CBSE Class 10 Maths Standard 2026 Question Paper (Set 30/2/1) with Solutions

1. $960 = 4 \times 240$, so $240$ divides $960$.
2. When one number is a multiple of the other, the LCM is the larger number: $960$. (The HCF is $240$.)

Answer: (A) 960

1. A prime has exactly two factors and a composite number has more than two. $1$ has only one factor (itself), so it is neither.

Answer: (D) neither prime nor composite

1. $5, 25, 125, 625, \dots$: a number ending in $5$ times $5$ always ends in $5$. So $5^n$ ends with $5$.

Answer: (B) 5

1. The zeroes are the $x$-values where the curve meets the $x$-axis.
2. It crosses at $A$ on the left, and it touches the axis once more at the bottom of the dip on the right. That is $2$ distinct points ($A$ and the touching point).

Answer: (C) 2

Common mistake: Counting the peak or counting the touching point twice. Each meeting point counts once, so there are two distinct zeroes.

1. $\alpha + \beta = -\dfrac{-2}{p} = \dfrac{2}{p}$ and $\alpha\beta = \dfrac{3p}{p} = 3$.
2. $\alpha + \beta = \alpha\beta$ gives $\dfrac{2}{p} = 3$, so $p = \dfrac{2}{3}$.

Answer: (B) $\dfrac{2}{3}$

1. Dependent (coincident) lines have infinitely many solutions, which needs all three ratios equal: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$. (Option (A) is a unique solution and (C) is parallel lines, no solution.)

Answer: (D) $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

1. (A) common difference $\dfrac{1}{2}$ ✓. (B) common difference $-2$ ✓. (C) $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}$: common difference $\sqrt{2}$ ✓.
2. (D) $1, 9, 25, 49$: differences $8, 16, 24$ are not equal. Not an A.P.

Answer: (D) $1^2, 3^2, 5^2, 7^2, \dots$

1. $\angle A = \angle Q$ and $\angle B = \angle R$, so the correspondence is $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$, and $\triangle ABC \sim \triangle QRP$.
2. $AB$ matches $QR$ and $AC$ matches $QP$, so $AB : AC = QR : QP$.

Answer: (C) $QR : QP$

1. The distance from the $x$-axis is the absolute value of the $y$-coordinate, $|3a|$, which is $3a$ (taking $a > 0$, as the options expect). Option (B) and (D) are negative, and $4a$ is the distance from the $y$-axis.

Answer: (A) 3a

1. Adjacent $= 4$, hypotenuse $= 5$, so opposite $= \sqrt{25 - 16} = 3$ (the $3$-$4$-$5$ triangle). $\tan A = \dfrac{3}{4}$.

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

1. $\sin A = \dfrac{1}{2}$, so $A = 30^\circ$.
2. $\tan 30^\circ + \cot 30^\circ = \dfrac{1}{\sqrt{3}} + \sqrt{3} = \dfrac{1 + 3}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}$.

Answer: (B) $\dfrac{4}{\sqrt{3}}$

1. $\tan 45^\circ = \dfrac{\text{height}}{60} = 1$, so the height is $60$ m.

Answer: (C) 60

1. $OA \perp PA$, so $\angle PAB = 90^\circ - \angle OAB = 90^\circ - 15^\circ = 75^\circ$.
2. Tangents from $P$ are equal, so $PA = PB$ and $\angle PBA = \angle PAB = 75^\circ$.
3. $\angle APB = 180^\circ - 75^\circ - 75^\circ = 30^\circ$.

Answer: (A) $30^\circ$

1. $OA \perp PA$ and $OB \perp PB$, so in the quadrilateral $OAPB$: $\angle APB = 360^\circ - 90^\circ - 90^\circ - 130^\circ = 50^\circ$. (The angle at the centre and the angle between the tangents add up to $180^\circ$.)

Answer: (B) $50^\circ$

1. Segment $=$ sector $-$ triangle. Sector $= \dfrac{60}{360}\pi r^2 = \dfrac{\pi r^2}{6}$.
2. The triangle with two sides $r$ and angle $60^\circ$ is equilateral: area $\dfrac{\sqrt{3}}{4}r^2$.
3. Segment $= \dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4}r^2$.

Answer: (C) $\dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4}r^2$

1. Outer radius $R = 6$ cm and inner radius $r = 6 - 1 = 5$ cm.
2. Steel $= \dfrac{2}{3}\pi(R^3 - r^3) = \dfrac{2}{3}\pi(216 - 125) = \dfrac{2}{3}\pi \times 91 = \dfrac{182}{3}\pi$ cm³.

Answer: (B) $\dfrac{182}{3}\pi$

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 3(43.4) - 2(43) = 130.2 - 86 = 44.2$.

Answer: (C) 44.2

1. Primes up to $25$: $2, 3, 5, 7, 11, 13, 17, 19, 23$, which is $9$. The number $1$ is neither prime nor composite.
2. Composite numbers $= 25 - 9 - 1 = 15$, so $P = \dfrac{15}{25}$.

Answer: (A) $\dfrac{15}{25}$

Common mistake: Counting 1 as composite (giving 16) or as prime. It is neither.

1. (A) The cuboid is $8 \times 4 \times 4$: surface area $= 2(32 + 16 + 32) = 2 \times 80 = 160$ cm². True. (Also: $10$ outer faces of $16$ cm² each $= 160$ cm².)
2. (R) The correct formula is $2(lb + bh + hl)$; the factor $2$ is missing. False.

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

1. (R) $1 + 2 + \dots + n = \dfrac{n(n + 1)}{2}$. True.
2. (A) Mean $= \dfrac{n(n + 1)}{2} \div n = \dfrac{n + 1}{2}$, not $\dfrac{n - 1}{2}$. False. (For $n = 3$: mean of $1, 2, 3$ is $2 = \dfrac{3 + 1}{2}$.)

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

1. $\alpha + \beta = -\dfrac{q}{p}$ and $\alpha\beta = \dfrac{r}{p}$.
2. $\alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha^2 + \beta^2) = \alpha\beta\left[(\alpha + \beta)^2 - 2\alpha\beta\right]$.
3. $= \dfrac{r}{p}\left[\dfrac{q^2}{p^2} - \dfrac{2r}{p}\right] = \dfrac{r}{p} \times \dfrac{q^2 - 2pr}{p^2} = \dfrac{r(q^2 - 2pr)}{p^3}$.

Answer: $\dfrac{r}{p^3}(q^2 - 2pr)$

1. (a) $\triangle AHK \sim \triangle ABC$, so $\dfrac{AK}{AC} = \dfrac{HK}{BC}$: $\dfrac{10}{AC} = \dfrac{7}{3.5} = 2$, so $AC = 5$ cm.
2. (b) $XY \parallel QR$, so $\dfrac{PQ}{XQ} = \dfrac{PR}{YR}$ (a line parallel to a side cuts the other two sides in the same ratio, so the whole-to-part ratios are equal).
3. $\dfrac{7}{3} = \dfrac{6.3}{YR}$, so $YR = \dfrac{6.3 \times 3}{7} = 2.7$ cm.

Answer: (a) $5$ cm   (b) $2.7$ cm

1. The diagonals of a parallelogram bisect each other, so the mid-points of $AC$ and $BD$ coincide.
2. Mid-point of $AC = \left(\dfrac{4 + 4}{2}, \dfrac{5 + 3}{2}\right) = (4, 4)$. Mid-point of $BD = \left(\dfrac{m + 1}{2}, \dfrac{6 + n}{2}\right)$.
3. $\dfrac{m + 1}{2} = 4$ gives $m = 7$, and $\dfrac{6 + n}{2} = 4$ gives $n = 2$.

Answer: $m = 7$ and $n = 2$

1. (a) Square both sides: $\tan^2\theta + 2 + \dfrac{1}{\tan^2\theta} = 4$, so $\tan^2\theta + \dfrac{1}{\tan^2\theta} = 2$.
2. (b) Multiply the top and bottom inside the root by $(1 - \sin\theta)$: LHS $= \sqrt{\dfrac{(1 - \sin\theta)^2}{1 - \sin^2\theta}} = \sqrt{\dfrac{(1 - \sin\theta)^2}{\cos^2\theta}}$.
3. $= \dfrac{1 - \sin\theta}{\cos\theta} = \sec\theta - \tan\theta =$ RHS.

Answer: (a) $2$   (b) proved

1. The radius is perpendicular to the tangent at the point of contact: $\angle PQO = 90^\circ$ and $\angle PRO = 90^\circ$.
2. $\angle PQO + \angle PRO = 180^\circ$. A pair of opposite angles of $PQOR$ is supplementary, so the quadrilateral $PQOR$ is cyclic.

Answer: Shown

1. Suppose $\sqrt{5} = \dfrac{p}{q}$ with $p, q$ co-prime integers and $q \neq 0$.
2. $5q^2 = p^2$, so $5$ divides $p^2$ and therefore $5$ divides $p$. Write $p = 5a$.
3. $25a^2 = 5q^2$, so $q^2 = 5a^2$; $5$ divides $q^2$ and so $5$ divides $q$.
4. $5$ divides both $p$ and $q$, contradicting that they are co-prime. So $\sqrt{5}$ is irrational.

Answer: Proved

1. Let $P$ and $Q$ be the points of trisection, with $AP : PB = 1 : 2$ and $AQ : QB = 2 : 1$.
2. $P = \left(\dfrac{1(-3) + 2(-1)}{3}, \dfrac{1(-2) + 2(4)}{3}\right) = \left(-\dfrac{5}{3}, 2\right)$.
3. $Q = \left(\dfrac{2(-3) + 1(-1)}{3}, \dfrac{2(-2) + 1(4)}{3}\right) = \left(-\dfrac{7}{3}, 0\right)$. (Check: $Q$ is the mid-point of $PB$ ✓.)

Answer: $\left(-\dfrac{5}{3}, 2\right)$ and $\left(-\dfrac{7}{3}, 0\right)$

1. (a) $\sec^2\theta - 1 = \tan^2\theta$ and $\operatorname{cosec}^2\theta - 1 = \cot^2\theta$. LHS $= \dfrac{\sec^3\theta}{\tan^2\theta} + \dfrac{\operatorname{cosec}^3\theta}{\cot^2\theta}$.
2. $\dfrac{\sec^3\theta}{\tan^2\theta} = \dfrac{1}{\cos^3\theta} \times \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\cos\theta\sin^2\theta}$ and, in the same way, $\dfrac{\operatorname{cosec}^3\theta}{\cot^2\theta} = \dfrac{1}{\sin\theta\cos^2\theta}$.
3. Sum $= \dfrac{1}{\sin\theta\cos\theta}\left(\dfrac{1}{\sin\theta} + \dfrac{1}{\cos\theta}\right) = \sec\theta\operatorname{cosec}\theta(\operatorname{cosec}\theta + \sec\theta) =$ RHS.
4. (b) $p^2 - q^2 = \dfrac{\sec^2\alpha - \tan^2\alpha}{\operatorname{cosec}^2\beta} = \dfrac{1}{\operatorname{cosec}^2\beta}$.
5. $(p^2 - q^2)\sec^2\alpha = \dfrac{\sec^2\alpha}{\operatorname{cosec}^2\beta} = \left(\dfrac{\sec\alpha}{\operatorname{cosec}\beta}\right)^2 = p^2$.

Answer: Proved (both parts)

1. (a) **Given:** a circle with centre $O$ and tangents $TP$, $TQ$ from the external point $T$ (touching at $P$ and $Q$). **To prove:** $TP = TQ$. Join $OP$, $OQ$ and $OT$.
2. The radius is perpendicular to the tangent: $\angle OPT = \angle OQT = 90^\circ$. In the right triangles $OPT$ and $OQT$: $OP = OQ$ (radii), $OT$ is common and the angles at $P, Q$ are $90^\circ$.
3. So $\triangle OPT \cong \triangle OQT$ (RHS) and $TP = TQ$.
4. (b) $OP = OQ$ (radii), so $\angle OPQ = \angle OQP$. In $\triangle OPQ$: $\angle POQ = 180^\circ - 2\angle OPQ$. (i)
5. In the quadrilateral $OPTQ$ the angles at $P$ and $Q$ are $90^\circ$, so $\angle POQ + \angle PTQ = 180^\circ$, that is $\angle POQ = 180^\circ - \angle PTQ$. (ii)
6. From (i) and (ii): $180^\circ - 2\angle OPQ = 180^\circ - \angle PTQ$, so $\angle PTQ = 2\angle OPQ$.

Answer: Proved (both parts)

1. Minor sector $= \dfrac{30}{360} \times \dfrac{22}{7} \times 42 \times 42 = \dfrac{1}{12} \times 5544 = 462$ cm².
2. The major sector has central angle $360^\circ - 30^\circ = 330^\circ$: $\dfrac{330}{360} \times \dfrac{22}{7} \times 42 \times 42 = \dfrac{11}{12} \times 5544 = 5082$ cm².
3. (Check: $462 + 5082 = 5544$, the whole circle ✓.)

Answer: Minor sector $= 462$ cm²; major sector $= 5082$ cm²

1. There are $6 \times 6 = 36$ equally likely outcomes.
2. (i) Sum $5$: $(1,4), (4,1), (2,3), (3,2)$, which is $4$ outcomes. $P = \dfrac{4}{36} = \dfrac{1}{9}$.
3. (ii) Difference $3$: $(1,4), (4,1), (2,5), (5,2), (3,6), (6,3)$, which is $6$ outcomes. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

Answer: (i) $\dfrac{1}{9}$   (ii) $\dfrac{1}{6}$

1. Let a pencil cost ₹ $x$ and a chocolate cost ₹ $y$. The equations are $2x + 3y = 11$ and $x + 2y = 7$.
2. Line 1, $2x + 3y = 11$: $y = \dfrac{11 - 2x}{3}$ gives the points $(1, 3)$, $(4, 1)$ and $(-2, 5)$.
3. Line 2, $x + 2y = 7$: $y = \dfrac{7 - x}{2}$ gives the points $(1, 3)$, $(3, 2)$ and $(5, 1)$.
4. The two lines meet at $(1, 3)$, so $x = 1$ and $y = 3$. (Check: $2 + 9 = 11$ ✓ and $1 + 6 = 7$ ✓.)

Answer: A pencil costs ₹ $1$ and a chocolate costs ₹ $3$

1. (a) Let the tour be $x$ days. Daily expense $= \dfrac{4200}{x}$, and after the extension $\dfrac{4200}{x + 3}$. $\dfrac{4200}{x} - \dfrac{4200}{x + 3} = 70$.
2. $4200 \times 3 = 70x(x + 3)$, so $x^2 + 3x - 180 = 0$ and $(x + 15)(x - 12) = 0$. Days cannot be negative, so $x = 12$.
3. (Check: $\dfrac{4200}{12} = 350$ and $\dfrac{4200}{15} = 280$; the difference is $70$ ✓.)
4. (b) Let the altitude be $x$ cm and the base $(x + 10)$ cm. $\dfrac{1}{2}x(x + 10) = 600$, so $x^2 + 10x - 1200 = 0$ and $(x + 40)(x - 30) = 0$.
5. $x = 30$ (reject $-40$). Altitude $= 30$ cm, base $= 40$ cm and hypotenuse $= \sqrt{30^2 + 40^2} = 50$ cm.

Answer: (a) $12$ days   (b) $30$ cm, $40$ cm and $50$ cm

1. (a) **Given:** in $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** join $BE$ and $CD$, and draw $DM \perp AC$ and $EN \perp AB$.
2. $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EN}{\frac{1}{2} \cdot DB \cdot EN} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{\frac{1}{2} \cdot AE \cdot DM}{\frac{1}{2} \cdot EC \cdot DM} = \dfrac{AE}{EC}$.
3. $\triangle BDE$ and $\triangle CDE$ have the same base $DE$ and lie between the same parallels $DE$ and $BC$, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (b) She walks $BD = 1.2 \times 4 = 4.8$ m. Let her shadow be $DE$. $AB \parallel CD$ (both vertical), so $\triangle ABE \sim \triangle CDE$ (AA).
5. $\dfrac{BE}{DE} = \dfrac{AB}{CD}$: $\dfrac{4.8 + DE}{DE} = \dfrac{3.6}{0.9} = 4$, so $4.8 + DE = 4DE$ and $DE = 1.6$ m.

Answer: (a) proved   (b) the shadow is $1.6$ m long

1. Cumulative frequencies: $2, 6, 24, 45, 78, 89, 92, 98, 100$.
2. Modal class: the highest frequency is $33$, so it is $35 - 40$: $l = 35$, $f_1 = 33$, $f_0 = 21$, $f_2 = 11$, $h = 5$.
3. Mode $= 35 + \dfrac{33 - 21}{2(33) - 21 - 11} \times 5 = 35 + \dfrac{12}{34} \times 5 = \dfrac{625}{17} \approx 36.76$ years.
4. Median: $\dfrac{n}{2} = 50$ lies in the class $35 - 40$ (the previous $cf$ is $45$, the next is $78$): $l = 35$, $cf = 45$, $f = 33$.
5. Median $= 35 + \dfrac{50 - 45}{33} \times 5 = 35 + \dfrac{25}{33} = \dfrac{1180}{33} \approx 35.76$ years.
6. (The marking scheme gives these as $36.7$ and $35.7$ years, approximately; rounding to one decimal place they are $36.8$ and $35.8$.)

Answer: Modal age $\approx 36.76$ years; median age $\approx 35.76$ years

1. The instalments form an A.P. with $a = 1000$ and $d = 100$.
2. (i) $a_{30} = 1000 + 29 \times 100 = 3900$.
3. (ii) $a_{40} = 1000 + 39 \times 100 = 4900$.
4. (iii)(a) $S_{30} = \dfrac{30}{2}[2(1000) + 29 \times 100] = 15 \times 4900 = 73500$. Left to pay $= 118000 - 73500 = ₹44500$.
5. (iii)(b) $a_{10} = 1000 + 900 = 1900$. $\dfrac{a_{10}}{a_{40}} = \dfrac{1900}{4900} = \dfrac{19}{49}$.
6. (Check: $S_{40} = 20(2000 + 3900) = 118000$, the full loan ✓.)