CBSE Class 10 Maths Standard 2025 Question Paper (Set 30/4/1) with Solutions

1. $x = a \cdot b^3$ and $y = a^3 \cdot b$. HCF takes the smaller power of each prime: $\mathrm{HCF} = a^1 b^1 = ab$.
2. LCM takes the greater power: $\mathrm{LCM} = a^3 b^3$.
3. $\mathrm{HCF} - \mathrm{LCM} = ab - a^3b^3 = ab(1 - a^2b^2) = ab(1 - ab)(1 + ab)$.

Answer: (D) $ab(1 - ab)(1 + ab)$

Common mistake: Stopping at $ab(1 - a^2b^2)$. Option (D) has this factorised further using $1 - a^2b^2 = (1 - ab)(1 + ab)$.

1. Use $a^2 - b^2 = (a + b)(a - b)$ with $a = 1 + \sqrt{3}$ and $b = 1 - \sqrt{3}$.
2. $a + b = 2$ and $a - b = 2\sqrt{3}$, so the value is $2 \times 2\sqrt{3} = 4\sqrt{3}$.
3. $4\sqrt{3}$ is positive, and irrational because $\sqrt{3}$ is irrational.

Answer: (C) a positive irrational number

1. Equal roots need $D = b^2 - 4ac = 0$: $1 - 4a^2 = 0$, so $a^2 = \dfrac{1}{4}$ and $a = \pm\dfrac{1}{2}$.
2. The equal root is $x = -\dfrac{b}{2a} = -\dfrac{1}{2a}$. For this to be positive, $a$ must be negative.
3. So $a = -\dfrac{1}{2}$, and then $x = -\dfrac{1}{2 \times (-1/2)} = 1 > 0$ ✓.

Answer: (D) $-\dfrac{1}{2}$

Common mistake: Choosing $a = \dfrac{1}{2}$. Then the root is $-1$, which is equal but not positive.

1. The distance of a point from the $x$-axis is $|y|$, the absolute value of its $y$-coordinate. So we need $|y| = 3$.
2. $(1, 3)$: $y = 3$ ✓. $(-3, -3)$: $|y| = 3$ ✓. $(-3, 3)$: $y = 3$ ✓. $(3, 1)$: $|y| = 1 \neq 3$ ✗.

Answer: (D) $(3, 1)$

Common mistake: Using the $x$-coordinate. The distance from the $x$-axis is the $y$-coordinate.

1. Let the black balls be $b$. Red balls $= b + 10$ and total $= 2b + 10$.
2. $\dfrac{b + 10}{2b + 10} = \dfrac{3}{5}$, so $5b + 50 = 6b + 30$ and $b = 20$.
3. Total $= 2 \times 20 + 10 = 50$. (Check: red $= 30$ and $\dfrac{30}{50} = \dfrac{3}{5}$ ✓.)

Answer: (A) 50

1. Infinitely many solutions need $\dfrac{p}{12} = \dfrac{3}{p} = \dfrac{p - 3}{p}$.
2. From the first two: $p^2 = 36$, so $p = \pm 6$.
3. From the last two: $\dfrac{3}{p} = \dfrac{p - 3}{p}$ gives $3 = p - 3$, so $p = 6$.
4. Only $p = 6$ satisfies all three. (For $p = -6$ the third ratio is $\dfrac{-9}{-6} = \dfrac{3}{2}$, not $-\dfrac{1}{2}$.)

Answer: (B) $6$ only

Common mistake: Stopping at $p = \pm 6$. Use the third ratio to reject $p = -6$.

1. Compare the sides: $\dfrac{AB}{RQ} = \dfrac{3.8}{7.6} = \dfrac{1}{2}$, $\dfrac{AC}{RP} = \dfrac{3\sqrt{3}}{6\sqrt{3}} = \dfrac{1}{2}$ and $\dfrac{BC}{QP} = \dfrac{6}{12} = \dfrac{1}{2}$.
2. All three ratios are equal, so $\triangle ABC \sim \triangle RQP$ (SSS), with $A \leftrightarrow R$, $B \leftrightarrow Q$ and $C \leftrightarrow P$.
3. So $\angle C = \angle P = 180^\circ - 80^\circ - 60^\circ = 40^\circ$.

Answer: (D) $40^\circ$

Common mistake: Matching $C$ with $R$ just because they are in the same corner of the figure. Match the sides by their ratios first.

1. Let $\tan A = u$. $\tan 2A = \dfrac{2u}{1 - u^2}$, so $\dfrac{2u}{1 - u^2} = 3u$.
2. For $u \neq 0$: $2 = 3(1 - u^2)$, so $u^2 = \dfrac{1}{3}$ and $u = \dfrac{1}{\sqrt{3}}$.
3. $\tan A = \dfrac{1}{\sqrt{3}}$ gives $A = 30^\circ$. Check: $\tan 60^\circ = \sqrt{3} = 3 \times \dfrac{1}{\sqrt{3}}$ ✓.

Answer: (D) $30^\circ$

Common mistake: Using $\tan 2A = 2\tan A$. The double-angle formula is $\dfrac{2\tan A}{1 - \tan^2 A}$. Or simply test the options.

1. Between $0^\circ$ and $90^\circ$, sine and tangent increase with the angle, and cosine decreases.
2. (A) $\sin 20^\circ < \sin 70^\circ$, so (A) is false. (D) $\tan 20^\circ < \tan 70^\circ$, so (D) is false.
3. (B) $\cos 20^\circ = \sin 70^\circ > \sin 20^\circ$, so (B) is false.
4. (C) $\cos 20^\circ > \cos 70^\circ$ is true, because cosine falls as the angle grows.

Answer: (C) $\cos 20^\circ > \cos 70^\circ$

1. The pole, the ground and the rope form a right triangle. The rope is the hypotenuse ($30$ m) and the pole is opposite the $60^\circ$ angle.
2. $\sin 60^\circ = \dfrac{h}{30}$, so $h = 30 \times \dfrac{\sqrt{3}}{2} = 15\sqrt{3}$ m.

Answer: (D) $15\sqrt{3}$ m

Common mistake: Using $\tan 60^\circ$ or $\cos 60^\circ$. The rope is the hypotenuse, so the height is found with sine.

1. Each depression has diameter $2 \times 0.35 = 0.7$ cm.
2. Along one edge of $7$ cm, at most $\dfrac{7}{0.7} = 10$ depressions fit side by side.
3. On the square face: $10 \times 10 = 100$.

Answer: (B) 100

Common mistake: Dividing the face area by the circle area ($\dfrac{49}{\pi \times 0.35^2} \approx 127$). Circles cannot fill a square without gaps, so count them along the sides.

1. In the median formula, $cf$ is the cumulative frequency of the class just before the median class. It counts the frequencies of the classes preceding the median class.
2. The median class itself is not included. Its own frequency is $f$ in the formula.

Answer: (C) classes preceding the median class

1. The empirical relation is $\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$.
2. $\text{Mode} = 3 \times 11 - 2 \times 10 = 33 - 20 = 13$.

Answer: (C) 13

1. $AB$ touches the smaller circle at $P$, so $OP \perp AB$. The perpendicular from the centre bisects a chord: $AP = \dfrac{AB}{2} = r$.
2. In right triangle $OPA$: $OA^2 = OP^2 + AP^2 = r^2 + r^2 = 2r^2$.
3. The radius of the larger circle is $OA = \sqrt{2}\,r$.

Answer: (D) $\sqrt{2}\,r$

1. A parallelogram that circumscribes a circle is a rhombus (opposite sides are equal, and $AB + CD = AD + BC$ forces all four sides to be equal).
2. All sides are $5$ cm, so the perimeter is $4 \times 5 = 20$ cm.

Answer: (A) 20 cm

1. $\dfrac{AE}{EB} = \dfrac{AF}{FC}$, so by the converse of the Basic Proportionality Theorem $EF \parallel BC$.
2. So $\triangle AEF \sim \triangle ABC$. $\dfrac{AE}{EB} = \dfrac{1}{2}$ means $AE$ is $1$ part out of $3$ parts of $AB$: $\dfrac{AE}{AB} = \dfrac{1}{3}$.
3. $\dfrac{EF}{BC} = \dfrac{AE}{AB} = \dfrac{1}{3}$, so $BC = 3EF$.

Answer: (D) $BC = 3\,EF$

Common mistake: Using $\dfrac{1}{2}$ as the ratio $EF : BC$. The ratio $1 : 2$ is $AE : EB$, so $EF : BC$ is $1 : 3$.

1. A polynomial of degree $n$ has at most $n$ zeroes. For degree $3$, at most $3$ distinct zeroes.
2. It can have fewer: for example $x^3$ has only one zero and $x^3 - x$ has three. So "at most two", "at least two" and "exactly three" are not always true.

Answer: (D) $p(x)$ has at most three distinct zeroes.

1. Two dice have $6 \times 6 = 36$ equally likely outcomes.
2. The sum is more than $10$ only for a sum of $11$: $(5, 6), (6, 5)$, or $12$: $(6, 6)$. That is $3$ outcomes.
3. $P(\text{sum} \le 10) = 1 - \dfrac{3}{36} = \dfrac{33}{36} = \dfrac{11}{12}$.

Answer: (D) $\dfrac{11}{12}$

Common mistake: Counting only the sum $11$ and forgetting the sum $12$.

1. (A) $4^n = 2^{2n}$ has only the prime factor $2$, so it is not divisible by $5$ and can never end in $0$. (Its last digit is $4$ or $6$.) The assertion is false.
2. (R) If $x$ has $2$ and $5$ among its prime factors, then $x$ is a multiple of $10$, and so is every power $x^n$. It ends in $0$. The reason is true.

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

1. (A) Each tangent is perpendicular to the diameter at its end, so the two tangents make equal alternate angles of $90^\circ$ with the diameter: they are parallel. True.
2. (R) Two tangents from an external point are equal in length. True.
3. The parallel tangents in (A) come from "tangent $\perp$ radius", not from the equal-length property. So R does not explain A.

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

1. Multiply the first equation by $4$ and the second by $3$ to make the $x$ coefficients equal: $120x + 176y = 40$ … (i) and $120x + 165y = 39$ … (ii).
2. Subtract (ii) from (i): $11y = 1$, so $y = \dfrac{1}{11}$.
3. Put in the first equation: $30x + 44 \times \dfrac{1}{11} = 10$, so $30x + 4 = 10$ and $x = \dfrac{6}{30} = \dfrac{1}{5}$.
4. Check in the second: $40 \times \dfrac{1}{5} + 55 \times \dfrac{1}{11} = 8 + 5 = 13$ ✓.

Answer: $x = \dfrac{1}{5}$, $y = \dfrac{1}{11}$

1. (A) In $3$ seconds the boy walks $2.5 \times 3 = 7.5$ m from the base of the post. Let his shadow be $x$ m long.
2. $\triangle ABE \sim \triangle CDE$ (AA: both have a right angle and share the angle at $E$), so $\dfrac{CD}{AB} = \dfrac{DE}{BE}$.
3. $\dfrac{1.5}{12} = \dfrac{x}{7.5 + x}$, so $12x = 1.5(7.5 + x) = 11.25 + 1.5x$ and $10.5x = 11.25$.
4. $x = \dfrac{11.25}{10.5} = \dfrac{15}{14} \approx 1.07$ m.
5. (B) In $\triangle ABE$ and $\triangle CFB$: $\angle A = \angle C$ (opposite angles of a parallelogram).
6. $AE \parallel BC$ and $BE$ is a transversal, so $\angle AEB = \angle CBF$ (alternate angles).
7. So $\triangle ABE \sim \triangle CFB$ by AA similarity.

Answer: (A) about $1.07$ m   (B) proved

Common mistake: In (A), writing $\dfrac{1.5}{12} = \dfrac{x}{7.5}$. The big triangle's base is the boy's distance plus the shadow, $7.5 + x$.

1. $C$ is on $AB$ produced (beyond $B$) with $AC = 2BC$. Since $AC = AB + BC$, we get $AB = BC$, so $B$ is the mid-point of $AC$.
2. Let $C = (x, y)$. Mid-point of $AC = \left(\dfrac{-1 + x}{2},\ \dfrac{7 + y}{2}\right) = (4, -3)$.
3. $\dfrac{-1 + x}{2} = 4$ gives $x = 9$, and $\dfrac{7 + y}{2} = -3$ gives $y = -13$.

Answer: $C = (9, -13)$

Common mistake: Using the section formula with ratio $2 : 1$ internally. $C$ lies outside the segment, beyond $B$.

1. (A) Expand: $\sin^2 A + \mathrm{cosec}^2 A + 2 + \cos^2 A + \sec^2 A + 2$.
2. $\sin^2 A + \cos^2 A = 1$, $\mathrm{cosec}^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$.
3. LHS $= 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4 = 7 + \tan^2 A + \cot^2 A$. So $x = 7$.
4. (B) Numerator: $3 \times \dfrac{1}{2} - 4 \times \left(\dfrac{1}{2}\right)^3 = \dfrac{3}{2} - \dfrac{1}{2} = 1$.
5. Denominator: $2(\sin^2 50^\circ + \cos^2 50^\circ) = 2 \times 1 = 2$. The value is $\dfrac{1}{2}$.

Answer: (A) $x = 7$   (B) $\dfrac{1}{2}$

1. December has $31$ days, so Ashraf's birthday can be any of $31$ dates, all equally likely.
2. (i) For the same date as Anil, only $1$ date works: $P = \dfrac{1}{31}$.
3. (ii) $P(\text{different}) = 1 - \dfrac{1}{31} = \dfrac{30}{31}$.

Answer: (i) $\dfrac{1}{31}$   (ii) $\dfrac{30}{31}$

Common mistake: Using 365 or 30 for the number of days. The month is December, so 31.

1. (A) Suppose $\sqrt{2}$ is rational: $\sqrt{2} = \dfrac{p}{q}$ with integers $p, q$, $q \neq 0$, having no common factor.
2. Squaring: $2q^2 = p^2$. So $2$ divides $p^2$, and as $2$ is prime, $2$ divides $p$. Write $p = 2a$.
3. Then $4a^2 = 2q^2$, so $q^2 = 2a^2$. So $2$ divides $q^2$ and hence $q$.
4. Now $2$ is a common factor of $p$ and $q$, which contradicts our assumption. So $\sqrt{2}$ is irrational.
5. (B) $p = x^2y^3$, $q = x^1y^4$, $r = x^5y^2$. HCF takes the smallest powers: $\mathrm{HCF} = x\,y^2$. LCM takes the greatest powers: $\mathrm{LCM} = x^5y^4$.
6. $\mathrm{HCF} \times \mathrm{LCM} = x^6y^6$, while $p \times q \times r = x^{2+1+5}y^{3+4+2} = x^8y^9$.
7. These are not equal, so $\mathrm{HCF}(p, q, r) \times \mathrm{LCM}(p, q, r) \neq p \times q \times r$.

Answer: (A) proved   (B) HCF $= xy^2$, LCM $= x^5y^4$; the product rule does not hold for three numbers

Common mistake: In (B), assuming $\mathrm{HCF} \times \mathrm{LCM} =$ product. That rule is only for two numbers.

1. Let the incomes be ₹ $9x$ and ₹ $7x$, and the expenditures ₹ $4y$ and ₹ $3y$.
2. Savings $=$ income $-$ expenditure $= 5000$ for each: $9x - 4y = 5000$ … (i) and $7x - 3y = 5000$ … (ii).
3. Multiply (i) by $3$ and (ii) by $4$: $27x - 12y = 15000$ and $28x - 12y = 20000$. Subtract: $x = 5000$.
4. Incomes: $9x = 9 \times 5000 = ₹45{,}000$ and $7x = 7 \times 5000 = ₹35{,}000$.
5. Check: $y = 10000$, so expenditures are $40000$ and $30000$, and the savings are $5000$ each ✓.

Answer: $9x - 4y = 5000,\ 7x - 3y = 5000$; incomes ₹ $45{,}000$ and ₹ $35{,}000$

1. Right angle at $P$, so by Pythagoras $PQ^2 + PR^2 = QR^2$.
2. $PQ^2 = (x + 2)^2 + (y + 3)^2$, $PR^2 = (x - 2)^2 + (y - 3)^2$ and $QR^2 = 4^2 + 6^2 = 52$.
3. Adding: $(x^2 + 4x + 4 + y^2 + 6y + 9) + (x^2 - 4x + 4 + y^2 - 6y + 9) = 52$, which gives $2x^2 + 2y^2 + 26 = 52$.
4. So $x^2 + y^2 = 13$.
5. For $y = 2$: $x^2 = 13 - 4 = 9$, so $x = 3$ or $x = -3$.

Answer: $x^2 + y^2 = 13$; $x = \pm 3$

Common mistake: Giving only $x = 3$. The equation $x^2 = 9$ has two roots, and both points are on the circle.

1. (A) Divide the numerator and denominator of the LHS by $\sin A$: $\dfrac{\cot A + 1 - \mathrm{cosec}\,A}{\cot A - 1 + \mathrm{cosec}\,A}$.
2. In the numerator write $1 = \mathrm{cosec}^2 A - \cot^2 A$: $\cot A - \mathrm{cosec}\,A + (\mathrm{cosec}\,A - \cot A)(\mathrm{cosec}\,A + \cot A)$.
3. $= (\mathrm{cosec}\,A - \cot A)\left[-1 + \mathrm{cosec}\,A + \cot A\right]$, and the bracket is the same as the denominator. So LHS $= \mathrm{cosec}\,A - \cot A =$ RHS.
4. (B) $p^2 - q^2 = (p + q)(p - q) = (2\cot\theta)(2\cos\theta) = 4\cot\theta\cos\theta$.
5. $pq = \cot^2\theta - \cos^2\theta = \cos^2\theta\left(\mathrm{cosec}^2\theta - 1\right) = \cos^2\theta\cot^2\theta$, so $4\sqrt{pq} = 4\cos\theta\cot\theta$.
6. Both sides are $4\cot\theta\cos\theta$, so $p^2 - q^2 = 4\sqrt{pq}$.

Answer: Both proved

1. For $px^2 + qx + 1$: $\alpha + \beta = -\dfrac{q}{p}$ and $\alpha\beta = \dfrac{1}{p}$.
2. Sum of the new zeroes $= \dfrac{2}{\alpha} + \dfrac{2}{\beta} = \dfrac{2(\alpha + \beta)}{\alpha\beta} = 2 \times \left(-\dfrac{q}{p}\right) \times p = -2q$.
3. Product of the new zeroes $= \dfrac{2}{\alpha} \times \dfrac{2}{\beta} = \dfrac{4}{\alpha\beta} = 4p$.
4. The polynomial is $x^2 - (\text{sum})x + \text{product} = x^2 + 2qx + 4p$.

Answer: $x^2 + 2qx + 4p$

Common mistake: Forgetting the minus sign in "sum $= -2q$", which turns into $+2qx$ in the polynomial.

1. Let the circle touch $AB, BC, CD, DA$ at $P, Q, R, S$. Tangents from a point are equal: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$.
2. Add the four: $AP + BP + CR + DR = AS + BQ + CQ + DS$, so $AB + CD = AD + BC$. … (i)
3. In a rectangle $AB = CD$ and $BC = AD$. Then (i) gives $2AB = 2AD$, so $AB = AD$.
4. All four sides are equal and each angle is $90^\circ$, so $ABCD$ is a square.
5. The side of the square equals the diameter of the circle: $2 \times 10 = 20$ cm. Perimeter $= 4 \times 20 = 80$ cm.

Answer: $ABCD$ is a square; perimeter $= 80$ cm