CBSE Class 10 Maths Standard 2026 Question Paper (Set 30/4/1) with Solutions

1. Equal roots need $D = 0$: $(8k)^2 - 4(9)(16) = 64k^2 - 576 = 0$, so $k^2 = 9$ and $k = \pm 3$.
2. Both $3$ and $-3$ are in the options, so options (A) and (B) are both correct. CBSE's key accepts either.

Answer: (A) 3 or (B) −3

Common mistake: Picking only one sign. The equation gives k² = 9, so k can be 3 or −3, and both are among the options.

1. The correspondence is $A \leftrightarrow E$, $B \leftrightarrow D$, $C \leftrightarrow F$.
2. (A), (B) and (D) are true: the perimeters and all pairs of corresponding sides are in the same ratio, and $AB + BC$ matches $ED + DF$ while $AC$ matches $EF$.
3. (C) says $\angle A = \angle D$, but $\angle A$ corresponds to $\angle E$, so this is not true (the true statement is $\angle A = \angle E$, $\angle C = \angle F$).

Answer: (C) $\angle A = \angle D,\ \angle C = \angle F$

1. $a_{n+1} - a_n = (\sqrt{2}(n + 1) + 1) - (\sqrt{2}n + 1) = \sqrt{2}$.

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

1. $OQ \perp PQ$, $OR \perp PR$ and $\angle QPR = 90^\circ$, so $OQPR$ has four right angles. $OQ = OR = 4$ and $PQ = PR$, so it is a square of side $4$.
2. $OP$ is its diagonal: $OP = 4\sqrt{2}$ cm.

Answer: (B) $4\sqrt{2}$ cm

1. The zeroes are the points where the graph meets the $x$-axis. The curve crosses the axis three times (up, down, up), so there are $3$ zeroes.

Answer: (D) 3

1. $DE \parallel BC$, so $\dfrac{AE}{EC} = \dfrac{AD}{DB} = \dfrac{1}{3}$. So $AE : AC = 1 : 4$.
2. $AE = \dfrac{6}{4} = 1.5$ cm.

Answer: (A) 1.5 cm

1. The circle passes through the origin, so the radius is the distance from the centre $(1, 2)$ to $(0, 0)$: $\sqrt{1^2 + 2^2} = \sqrt{5}$.
2. Diameter $= 2\sqrt{5}$.

Answer: (D) $2\sqrt{5}$

1. The first equation is $3x + 2y = 30$. Inconsistent (parallel lines) needs $\dfrac{3}{2} = \dfrac{2}{k} \neq \dfrac{30}{7}$.
2. $\dfrac{3}{2} = \dfrac{2}{k}$ gives $k = \dfrac{4}{3}$, and then $\dfrac{2}{k} = \dfrac{3}{2} \neq \dfrac{30}{7}$ ✓.

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

1. Area of the circle $= \dfrac{22}{7} \times 7 \times 7 = 154$ cm². Each of $16$ sectors: $\dfrac{154}{16} = \dfrac{77}{8}$ cm².

Answer: (D) $\dfrac{77}{8}$ cm²

1. Each die must show $1, 2$ or $3$: $3 \times 3 = 9$ outcomes out of $36$. $P = \dfrac{9}{36} = \dfrac{1}{4}$.

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

1. Opposite $= 1$, hypotenuse $= 3$, adjacent $= \sqrt{9 - 1} = 2\sqrt{2}$. $\cot A = \dfrac{\text{adjacent}}{\text{opposite}} = 2\sqrt{2}$.

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

1. $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \operatorname{cosec}^2 A$. The ratio is $\dfrac{\sec^2 A}{\operatorname{cosec}^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A$.

Answer: (A) $\tan^2 A$

1. The jar has radius $r$ and height $3 \times 2r = 6r$: volume $\pi r^2 \times 6r = 6\pi r^3$.
2. The three balls: $3 \times \dfrac{4}{3}\pi r^3 = 4\pi r^3$. Air $= 6\pi r^3 - 4\pi r^3 = 2\pi r^3$.

Answer: (A) $2\pi r^3$

1. $SP$ is a diameter, and the angle at the circle is half the angle at the centre on the same arc $PR$: $\angle PSR = \dfrac{65^\circ}{2} = 32.5^\circ$.
2. $SP \perp PT$ (radius and tangent), so in the right triangle $SPT$: $\angle PTR = 90^\circ - 32.5^\circ = 57.5^\circ$.

Answer: (C) $57.5^\circ$

1. $d^2 = (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2$.
2. $= a^2\cos^2\theta + 2ab\sin\theta\cos\theta + b^2\sin^2\theta + a^2\sin^2\theta - 2ab\sin\theta\cos\theta + b^2\cos^2\theta = a^2 + b^2$. So $d = \sqrt{a^2 + b^2}$.

Answer: (A) $\sqrt{a^2 + b^2}$

1. Two scoops: $2 \times \dfrac{4}{3}\pi\left(\dfrac{r}{2}\right)^3 = \dfrac{8}{3}\pi \times \dfrac{r^3}{8} = \dfrac{\pi r^3}{3}$.
2. The cone: $\dfrac{1}{3}\pi r^2 h$. Setting them equal, $h = r$, so $h : 2r = r : 2r = 1 : 2$.

Answer: (B) 1 : 2

1. Arc $= \dfrac{\theta}{360} \times 2\pi r$: $11 = \dfrac{\theta}{360} \times 2 \times \dfrac{22}{7} \times 6.3 = \dfrac{\theta}{360} \times 39.6$.
2. $\theta = \dfrac{11 \times 360}{39.6} = 100^\circ$.

Answer: (D) $100^\circ$

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 3(40) - 2(43) = 120 - 86 = 34$.

Answer: (A) 34

1. (A) A probability cannot exceed $1$, so $0.2p \leq 1$ and $p \leq 5$. True.
2. (R) The complement rule is true. But the bound on $p$ comes from probability being at most $1$, not from the complement rule, so R does not explain A.

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

1. (A) $\sqrt{3} + \sqrt{5}$ is irrational (if it were rational $= r$, then squaring $8 + 2\sqrt{15} = r^2$ would make $\sqrt{15}$ rational, which it is not). True.
2. (R) False: $\sqrt{2}$ and $-\sqrt{2}$ are irrational but their sum is $0$, a rational number.

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

1. $D = (q - r)^2 - 4(p - q)(r - p)$. Expand: $(q - r)^2 - 4(pr - p^2 - qr + pq) = q^2 - 2qr + r^2 - 4pr + 4p^2 + 4qr - 4pq = q^2 + 2qr + r^2 + 4p^2 - 4pr - 4pq = (q + r - 2p)^2$.
2. If $q + r = 2p$, then $D = 0$, so the roots are real and equal.

Answer: Verified

1. In $\triangle CDA$ and $\triangle CAB$: $\angle C$ is common and $\angle CDA = \angle CAB$ (given).
2. So $\triangle CDA \sim \triangle CAB$ by AA, and $\dfrac{CD}{CA} = \dfrac{CA}{CB}$.
3. Cross-multiplying, $CA^2 = CB \times CD$.

Answer: Shown

1. (a) Multiply the top and bottom by $(1 + \sin A)$: $\sqrt{\dfrac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\dfrac{(1 + \sin A)^2}{\cos^2 A}} = \dfrac{1 + \sin A}{\cos A} = \sec A + \tan A$.
2. (b) $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, $\operatorname{cosec} 30^\circ = 2$, $\tan 60^\circ = \sqrt{3}$.
3. $\dfrac{3 \times \frac{3}{4} - 6 \times 4}{3} = \dfrac{\frac{9}{4} - 24}{3} = \dfrac{-\frac{87}{4}}{3} = -\dfrac{29}{4}$.

Answer: (a) proved   (b) $-\dfrac{29}{4}$

1. (a) Suppose $2 + 3\sqrt{5} = \dfrac{p}{q}$ with integers $p, q$ and $q \neq 0$. Then $\sqrt{5} = \dfrac{p - 2q}{3q}$, a ratio of integers, so $\sqrt{5}$ would be rational. That contradicts the given fact, so $2 + 3\sqrt{5}$ is irrational.
2. (b) $210 = 2 \times 3 \times 5 \times 7$ and $55 = 5 \times 11$, so HCF $= 5$. $5 = 1050 + 55m$ gives $55m = -1045$ and $m = -19$.

Answer: (a) proved   (b) $m = -19$

1. $\alpha + \beta = \dfrac{16}{5}$ and $\alpha\beta = \dfrac{-10}{5} = -2$.
2. $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \dfrac{\frac{256}{25} + 4}{-2} = \dfrac{\frac{356}{25}}{-2} = -\dfrac{178}{25}$.

Answer: $-\dfrac{178}{25}$

1. (a) Let Veer have $V$ and Kevin $K$. $V = 2K + 6$ and $V + K + 4 = 40$ (one of them having $4$ more adds $4$ to the total).
2. $2K + 6 + K = 36$, so $3K = 30$ and $K = 10$, $V = 26$. (Check: $26 = 2 \times 10 + 6$ ✓ and $26 + 10 + 4 = 40$ ✓.)
3. (b) $3x + y = 14$: points $(4, 2)$, $(3, 5)$ and $(2, 8)$. $y = 2$ is the horizontal line through $(0, 2)$. They meet at $(4, 2)$: $x = 4$, $y = 2$.

Answer: (a) Veer $26$, Kevin $10$   (b) $x = 4$, $y = 2$

1. The shaded region is the minor segment, bounded by the minor arc $AB$ and the chord $AB$.
2. Arc $AB = \dfrac{120}{360} \times 2 \times \dfrac{22}{7} \times 21 = 44$ mm.
3. Chord: draw $OC \perp AB$; it bisects the angle and the chord, so $\angle AOC = 60^\circ$ and $AC = 21\sin 60^\circ = \dfrac{21\sqrt{3}}{2}$. $AB = 21\sqrt{3} = 21 \times 1.73 = 36.33$ mm.
4. Perimeter $= 44 + 36.33 = 80.33$ mm.

Answer: $44 + 21\sqrt{3} \approx 80.33$ mm

1. Divide the top and bottom of the LHS by $\cos\theta$: $\dfrac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta}$.
2. Write $1 = \sec^2\theta - \tan^2\theta$ in the top: $\tan\theta + \sec\theta - (\sec^2\theta - \tan^2\theta) = (\sec\theta + \tan\theta)(1 - \sec\theta + \tan\theta)$.
3. So the LHS $= \dfrac{(\sec\theta + \tan\theta)(1 - \sec\theta + \tan\theta)}{\tan\theta + 1 - \sec\theta} = \sec\theta + \tan\theta$.
4. Multiply and divide by $(\sec\theta - \tan\theta)$: $\dfrac{\sec^2\theta - \tan^2\theta}{\sec\theta - \tan\theta} = \dfrac{1}{\sec\theta - \tan\theta} =$ RHS.

Answer: Proved

1. (a) The cloth covers the four walls and the semi-cylindrical roof with its two semicircular ends (the floor is not covered).
2. Four walls: $2(14 \times 16 + 25 \times 16) = 2(224 + 400) = 1248$ m².
3. The semi-cylinder has radius $7$ m and length $25$ m. Curved area $= \pi r l = \dfrac{22}{7} \times 7 \times 25 = 550$ m². The two semicircular ends: $\pi r^2 = \dfrac{22}{7} \times 49 = 154$ m². Roof $= 704$ m².
4. Total cloth $= 1248 + 704 = 1952$ m².
5. (b) Cone slant height $l = \sqrt{(5\sqrt{2})^2 + (5\sqrt{7})^2} = \sqrt{50 + 175} = 15$ cm.
6. The surface of the object: the outer curved area of the hemisphere $2\pi R^2$, the cone's curved area $\pi r l$, and the ring-shaped rim of the hollow hemisphere $\pi(R^2 - r^2)$.
7. $= \pi(2 \times 100 + 5\sqrt{2} \times 15 + 100 - 50) = \pi(200 + 75\sqrt{2} + 50) = \pi(250 + 105) = 355\pi$ cm² (with $\sqrt{2} = 1.4$).

Answer: (a) $1952$ m²   (b) $355\pi$ cm²

1. (i) $P(\text{not blue}) = \dfrac{30 - m}{30} = 1 - \dfrac{m}{30}$.
2. (ii) Now there are $36$ balls and $m + 6$ blue, so $P(\text{blue}) = \dfrac{m + 6}{36}$. The condition: $\dfrac{m + 6}{36} = \dfrac{5}{4} \times \dfrac{m}{30} = \dfrac{m}{24}$.
3. $24(m + 6) = 36m$, so $24m + 144 = 36m$ and $m = 12$. (Check: $\dfrac{18}{36} = \dfrac{1}{2}$ and $\dfrac{5}{4} \times \dfrac{12}{30} = \dfrac{1}{2}$ ✓.)

Answer: (i) $\dfrac{30 - m}{30}$   (ii) $m = 12$

1. $48 = 2^4 \times 3$, $60 = 2^2 \times 3 \times 5$, $65 = 5 \times 13$. LCM $= 2^4 \times 3 \times 5 \times 13 = 3120$.
2. Numbers divisible by all three are multiples of $3120$: $3120, 6240, 9360, 12480, \dots$. The greatest below $10000$ is $3 \times 3120 = 9360$.

Answer: $9360$

1. (a) Let the tour be $x$ days: daily expense $\dfrac{5400}{x}$, and after the extension $\dfrac{5400}{x + 5}$. $\dfrac{5400}{x} - \dfrac{5400}{x + 5} = 180$.
2. $5400 \times 5 = 180x(x + 5)$, so $x^2 + 5x - 150 = 0$ and $(x + 15)(x - 10) = 0$. Days cannot be negative: $x = 10$.
3. The tour was $10$ days and the daily expense ₹ $\dfrac{5400}{10} = 540$.
4. (b) Let the length be $x$ m, so the price per metre is $\dfrac{2100}{x}$. With $2$ m extra the price is $\dfrac{2100}{x + 2}$, and $\dfrac{2100}{x} - \dfrac{2100}{x + 2} = 120$.
5. $2100 \times 2 = 120x(x + 2)$, so $x^2 + 2x - 35 = 0$ and $(x + 7)(x - 5) = 0$. $x = 5$ m (reject $-7$).
6. The original price per metre is $\dfrac{2100}{5} = ₹420$. (Check: $\dfrac{2100}{7} = 300$ and $420 - 300 = 120$ ✓.)

Answer: (a) $10$ days, ₹ $540$ a day   (b) ₹ $420$ per metre, $5$ m

1. (i) $MQ \perp QT$ and $NB \perp QT$ (radius to the point of contact), so $MQ \parallel NB$. $MN$ is not parallel to $QB$ (the radii differ), so exactly one pair of opposite sides is parallel: $MQBN$ is a trapezium.
2. (ii) No. $PM \perp PA$ and $NA \perp PA$, so $PM \parallel NA$. If $MN \parallel PA$ too, then $PMNA$ would be a parallelogram and $PM = NA$, but $PM = 13 \neq 8 = NA$.
3. (iii) Tangents from $T$ to the circle with centre $M$ are equal: $TQ = TP = 80$ cm. $TB = TQ - QB = 80 - 35 = 45$ cm.
4. (iv) Draw $NS \perp MQ$ with $S$ on $MQ$. Then $NS = QB = 35$ and $MS = MQ - NB = 13 - 8 = 5$. $MN^2 = NS^2 + MS^2 = 1225 + 25 = 1250$, so $MN = 25\sqrt{2}$ cm.
5. (Note: the data are not fully consistent; if $M$, $N$, $T$ were assumed collinear, similar triangles would give $TB = 56$ cm, which CBSE's key also accepts.)

Answer: (i) trapezium   (ii) no   (iii) $45$ cm   (iv) $25\sqrt{2}$ cm

1. Let the kite be $K$, the roof $R$ and the foot of the building $G$. From $G$: $\tan 45^\circ = \dfrac{60}{GT} = 1$, so the kite is $GT = 60$ m away horizontally.
2. From the roof, the horizontal distance is the same $60$ m. The kite is $KS$ above the roof level: $\tan 30^\circ = \dfrac{KS}{60}$, so $KS = \dfrac{60}{\sqrt{3}} = 20\sqrt{3} = 34.6$ m.
3. Height of the roof $= 60 - 20\sqrt{3} = 60 - 34.6 = 25.4$ m.
4. String: $\sin 30^\circ = \dfrac{KS}{KR}$, so $KR = 2 \times 20\sqrt{3} = 40\sqrt{3} = 69.2$ m.

Answer: String $= 40\sqrt{3} \approx 69.2$ m; roof height $\approx 25.4$ m

1. (a) Cumulative frequencies: $p$, $p + 15$, $p + 40$, $p + 60$, $p + q + 60$, $p + q + 68$, $p + q + 78$. Total $p + q + 78 = 90$, so $p + q = 12$.
2. The median is $50$, which is the lower limit of the class $50 - 60$, so that is the median class: $l = 50$, $cf = p + 40$, $f = 20$, $h = 10$, $\dfrac{N}{2} = 45$. $50 + \dfrac{45 - (p + 40)}{20} \times 10 = 50$ gives $45 - p - 40 = 0$, so $p = 5$.
3. $q = 12 - 5 = 7$.
4. (b) $N = 60$. Take $a = 52.5$, $h = 15$: $u = \dfrac{x - 52.5}{15} = -3, -2, -1, 0, 1, 2, 3$ for the class marks $7.5, 22.5, 37.5, 52.5, 67.5, 82.5, 97.5$. $\sum f_iu_i = -12 - 16 - 11 + 0 + 10 + 14 + 18 = 3$.
5. Mean $= 52.5 + \dfrac{3}{60} \times 15 = 52.5 + 0.75 = 53.25$.
6. Modal class $45 - 60$ ($f_1 = 14$, $f_0 = 11$, $f_2 = 10$): Mode $= 45 + \dfrac{14 - 11}{28 - 11 - 10} \times 15 = 45 + \dfrac{45}{7} \approx 51.4$.

Answer: (a) $p = 5$, $q = 7$   (b) mean $= 53.25$, mode $\approx 51.4$

1. (i) $TR \perp AB$, so $\angle PRT = 90^\circ - \alpha$ and $\angle CRT = 90^\circ - \theta$ (the angles on a straight line at $R$ with the normal $TR$). The law of reflection gives $\angle PRT = \angle CRT$, so $\alpha = \theta$.
2. (ii) In $\triangle PQR$ and $\triangle CBR$: $\angle PRQ = \alpha = \theta = \angle CRB$ and $\angle PQR = \angle CBR = 90^\circ$. So $\triangle PQR \sim \triangle CBR$ by AA.
3. (iii)(a) $QR = AB - AQ - BR = 65 - 9 - x = 56 - x$ (since $AQ = PS = 9$). From the similarity, $\dfrac{PQ}{CB} = \dfrac{QR}{BR}$: $\dfrac{35}{65} = \dfrac{56 - x}{x}$.
4. $35x = 65(56 - x) = 3640 - 65x$, so $100x = 3640$ and $x = 36.4$ cm.
5. (iii)(b) The triangles are similar, so the ratio of their areas is the square of the ratio of corresponding sides: $\dfrac{\text{Area }\triangle PQR}{\text{Area }\triangle CBR} = \left(\dfrac{QR}{BR}\right)^2$. Equating with $\dfrac{PQ^2}{CB^2}$ gives $\dfrac{QR}{BR} = \dfrac{PQ}{CB} = \dfrac{35}{65}$, so $\dfrac{56 - x}{x} = \dfrac{35}{65}$ and $x = 36.4$ cm.