CBSE Class 10 Maths Standard 2025 Question Paper (Set 30/3/1) with Solutions

1. $\tan 60^\circ = \sqrt{3}$, so $3\theta = 60^\circ$ and $\theta = 20^\circ$.
2. The question asks for $\dfrac{\theta}{2} = \dfrac{20^\circ}{2} = 10^\circ$.

Answer: (D) $10^\circ$

Common mistake: Stopping at $\theta = 20^\circ$. Read what is asked: $\dfrac{\theta}{2}$.

1. $4 = 2^2$, $6 = 2 \times 3$, $8 = 2^3$, so $x = \text{LCM} = 2^3 \times 3 = 24$.
2. $3, 5, 7$ are primes, so $y = 3 \times 5 \times 7 = 105$.
3. $p = \text{LCM}(24, 105)$. $24 = 2^3 \times 3$ and $105 = 3 \times 5 \times 7$, so $p = 2^3 \times 3 \times 5 \times 7 = 840$.
4. $\dfrac{840}{24} = 35$, so $p = 35x$.

Answer: (A) $p = 35x$

1. Infinitely many solutions need $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
2. $\dfrac{6}{36} = \dfrac{1}{6} = \dfrac{3k}{3}$, so $k = \dfrac{1}{6}$.

Answer: (B) $\dfrac{1}{6}$

1. For $x^2 - ax - b$: sum of zeroes $\alpha + \beta = -\dfrac{-a}{1} = a$.
2. Product of zeroes $\alpha\beta = \dfrac{-b}{1} = -b$.
3. $\alpha + \beta + \alpha\beta = a + (-b) = a - b$.

Answer: (C) $a - b$

Common mistake: Using $\alpha\beta = b$. The constant term is $-b$, so the product is $-b$.

1. $\dfrac{x}{12} = \dfrac{3}{x}$. Cross-multiply: $x^2 = 36$.
2. So $x = \pm 6$.

Answer: (A) $\pm 6$

1. On the $x$-axis $y = 0$: $\dfrac{x}{4} = 1$, so $P = (4, 0)$. On the $y$-axis $x = 0$: $\dfrac{y}{6} = 1$, so $Q = (0, 6)$.
2. Mid-point of $PQ = \left(\dfrac{4 + 0}{2},\ \dfrac{0 + 6}{2}\right) = (2, 3)$.

Answer: (A) $(2, 3)$

1. Section formula with $m : n = 2 : 1$: $x = \dfrac{2 \times 5 + 1 \times (-1)}{2 + 1} = \dfrac{9}{3} = 3$.
2. $y = \dfrac{2 \times 2 + 1 \times 5}{3} = \dfrac{9}{3} = 3$. The point is $(3, 3)$.

Answer: (C) $(3, 3)$

Common mistake: Swapping the ratio. For $2 : 1$ from $P$, the weight $2$ goes with the coordinates of $Q$.

1. $\angle APB = 80^\circ$. The line $PO$ bisects it, so $\angle OPA = 40^\circ$.
2. $OA \perp PA$ (radius to the point of contact), so $\angle OAP = 90^\circ$.
3. In $\triangle OAP$: $\angle POA = 180^\circ - 90^\circ - 40^\circ = 50^\circ$.

Answer: (B) $50^\circ$

Common mistake: Answering $40^\circ$, the angle at $P$. The question asks for the angle at the centre.

1. $\cot\theta + \tan\theta = \dfrac{\cos\theta}{\sin\theta} + \dfrac{\sin\theta}{\cos\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta}$.
2. $= \dfrac{1}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta} \cdot \dfrac{1}{\cos\theta} = \mathrm{cosec}\,\theta \sec\theta$.

Answer: (A) $\mathrm{cosec}\,\theta \sec\theta$

1. Match the equal angles: $D \leftrightarrow Q$ and $E \leftrightarrow R$, so $F \leftrightarrow P$. By AA, $\triangle DEF \sim \triangle QRP$.
2. Corresponding sides are in proportion: $\dfrac{DE}{QR} = \dfrac{EF}{RP} = \dfrac{DF}{QP}$.
3. Options (A), (B) and (C) are all taken from these equal ratios, so they are true.
4. In (D), $DE$ is paired with $PQ$, but $DE$ corresponds to $QR$. So (D) is not true.

Answer: (D) $\dfrac{DE}{PQ} = \dfrac{EF}{RP}$

Common mistake: Pairing sides in the order the letters are written. Match the vertices by the equal angles first.

1. In $\triangle LMN$: $\angle L = 180^\circ - 130^\circ - 28^\circ = 22^\circ$.
2. In $\triangle ABC$: $\angle A = 22^\circ$ and $\angle B = 130^\circ$, so $\angle C = 28^\circ$.
3. So $\triangle LMN \sim \triangle ABC$ by AA with $L \leftrightarrow A$, $M \leftrightarrow B$, $N \leftrightarrow C$.
4. $\dfrac{LM}{AB} = \dfrac{LN}{AC}$: $\dfrac{45}{5} = \dfrac{72}{AC}$, so $9 = \dfrac{72}{AC}$ and $AC = 8$ cm.

Answer: (C) $8$ cm

Common mistake: Matching $LN$ with $BC$. $AC$ is opposite $\angle B = 130^\circ$, so it matches $LN$, which is opposite $\angle M = 130^\circ$.

1. Volume of a cube $= a^3$, so $\dfrac{a_1^3}{a_2^3} = \dfrac{8}{125}$ and $\dfrac{a_1}{a_2} = \dfrac{2}{5}$.
2. Surface area $= 6a^2$, so the ratio of surface areas $= \left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}$.

Answer: (B) 4 : 25

Common mistake: Squaring the volume ratio. First take the cube root to get the ratio of sides.

1. Area of a sector $= \dfrac{1}{2} \times \text{arc length} \times r$.
2. $54\pi = \dfrac{1}{2} \times l \times 36 = 18\,l$, so $l = \dfrac{54\pi}{18} = 3\pi$ cm.

Answer: (D) $3\pi$ cm

1. Factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$. Of the numbers $1$ to $6$ on a die, $1, 2, 3, 4, 6$ are factors.
2. The only number that is not a factor is $5$, so there is $1$ favourable outcome out of $6$.
3. $P = \dfrac{1}{6}$.

Answer: (C) $\dfrac{1}{6}$

1. There are $7$ numbers and the mean is $7$, so their sum is $7 \times 7 = 49$.
2. Sum $= 2 + 9 + (x + 6) + (2x + 3) + 5 + 10 + 5 = 40 + 3x$.
3. $40 + 3x = 49$, so $3x = 9$ and $x = 3$.

Answer: (D) 3

1. In rectangle $AOBC$ the diagonals are $AB$ and $OC$, and they are equal. Use $AB$, since both $A$ and $B$ are given.
2. $AB^2 = (4 - 0)^2 + (0 - 2)^2 = 16 + 4 = 20$.

Answer: (B) 20

Common mistake: Giving the side length ($4$ or $2$). The question asks for the square of the diagonal.

1. Split the middle term: $x^2 - 2\sqrt{2}\,x - \sqrt{2}\,x + 4$, because $2\sqrt{2} \times \sqrt{2} = 4$ and $2\sqrt{2} + \sqrt{2} = 3\sqrt{2}$.
2. $= x(x - 2\sqrt{2}) - \sqrt{2}(x - 2\sqrt{2}) = (x - 2\sqrt{2})(x - \sqrt{2})$.
3. The zeroes are $2\sqrt{2}$ and $\sqrt{2}$. Check: sum $= 3\sqrt{2}$ ✓ and product $= 4$ ✓.

Answer: (B) $2\sqrt{2},\ \sqrt{2}$

1. In $\triangle ADC$ and $\triangle BAC$: $\angle ADC = \angle BAC = 90^\circ$ and $\angle C$ is common. So $\triangle ADC \sim \triangle BAC$ (AA).
2. $\dfrac{DC}{AC} = \dfrac{AC}{BC}$, so $AC^2 = BC \times DC = 16 \times 4 = 64$.
3. $x = AC = 8$ cm.

Answer: (C) $8$ cm

1. (A) Ladder $= \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ m. The assertion is true.
2. (R) The angle with the ground $\theta$ has $\tan\theta = \dfrac{8}{6} = \dfrac{4}{3} \neq \sqrt{3} = \tan 60^\circ$. So the angle is not $60^\circ$ and the reason is false.

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

Common mistake: Accepting the reason because it sounds plausible. Check it with the given 6 m and 8 m.

1. (A) From an external point $P$, $PA = PB$, $OA = OB$ and $OP$ is common, so $\triangle OAP \cong \triangle OBP$ (SSS). So $\angle AOP = \angle BOP$: equal angles at the centre. True.
2. (R) A parallelogram circumscribing a circle has $AB + CD = AD + BC$, which forces all sides equal, so it is a rhombus. True.
3. The reason is a separate result about parallelograms. It does not explain the assertion, so R is not the explanation of A.

Answer: (B) Both true, but R is not the correct explanation of A

1. $\tan 60^\circ = \sqrt{3}$, so $\tan^2 60^\circ = 3$. $\mathrm{cosec}\,30^\circ = 2$, so $\mathrm{cosec}^2 30^\circ = 4$. $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, so $\tan^2 30^\circ = \dfrac{1}{3}$.
2. $4k = 3 - 2 \times 4 - 2 \times \dfrac{1}{3} = 3 - 8 - \dfrac{2}{3} = -5 - \dfrac{2}{3} = -\dfrac{17}{3}$.
3. $k = -\dfrac{17}{12}$.

Answer: $k = -\dfrac{17}{12}$

Common mistake: Forgetting the squares: the question has $\mathrm{cosec}^2 30^\circ$ and $\tan^2 30^\circ$, not $\mathrm{cosec}\,30^\circ$ and $\tan 30^\circ$.

1. $P(E) + P(\text{not } E) = 1$.
2. $\dfrac{x}{12} + \dfrac{5}{6} = 1$, so $\dfrac{x}{12} = 1 - \dfrac{5}{6} = \dfrac{1}{6}$.
3. $x = \dfrac{12}{6} = 2$.

Answer: $x = 2$

1. (a) The smallest number divisible by both is their LCM. $644 = 2^2 \times 7 \times 23$ and $462 = 2 \times 3 \times 7 \times 11$.
2. LCM $= 2^2 \times 3 \times 7 \times 11 \times 23 = 12 \times 7 \times 253 = 84 \times 253 = 21252$.
3. (b) Let the numbers be $4x$ and $5x$, where $x$ is their HCF, so $x = 11$. The numbers are $44$ and $55$.
4. LCM $= 4 \times 5 \times 11 = 220$ (check: $44 \times 55 = 2420 = 11 \times 220$ ✓).

Answer: (a) $21252$   (b) $220$

Common mistake: In (b), using $4 \times 5 = 20$ as the LCM. The common factor $11$ must be multiplied back.

1. (a) Real and equal roots need the discriminant $D = b^2 - 4ac = 0$.
2. $k^2 - 4 \times 4 \times 1 = 0$, so $k^2 = 16$ and $k = \pm 4$.
3. (b) For $y^2 - 5y + 3$: $\alpha + \beta = 5$ and $\alpha\beta = 3$.
4. $\alpha^4\beta^3 + \alpha^3\beta^4 = \alpha^3\beta^3(\alpha + \beta) = (\alpha\beta)^3(\alpha + \beta) = 27 \times 5 = 135$.

Answer: (a) $k = \pm 4$   (b) $135$

Common mistake: In (a), giving only $k = 4$. Both $k = 4$ and $k = -4$ work.

1. Let $AB$ be a diameter of a circle with centre $O$, and let $PQ$ and $RS$ be the tangents at $A$ and $B$.
2. A tangent is perpendicular to the radius through the point of contact: $OA \perp PQ$ so $\angle OAP = 90^\circ$, and $OB \perp RS$ so $\angle OBR = 90^\circ$.
3. $OA$ and $OB$ lie along the same line $AB$, so $AB$ is a transversal to $PQ$ and $RS$. Then $\angle BAP = 90^\circ$ and $\angle ABS = 90^\circ$ are alternate angles.
4. Alternate angles are equal, so $PQ \parallel RS$.

Answer: Proved

Common mistake: Not stating why the angle is 90°: the radius is perpendicular to the tangent at the point of contact.

1. (a) Mid-point $P = \left(\dfrac{3 + k}{2},\ \dfrac{4 + 6}{2}\right) = \left(\dfrac{3 + k}{2},\ 5\right)$.
2. Put in $x + y - 10 = 0$: $\dfrac{3 + k}{2} + 5 - 10 = 0$, so $\dfrac{3 + k}{2} = 5$, $3 + k = 10$ and $k = 7$.
3. (b) Let the three points be $P_1, P_2, P_3$. $P_2$ is the mid-point of $AB$: $\left(\dfrac{-2 + 2}{2},\ \dfrac{2 + 8}{2}\right) = (0, 5)$.
4. $P_1$ is the mid-point of $A$ and $P_2$: $\left(\dfrac{-2 + 0}{2},\ \dfrac{2 + 5}{2}\right) = (-1,\ 3.5)$.
5. $P_3$ is the mid-point of $P_2$ and $B$: $\left(\dfrac{0 + 2}{2},\ \dfrac{5 + 8}{2}\right) = (1,\ 6.5)$.

Answer: (a) $k = 7$   (b) $(-1, \tfrac{7}{2}),\ (0, 5),\ (1, \tfrac{13}{2})$

Common mistake: In (b), using the section formula with a wrong ratio. Three points make four equal parts: use mid-points, or ratios $1 : 3$, $1 : 1$, $3 : 1$.

1. Suppose $5\sqrt{3} + \dfrac{2}{3}$ is rational, equal to $\dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$.
2. Then $5\sqrt{3} = \dfrac{p}{q} - \dfrac{2}{3} = \dfrac{3p - 2q}{3q}$, so $\sqrt{3} = \dfrac{3p - 2q}{15q}$.
3. $p$ and $q$ are integers, so $\dfrac{3p - 2q}{15q}$ is a rational number. That would make $\sqrt{3}$ rational.
4. But $\sqrt{3}$ is irrational (given). This is a contradiction, so our assumption was wrong. Hence $5\sqrt{3} + \dfrac{2}{3}$ is irrational.

Answer: Proved by contradiction

Common mistake: Writing the contradiction without saying which given fact it contradicts (that √3 is irrational).

1. (a) Put $s = \sec A$. Over a common root: $\dfrac{(s - 1) + (s + 1)}{\sqrt{(s + 1)(s - 1)}} = \dfrac{2s}{\sqrt{s^2 - 1}}$.
2. $s^2 - 1 = \sec^2 A - 1 = \tan^2 A$, so this is $\dfrac{2\sec A}{\tan A} = 2 \times \dfrac{1}{\cos A} \times \dfrac{\cos A}{\sin A} = \dfrac{2}{\sin A} = 2\,\mathrm{cosec}\,A$.
3. (b) $\dfrac{1}{\cos A} - \cos A = \dfrac{1 - \cos^2 A}{\cos A} = \dfrac{\sin^2 A}{\cos A}$ and $\dfrac{1}{\sin A} - \sin A = \dfrac{\cos^2 A}{\sin A}$.
4. LHS $= \dfrac{\sin^2 A}{\cos A} \times \dfrac{\cos^2 A}{\sin A} = \sin A \cos A$.
5. RHS: $\tan A + \cot A = \dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \dfrac{1}{\sin A \cos A}$, so $\dfrac{1}{\tan A + \cot A} = \sin A \cos A$. LHS = RHS.

Answer: Both identities proved

1. Area of minor segment $=$ area of sector $-$ area of the triangle formed by the radii and the chord.
2. Sector with $\theta = 90^\circ$: $\dfrac{90}{360} \times 3.14 \times 10 \times 10 = \dfrac{314}{4} = 78.5$ cm².
3. Right triangle $OAB$ with both legs $10$ cm: $\dfrac{1}{2} \times 10 \times 10 = 50$ cm².
4. Minor segment $= 78.5 - 50 = 28.5$ cm².

Answer: $28.5$ cm²

Common mistake: Giving the sector area (78.5) and forgetting to subtract the triangle.

1. All outcomes: $HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$. That is $8$ equally likely outcomes.
2. (a) Exactly two tails: $HTT, THT, TTH$ → $3$ outcomes. $P = \dfrac{3}{8}$.
3. (b) At least one head: every outcome except $TTT$ → $7$ outcomes. $P = \dfrac{7}{8}$.
4. (c) At most two heads: every outcome except $HHH$ → $7$ outcomes. $P = \dfrac{7}{8}$.

Answer: (a) $\dfrac{3}{8}$   (b) $\dfrac{7}{8}$   (c) $\dfrac{7}{8}$

Common mistake: Confusing “at least one” with “exactly one”. At least one head means 1, 2 or 3 heads.

1. $AB$ is a diameter, so the angle in a semicircle $\angle ACB = 90^\circ$.
2. $\angle PCA = \angle PCB + \angle BCA$, so $\angle PCB = 110^\circ - 90^\circ = 20^\circ$.
3. $OC \perp PC$ (radius to the tangent), so $\angle PCB + \angle BCO = 90^\circ$ and $\angle BCO = 90^\circ - 20^\circ = 70^\circ$.
4. $OB = OC$ (radii), so $\triangle OBC$ is isosceles and $\angle CBA = \angle OBC = \angle OCB = 70^\circ$.

Answer: $\angle CBA = 70^\circ$ and $\angle BCO = 70^\circ$

Common mistake: Forgetting that $B$ lies between $P$ and $O$, so $\angle PCA$ is made of two angles: $\angle PCB$ and $\angle BCA = 90^\circ$.

1. Let each equal side be $x$ cm and the base $y$ cm. Perimeter: $2x + y = 32$. … (i)
2. Each equal side is $\dfrac{5}{6}$ of the base: $x = \dfrac{5}{6}y$. … (ii)
3. Put (ii) in (i): $\dfrac{5}{3}y + y = 32$, so $\dfrac{8}{3}y = 32$ and $y = 12$. Then $x = \dfrac{5}{6} \times 12 = 10$.
4. The sides are $10$ cm, $10$ cm and $12$ cm. Semi-perimeter $s = \dfrac{32}{2} = 16$ cm.
5. Area $= \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{16 \times 6 \times 6 \times 4} = 4 \times 6 \times 2 = 48$ cm².
6. Check by height: the height on the base is $\sqrt{10^2 - 6^2} = 8$, so area $= \dfrac{1}{2} \times 12 \times 8 = 48$ cm² ✓.

Answer: Area $= 48$ cm²

Common mistake: Taking the equal side as $\dfrac{6}{5}$ of the base. "$\dfrac{5}{6}$th of the base" means $x = \dfrac{5}{6}y$.

1. (a) $a_3 + a_7 = (a + 2d) + (a + 6d) = 2a + 8d = 6$, so $a + 4d = 3$ and $a = 3 - 4d$.
2. $a_3 \cdot a_7 = (a + 2d)(a + 6d) = (3 - 2d)(3 + 2d) = 9 - 4d^2 = 8$, so $d^2 = \dfrac{1}{4}$ and $d = \pm\dfrac{1}{2}$.
3. If $d = \dfrac{1}{2}$: $a = 1$ and $S_{16} = \dfrac{16}{2}\left[2 \times 1 + 15 \times \dfrac{1}{2}\right] = 8 \times 9.5 = 76$.
4. If $d = -\dfrac{1}{2}$: $a = 5$ and $S_{16} = 8\left[2 \times 5 + 15 \times \left(-\dfrac{1}{2}\right)\right] = 8 \times 2.5 = 20$.
5. (b) Ages in years: $8,\ 8\dfrac{1}{3},\ 8\dfrac{2}{3},\ 9, \dots$ (since $4$ months $= \dfrac{1}{3}$ year). So $a = 8$ and $d = \dfrac{1}{3}$.
6. $S_n = \dfrac{n}{2}\left[16 + (n - 1)\dfrac{1}{3}\right] = 168$, so $n(n + 47) = 1008$, that is $n^2 + 47n - 1008 = 0$.
7. $(n + 63)(n - 16) = 0$. $n$ cannot be negative, so $n = 16$.
8. Eldest participant: $a_{16} = 8 + 15 \times \dfrac{1}{3} = 13$ years.