$.

Answer: (B) 45

Common mistake: Stopping at $x = 22.5$. The question asks for $2x$.

1. $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, so $\tan^2 30^\circ = \dfrac{1}{3}$.
2. $\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{2/\sqrt{3}}{4/3} = \dfrac{\sqrt{3}}{2}$ (this is $\sin 60^\circ$).
3. $\dfrac{2\tan 30^\circ}{1 - \tan^2 30^\circ} = \dfrac{2/\sqrt{3}}{2/3} = \sqrt{3}$ (this is $\tan 60^\circ$).
4. So $x \cdot \dfrac{\sqrt{3}}{2} = y\sqrt{3}$, giving $x = 2y$ and $x : y = 2 : 1$.

Answer: (C) 2 : 1

1. In the right triangle formed by the tree, the ground and the line of sight: $\tan\theta = \dfrac{\text{height}}{\text{distance}} = \dfrac{10}{10\sqrt{3}} = \dfrac{1}{\sqrt{3}}$.
2. So $\theta = 30^\circ$ at the snake. The angle of depression at the peacock equals this angle of elevation (alternate angles), $30^\circ$.

Answer: (A) $30^\circ$

1. The greatest cone has the same base radius $r$ and the same height $h$ as the cylinder.
2. Cone $= \dfrac{1}{3}\pi r^2 h$ and cylinder $= \pi r^2 h$, so the remaining wood $= \pi r^2 h - \dfrac{1}{3}\pi r^2 h = \dfrac{2}{3}\pi r^2 h$.
3. Remaining wood : cone $= \dfrac{2}{3} : \dfrac{1}{3} = 2 : 1$.

Answer: (C) 2 : 1

1. Mode $= 3\,\text{Median} - 2\,\text{Mean}$, and Mean $+$ Median $= 25$, so Mean $= 25 -$ Median.
2. $10 = 3\,\text{Median} - 2(25 - \text{Median}) = 5\,\text{Median} - 50$, so Median $= 12$.
3. Mean $= 25 - 12 = 13$. So the mean is $13$ and the median is $12$.

Answer: (B) 13 and 12

Common mistake: Reading the options as (median, mean). The question asks for the mean first.

1. The mode is the value that occurs most often, that is the value obtained by the maximum number of students.
2. So $52$ is the mode of the data.

Answer: (C) 52 is the mode of the data.

1. $2x + 1 = 0$ gives $x = -\dfrac{1}{2}$ and $3y - 5 = 0$ gives $y = \dfrac{5}{3}$.
2. These are a vertical and a horizontal line, which always cross at exactly one point, $\left(-\dfrac{1}{2}, \dfrac{5}{3}\right)$. A unique solution.

Answer: (A) unique solution

1. $\sin B = \dfrac{AC}{BC} = \dfrac{1}{4}$. Take $AC = 1$ and $BC = 4$.
2. $AB = \sqrt{BC^2 - AC^2} = \sqrt{16 - 1} = \sqrt{15}$.
3. $\cos B = \dfrac{AB}{BC} = \dfrac{\sqrt{15}}{4}$, so $\sec B = \dfrac{1}{\cos B} = \dfrac{4}{\sqrt{15}}$.

Answer: (D) $\dfrac{4}{\sqrt{15}}$

Common mistake: Giving $\cos B = \dfrac{\sqrt{15}}{4}$ (option B). Secant is the reciprocal of cosine.

1. (A) For two different primes the HCF is $1$, but the LCM is $p \times q$, not $p + q$ (for $p = 2$, $q = 3$ the LCM is $6$, not $5$). The assertion is false.
2. (R) HCF $\times$ LCM $=$ product of the two numbers is a true standard result.

Answer: (D) A is false, R is true

1. (A) $E_1 = \{1, 2\}$ and $E_2 = \{4, 5, 6\}$. Together they miss the outcome $3$, so $E_2$ is not "not $E_1$". They are not complementary. The assertion is false.
2. (R) Complementary events satisfy $P(E) + P(F) = 1$. True.

Answer: (D) A is false, R is true

1. (a) Add the equations: $203x + 203y = 609$, so $x + y = 3$. … (i)
2. Subtract the first from the second: $x - y = 1$, that is $-x + y = -1$. … (ii)
3. From (i) and (ii): $x = 2$ and $y = 1$. (Check: $101 \times 2 + 102 = 304$ ✓.)
4. (b) Let the smaller angle be $x$ and the greater angle $y$. Supplementary: $x + y = 180$. The greater exceeds the smaller by $50$: $y = x + 50$.
5. Put in: $x + x + 50 = 180$, so $x = 65^\circ$ and $y = 115^\circ$.

Answer: (a) $x = 2,\ y = 1$   (b) $x + y = 180,\ y - x = 50$; angles $65^\circ$ and $115^\circ$

1. (a) $m^2 = a^2\sec^2\theta + b^2\tan^2\theta + 2ab\sec\theta\tan\theta$ and $n^2 = b^2\sec^2\theta + a^2\tan^2\theta + 2ab\sec\theta\tan\theta$.
2. Subtract: $m^2 - n^2 = a^2(\sec^2\theta - \tan^2\theta) + b^2(\tan^2\theta - \sec^2\theta) = a^2 - b^2$ (as $\sec^2\theta - \tan^2\theta = 1$).
3. So $m^2 - n^2 = a^2 - b^2$, that is $a^2 + n^2 = b^2 + m^2$.
4. (b) Divide $\sin^2 A + \cos^2 A = 1$ by $\cos^2 A$: $\tan^2 A + 1 = \sec^2 A$.
5. With $\sec A = \dfrac{5}{3}$: $\tan^2 A = \dfrac{25}{9} - 1 = \dfrac{16}{9}$, so $\tan A = \dfrac{4}{3}$ (positive, as $A$ is acute).

Answer: (a) proved   (b) $\tan A = \dfrac{4}{3}$

1. Let $P = (x, y)$ with $PA = PB$, so $PA^2 = PB^2$.
2. $(x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$.
3. $x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$, so $-8x + 8y = -16$.
4. $x - y = 2$, that is $x = y + 2$: the abscissa is $2$ more than the ordinate.

Answer: $x = y + 2$

1. In $\triangle APC$ and $\triangle BAC$: $\angle APC = \angle BAC$ (given) and $\angle C$ is common.
2. So $\triangle APC \sim \triangle BAC$ by AA similarity.
3. Corresponding sides are proportional: $\dfrac{AC}{BC} = \dfrac{CP}{AC}$, so $AC^2 = BC \cdot CP$.

Answer: $AC^2 = BC \cdot CP$

Common mistake: Writing $\dfrac{AC}{CP} = \dfrac{BC}{AC}$ with the wrong pairing. Match the vertices in order: $A \leftrightarrow B$, $P \leftrightarrow A$, $C \leftrightarrow C$.

1. Let the black balls be $x$. Red balls $= x + 3$ and the total $= 2x + 3$.
2. $\dfrac{x + 3}{2x + 3} = \dfrac{12}{23}$, so $23x + 69 = 24x + 36$ and $x = 33$.
3. Total $= 2 \times 33 + 3 = 69$. (Check: red $= 36$ and $\dfrac{36}{69} = \dfrac{12}{23}$ ✓.)

Answer: 69 balls

1. (a) Suppose $\sqrt{5} = \dfrac{p}{q}$, with $p, q$ co-prime and $q \neq 0$. Then $5q^2 = p^2$, so $5$ divides $p^2$ and so $5$ divides $p$. Write $p = 5a$.
2. Then $25a^2 = 5q^2$, so $q^2 = 5a^2$, so $5$ divides $q^2$ and hence $q$. Now $5$ divides both $p$ and $q$, contradicting that they are co-prime. So $\sqrt{5}$ is irrational.
3. (b) $pqr + q = q(pr + 1)$. It has the factor $q$ and the factor $(pr + 1) > 1$, so it has more than two factors: it is a composite number.
4. (i) Take $p = 3$, $q = 5$, $r = 7$: $pqr + 1 = 105 + 1 = 106 = 2 \times 53$, composite.
5. (ii) Take $p = 2$, $q = 3$, $r = 5$: $pqr + 1 = 30 + 1 = 31$, which is prime.

Answer: (a) proved   (b) composite; (i) $106$   (ii) $31$

Common mistake: In (b), assuming $pqr + 1$ is always prime or always composite. It can be either, so an example is needed for each.

1. $3x^2 - 4x - 4 = 3x^2 - 6x + 2x - 4 = 3x(x - 2) + 2(x - 2) = (3x + 2)(x - 2)$.
2. The zeroes are $x = -\dfrac{2}{3}$ and $x = 2$.
3. The new zeroes are each $2$ more: $-\dfrac{2}{3} + 2 = \dfrac{4}{3}$ and $2 + 2 = 4$.
4. Sum $= \dfrac{4}{3} + 4 = \dfrac{16}{3}$ and product $= \dfrac{4}{3} \times 4 = \dfrac{16}{3}$.
5. Polynomial: $x^2 - \dfrac{16}{3}x + \dfrac{16}{3}$, or multiplying by $3$: $3x^2 - 16x + 16$.

Answer: Zeroes $-\dfrac{2}{3}$ and $2$; new polynomial $3x^2 - 16x + 16$

1. Write as $x + 3y - 6 = 0$ and $-2x + 3y + 12 = 0$. $\dfrac{a_1}{a_2} = \dfrac{1}{-2}$ and $\dfrac{b_1}{b_2} = \dfrac{3}{3} = 1$.
2. $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so the lines intersect at one point: the pair is consistent.
3. First line: $(0, 2)$, $(3, 1)$, $(6, 0)$. Second line: $3y = 2x - 12$, points $(0, -4)$, $(3, -2)$, $(6, 0)$.
4. The two lines cross at $(6, 0)$. So $x = 6$, $y = 0$.

Answer: Consistent; $x = 6,\ y = 0$

1. The diagonals of a parallelogram bisect each other, so the mid-point of $AC$ equals the mid-point of $BD$.
2. Mid-point of $AC = \left(\dfrac{6 + 9}{2},\ \dfrac{1 + 4}{2}\right) = \left(\dfrac{15}{2},\ \dfrac{5}{2}\right)$. Mid-point of $BD = \left(\dfrac{p + 7}{2},\ \dfrac{2 + q}{2}\right)$.
3. $\dfrac{p + 7}{2} = \dfrac{15}{2}$ gives $p = 8$, and $\dfrac{2 + q}{2} = \dfrac{5}{2}$ gives $q = 3$.
4. Now $B = (8, 2)$ and $D = (7, 3)$. $AC = \sqrt{3^2 + 3^2} = 3\sqrt{2}$ and $BD = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.
5. The diagonals of a rectangle are equal. Here $AC \neq BD$, so $ABCD$ is not a rectangle.

Answer: $p = 8,\ q = 3$; $ABCD$ is not a rectangle

1. (a) Factorise: $\dfrac{\cos\theta(1 - 2\cos^2\theta)}{\sin\theta(1 - 2\sin^2\theta)}$. Write $1 = \sin^2\theta + \cos^2\theta$ in each bracket.
2. $1 - 2\cos^2\theta = \sin^2\theta - \cos^2\theta$ and $1 - 2\sin^2\theta = \cos^2\theta - \sin^2\theta$. So the fraction is $\cot\theta \times \dfrac{\sin^2\theta - \cos^2\theta}{\cos^2\theta - \sin^2\theta} = -\cot\theta$.
3. LHS $= -\cot\theta + \cot\theta = 0 =$ RHS.
4. (b) Square $\sin\theta + \cos\theta = x$: $1 + 2\sin\theta\cos\theta = x^2$, so $2\sin\theta\cos\theta = x^2 - 1$.
5. RHS $= \dfrac{2 - (2\sin\theta\cos\theta)^2}{2} = \dfrac{2 - 4\sin^2\theta\cos^2\theta}{2} = 1 - 2\sin^2\theta\cos^2\theta$.
6. $1 = (\sin^2\theta + \cos^2\theta)^2 = \sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta$, so RHS $= \sin^4\theta + \cos^4\theta =$ LHS.

Answer: Both proved

1. $OP \perp TP$, so $\angle OPT = 90^\circ$. Then $\angle TPQ = \angle OPT - \angle OPQ = 90^\circ - 15^\circ = 75^\circ$.
2. Tangents from $T$ are equal: $TP = TQ$, so $\angle TQP = \angle TPQ = 75^\circ$.
3. In $\triangle TPQ$: $\theta = 180^\circ - 75^\circ - 75^\circ = 30^\circ$.
4. $\sin 2\theta = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$.

Answer: $\sin 2\theta = \dfrac{\sqrt{3}}{2}$

Common mistake: Answering $\sin 30^\circ$. The question asks for $\sin 2\theta$, with $\theta = 30^\circ$.

1. (a) Let $PB = x$ m. Then $PA = (x + 35)$ m.
2. $AB$ is a diameter, so $\angle APB = 90^\circ$ (angle in a semicircle). By Pythagoras: $(x + 35)^2 + x^2 = 65^2$.
3. $2x^2 + 70x + 1225 = 4225$, so $x^2 + 35x - 1500 = 0$. Factorise: $(x + 60)(x - 25) = 0$.
4. A distance is positive, so $x = 25$. $PB = 25$ m and $PA = 25 + 35 = 60$ m.
5. (b) Real roots need $D \ge 0$: $[-2(p + 1)]^2 - 4 \times 1 \times p^2 \ge 0$, so $4(p^2 + 2p + 1) - 4p^2 \ge 0$ and $8p + 4 \ge 0$.
6. $p \ge -\dfrac{1}{2}$, so the smallest value is $p = -\dfrac{1}{2}$.
7. Put $p = -\dfrac{1}{2}$: $x^2 - 2\left(\dfrac{1}{2}\right)x + \dfrac{1}{4} = 0$, that is $x^2 - x + \dfrac{1}{4} = 0$ or $4x^2 - 4x + 1 = 0$. So $(2x - 1)^2 = 0$ and $x = \dfrac{1}{2}, \dfrac{1}{2}$.

Answer: (a) $PA = 60$ m, $PB = 25$ m   (b) $p = -\dfrac{1}{2}$; roots $\dfrac{1}{2}, \dfrac{1}{2}$