CBSE Class 10 Maths Standard 2026 Question Paper (Set 30/3/1) with Solutions

1. $x - 1 = \pm 4$, so $x = 5$ or $x = -3$.

Answer: (C) $5, -3$

1. $OP$ bisects $\angle QPR$, so $\angle OPQ = 30^\circ$. $OQ \perp PQ$ (radius and tangent).
2. In the right triangle $OQP$: $\tan 30^\circ = \dfrac{OQ}{PQ}$, so $PQ = \dfrac{3}{1/\sqrt{3}} = 3\sqrt{3}$ cm.

Answer: (A) $3\sqrt{3}$ cm

1. $AB \parallel EF$, so $\dfrac{DA}{AE} = \dfrac{DB}{BF}$: $\dfrac{2x}{3x + 1} = \dfrac{x}{2x - \frac{1}{2}}$.
2. $2x\left(2x - \dfrac{1}{2}\right) = x(3x + 1)$, so $4x^2 - x = 3x^2 + x$ and $x^2 - 2x = 0$, giving $x = 0$ or $x = 2$.
3. $x = 0$ would make the sides $DA$ and $DB$ zero, which is impossible, so $x = 2$ only.

Answer: (B) 2 only

1. $3 \times 11 \times 13 + 3 = 3(143 + 1) = 3 \times 144 = 432$. It has factors other than $1$ and itself ($3$, $144$...), and it is even, so it is composite (not prime, not odd, and $432 \div 13$ is not whole).

Answer: (C) a composite number

1. The sum is $6$ or $12$. Sum $6$: $(1,5), (5,1), (2,4), (4,2), (3,3)$, which is $5$ outcomes. Sum $12$: $(6,6)$, $1$ outcome.
2. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

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

1. The multiples of $4$ strictly between $12$ and $250$ are $16, 20, \dots, 248$ (since $12$ and $250$ themselves are excluded and $250$ is not a multiple of $4$).
2. $248 = 16 + (n - 1) \times 4$, so $n - 1 = 58$ and $n = 59$.

Answer: (A) 59

1. The largest cone has its base circle inscribed in a face, so the radius is $\dfrac{l}{2}$, and its height is the edge $l$.
2. $V = \dfrac{1}{3}\pi\left(\dfrac{l}{2}\right)^2 l = \dfrac{\pi l^3}{12}$.

Answer: (A) $\dfrac{\pi l^3}{12}$

1. A parallel line has the same slope: $2x - 6y = 7$ gives $y = \dfrac{1}{3}x - \dfrac{7}{6}$, slope $\dfrac{1}{3}$.
2. (C) $x - 3y = 7$ gives $y = \dfrac{1}{3}x - \dfrac{7}{3}$: the same slope and a different intercept, so it is parallel (and not the same line). The others have slopes $3$, $-\dfrac{1}{3}$, $-\dfrac{1}{3}$.

Answer: (C) $x - 3y = 7$

1. $\tan\theta = \dfrac{\text{height}}{\text{shadow}} = \dfrac{1}{\sqrt{3}}$, so $\theta = 30^\circ$.

Answer: (B) $30^\circ$

1. Equal roots need $D = 0$: $k^2 - 4(\sqrt{3})(2\sqrt{3}) = k^2 - 24 = 0$, so $k = \pm\sqrt{24}$.

Answer: (A) $\pm\sqrt{24}$

1. $a = -\dfrac{1}{3}$ and $d = \dfrac{2}{3} + \dfrac{1}{3} = 1$. $a_n = -\dfrac{1}{3} + (n - 1) \times 1 = n - \dfrac{4}{3}$.

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

1. $OT \perp PT$, so in the right triangle $OTP$: $\cos 45^\circ = \dfrac{OT}{OP} = \dfrac{r}{OP}$, so $OP = \dfrac{r}{1/\sqrt{2}} = r\sqrt{2}$.

Answer: (A) $r\sqrt{2}$

1. $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$. So $AB \leftrightarrow QR$ and $BC \leftrightarrow RP$.
2. $\dfrac{AB}{QR} = \dfrac{BC}{RP}$: $\dfrac{9}{QR} = \dfrac{5}{2}$, so $QR = \dfrac{18}{5} = 3.6$ cm.

Answer: (D) 3.6 cm

1. $k[x^2 - (\text{sum})x + \text{product}] = k\left[x^2 + \dfrac{1}{3}x + 2\right]$.
2. With $k = -3$: $-3x^2 - x - 6$, which is option (D).

Answer: (D) $-3x^2 - x - 6$

1. $\sin 2\alpha = \dfrac{\sqrt{3}}{2}$ gives $2\alpha = 60^\circ$ for an acute angle, so $\alpha = 30^\circ$.
2. $\sin 3\alpha = \sin 90^\circ = 1$.

Answer: (C) 1

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$: $26.1 = 75.6 - 2\,\text{Mean}$, so $\text{Mean} = \dfrac{49.5}{2} = 24.75$.

Answer: (A) 24.75

1. In an equilateral triangle of side $s$, the median (also the altitude) is $\dfrac{\sqrt{3}}{2}s$. Here $AD = \dfrac{5\sqrt{3}}{2}$, so $s = 5$.
2. $D$ is the mid-point of $BC$ at the origin, so $B = \left(-\dfrac{5}{2}, 0\right)$ and $C = \left(\dfrac{5}{2}, 0\right)$.

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

1. $\tan 45^\circ = 1$ and $\cos 60^\circ = \dfrac{1}{2}$: $\dfrac{1}{2}(1) - \dfrac{1}{4} = \dfrac{1}{4}$.

Answer: (C) $\dfrac{1}{4}$

1. (R) The radius to the point of contact is perpendicular to the tangent. True.
2. (A) Every other point of the tangent is outside the circle, and the perpendicular is the shortest distance from the centre to a line. So the radius is the smallest distance. True, and it follows from (R).

Answer: (A) Both A and R are true and R is the correct explanation of A

1. (A) At $\theta = 45^\circ$, $\tan 2\theta = \tan 90^\circ = \dfrac{\sin 90^\circ}{\cos 90^\circ} = \dfrac{1}{0}$, which is not defined. True.
2. (R) $\sin 90^\circ = 1$ and $\cos 90^\circ = 0$, so they are unequal. True. But being unequal is not why $\tan 90^\circ$ is undefined (that is because $\cos 90^\circ = 0$), so R does not explain A.

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

1. In cm: $4$ m $20$ cm $= 420$ cm and $5$ m $4$ cm $= 504$ cm. The longest plank that measures both exactly takes the fewest placements, and it is the HCF.
2. $420 = 2^2 \times 3 \times 5 \times 7$ and $504 = 2^3 \times 3^2 \times 7$. HCF $= 2^2 \times 3 \times 7 = 84$.

Answer: $84$ cm

1. (a) $32 + (n - 1)(-2) = -10$, so $-2(n - 1) = -42$ and $n = 22$. $S_{22} = \dfrac{22}{2}(32 + (-10)) = 11 \times 22 = 242$.
2. (b) $a_1 = 3(1) - 2 = 1$ and $a_{28} = 3(28) - 2 = 82$. $S_{28} = \dfrac{28}{2}(1 + 82) = 14 \times 83 = 1162$.

Answer: (a) $22$ terms, sum $242$   (b) $1162$

1. (a)(i) The diagonals of a square bisect each other, so $P$ is the mid-point of $BD$: $P = \left(\dfrac{9 + 1}{2}, \dfrac{-2 + 6}{2}\right) = (5, 2)$.
2. (a)(ii) $BD = \sqrt{(9 - 1)^2 + (-2 - 6)^2} = \sqrt{128} = 8\sqrt{2}$. A square of side $s$ has diagonal $s\sqrt{2}$, so $s = 8$.
3. (b) Let the point be $(x, y)$. $(x - 6)^2 + (y - 4)^2 = (x - 5)^2 + (y - 2)^2$ gives $-12x - 8y + 52 = -10x - 4y + 29$, that is $2x + 4y = 23$.
4. With $x + y = 5$: $2(5 - y) + 4y = 23$, so $2y = 13$ and $y = \dfrac{13}{2}$, $x = -\dfrac{3}{2}$. The point is $\left(-\dfrac{3}{2}, \dfrac{13}{2}\right)$.

Answer: (a) $P(5, 2)$, side $= 8$ units   (b) $\left(-\dfrac{3}{2}, \dfrac{13}{2}\right)$

1. $\dfrac{AO}{OC} = \dfrac{BO}{OD}$ means $\dfrac{AO}{BO} = \dfrac{OC}{OD}$. Also $\angle AOB = \angle COD$ (vertically opposite).
2. So $\triangle AOB \sim \triangle COD$ by SAS, and $\angle OAB = \angle OCD$. These are alternate angles for $AB$ and $DC$ with transversal $AC$, so $AB \parallel DC$.
3. One pair of opposite sides is parallel, so $ABCD$ is a trapezium.

Answer: Shown

1. Height of the cone $= 31 - 7 = 24$ cm and its radius $= 7$ cm. Slant height $l = \sqrt{7^2 + 24^2} = 25$ cm.
2. The total surface area is the curved area of the cone plus the curved area of the hemisphere (the joining base is hidden): $\pi r l + 2\pi r^2 = \dfrac{22}{7} \times 7 \times 25 + 2 \times \dfrac{22}{7} \times 49 = 550 + 308 = 858$ cm².

Answer: $858$ cm²

1. (a) Square: $\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 3$, so $\sin\theta\cos\theta = 1$.
2. $\tan\theta + \cot\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta} = 1$.
3. (Note: $\sin\theta\cos\theta \leq \dfrac{1}{2}$ for every real angle, so no real $\theta$ satisfies the given condition; the result follows from the algebra, which is what the question asks.)
4. (b) LHS $= \sin^2 A + 2\sin A\sec A + \sec^2 A + \cos^2 A + 2\cos A\operatorname{cosec} A + \operatorname{cosec}^2 A$. Here $\sin A\sec A = \tan A$ and $\cos A\operatorname{cosec} A = \cot A$, so LHS $= 1 + 2(\tan A + \cot A) + \sec^2 A + \operatorname{cosec}^2 A$.
5. $\tan A + \cot A = \dfrac{1}{\sin A\cos A} = \sec A\operatorname{cosec} A$ and $\sec^2 A + \operatorname{cosec}^2 A = \dfrac{1}{\cos^2 A\sin^2 A} = \sec^2 A\operatorname{cosec}^2 A$.
6. LHS $= 1 + 2\sec A\operatorname{cosec} A + \sec^2 A\operatorname{cosec}^2 A = (1 + \sec A\operatorname{cosec} A)^2 =$ RHS.

Answer: Proved (both parts)

1. (i) The major sector has central angle $360^\circ - 120^\circ = 240^\circ$. Perimeter $=$ major arc $+ 2r = \dfrac{240}{360} \times 2 \times \dfrac{22}{7} \times 7 + 14 = \dfrac{88}{3} + 14 = \dfrac{130}{3} \approx 43.3$ cm.
2. (ii) Segment $=$ minor sector $-$ triangle $= \dfrac{120}{360} \times \dfrac{22}{7} \times 7 \times 7 - 21.2 = \dfrac{154}{3} - 21.2 \approx 51.33 - 21.2 = 30.1$ cm².

Answer: (i) $\dfrac{130}{3} \approx 43.3$ cm   (ii) $\approx 30.1$ cm²

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 the integers be $x$ and $x + 1$. $x^2 + (x + 1)^2 = 481$, so $2x^2 + 2x - 480 = 0$ and $x^2 + x - 240 = 0$.
2. $(x + 16)(x - 15) = 0$, so $x = -16$ or $x = 15$. We need negative integers, so $x = -16$ and the other is $-15$.
3. (Check: $256 + 225 = 481$ ✓.)

Answer: $-16$ and $-15$

1. Let $AP : PB = k : 1$. The $y$-coordinate: $\dfrac{9k + 4}{k + 1} = 7$, so $9k + 4 = 7k + 7$ and $k = \dfrac{3}{2}$. The ratio is $3 : 2$.
2. $x = \dfrac{3(7) + 2(-5)}{3 + 2} = \dfrac{21 - 10}{5} = \dfrac{11}{5}$.

Answer: Ratio $3 : 2$; $x = \dfrac{11}{5}$

1. (a) $OA \perp AP$, so $\angle PAB = 90^\circ - \angle OAB$. (i)
2. Tangents from $P$ are equal, $PA = PB$, so $\angle PAB = \angle PBA$ and $\angle APB = 180^\circ - 2\angle PAB$.
3. Using (i): $\angle APB = 180^\circ - 2(90^\circ - \angle OAB) = 2\angle OAB$.
4. (b) In the right triangle $OBA$: $OB = \sqrt{10^2 - 8^2} = 6$ cm. Let $PB = y$, so $OP = y + 6$.
5. $PA^2 = PB^2 + AB^2 = y^2 + 64$ (right angle at $B$) and $PA^2 = OP^2 - OA^2 = (y + 6)^2 - 100$ (right angle at $A$).
6. $y^2 + 64 = y^2 + 12y + 36 - 100$, so $12y = 128$ and $y = \dfrac{32}{3}$ cm $\approx 10.67$ cm.

Answer: (a) proved   (b) $PB = \dfrac{32}{3}$ cm

1. (a) Cumulative frequencies: $4, 9, 9 + x, 29 + x, 43 + x, 43 + x + y, 47 + x + y$. The total is $47 + x + y = 68$, so $x + y = 21$.
2. $\dfrac{N}{2} = 34$ and the median $137$ lies in $125 - 145$: $l = 125$, $cf = 9 + x$, $f = 20$, $h = 20$. Median $= 125 + \dfrac{34 - (9 + x)}{20} \times 20 = 125 + 25 - x = 150 - x$.
3. $150 - x = 137$, so $x = 13$ and $y = 21 - 13 = 8$.
4. (b) The table is: classes $0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70$ with frequencies $3, 6, 11, 10, 13, 3, 4$ (total $50$). Take $a = 35$, $h = 10$, $u = \dfrac{x - 35}{10} = -3, -2, -1, 0, 1, 2, 3$.
5. $\sum f_iu_i = -9 - 12 - 11 + 0 + 13 + 6 + 12 = -1$. Mean $= 35 + \dfrac{-1}{50} \times 10 = 34.8$.
6. Modal class $40 - 50$ ($f_1 = 13$, $f_0 = 10$, $f_2 = 3$): Mode $= 40 + \dfrac{13 - 10}{26 - 10 - 3} \times 10 = 40 + \dfrac{30}{13} \approx 42.3$.

Answer: (a) $x = 13$, $y = 8$   (b) mean $= 34.8$, mode $\approx 42.3$

1. Draw $DZ \parallel BY$ with $Z$ on $AC$ (the $Z$ lies between $Y$ and $C$).
2. In $\triangle CBY$, $D$ is the mid-point of $BC$ and $DZ \parallel BY$, so $Z$ is the mid-point of $CY$ and $DZ = \dfrac{1}{2}BY$, that is $BY = 2DZ$. (i)
3. In $\triangle ADZ$, $XY \parallel DZ$, so $\triangle AXY \sim \triangle ADZ$: $\dfrac{AX}{AD} = \dfrac{XY}{DZ}$. $AX : AD = 2 : 5$, so $DZ = \dfrac{5}{2}XY$. (ii)
4. From (i) and (ii): $BY = 2 \times \dfrac{5}{2}XY = 5XY$.
5. $BX = BY - XY = 5XY - XY = 4XY$.

Answer: Proved

1. (a) $x - 2y = 3$: points $(3, 0)$, $(5, 1)$, $(1, -1)$. $3x - 8y = 7$: points $(5, 1)$, $(-3, -2)$, $(1, -\tfrac{1}{2})$. The lines meet at $(5, 1)$, so $x = 5$, $y = 1$. (Check: $5 - 2 = 3$ ✓, $15 - 8 = 7$ ✓.)
2. (b)(i) Let Adil be $x$ and Bharat $y$ years old. Five years ago: $x - 5 = 3(y - 5)$, that is $x - 3y = -10$. Ten years later: $x + 10 = 2(y + 10)$, that is $x - 2y = 10$.
3. (ii) $\dfrac{a_1}{a_2} = \dfrac{1}{1} = 1$ and $\dfrac{b_1}{b_2} = \dfrac{-3}{-2} = \dfrac{3}{2}$. Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the system is consistent with a unique solution.
4. (iii) Subtracting: $(x - 2y) - (x - 3y) = 10 - (-10)$ gives $y = 20$, and then $x = 10 + 40 = 50$. Adil is $50$ and Bharat is $20$ years old.

Answer: (a) $x = 5$, $y = 1$   (b) Adil $50$ years, Bharat $20$ years