CBSE Class 10 Maths Standard 2026 Question Paper (Set 30/5/1) with Solutions

1. Equal roots need $D = 0$: $(-6)^2 - 4(k)(-4) = 36 + 16k = 0$, so $k = -\dfrac{36}{16} = -\dfrac{9}{4}$.

Answer: (C) $-\dfrac{9}{4}$

1. A point on the $y$-axis has $x = 0$. Let it divide $PQ$ in the ratio $k : 1$: $\dfrac{10k + (-4)}{k + 1} = 0$.
2. $10k = 4$, so $k = \dfrac{2}{5}$. The ratio is $2 : 5$.

Answer: (A) 2 : 5

1. A polynomial with zeroes $-3$ and $8$ has the form $k(x + 3)(x - 8)$ with $k \neq 0$.
2. Option (B) $(x + 3)(-x + 8) = -(x + 3)(x - 8)$ is of this form with $k = -1$. The others have different zeroes: $x^2 + 5x - 24 = (x + 8)(x - 3)$ has zeroes $-8$ and $3$.

Answer: (B) $(x + 3)(-x + 8)$

1. $P$, $Q$, $T$ are on a tangent line, so $\angle PQS = 180^\circ - 55^\circ = 125^\circ$. $OQ \perp PQ$ gives $\angle PQO = 90^\circ$, so $\angle OQS = 125^\circ - 90^\circ = 35^\circ$.
2. $OQ = OS$, so $\angle OSQ = \angle OQS = 35^\circ$.
3. In $\triangle PQS$ (with $P$, $O$, $S$ in a line): $\angle QPS = 180^\circ - 125^\circ - 35^\circ = 20^\circ$.

Answer: (B) $20^\circ$

1. For SAS with the equal angles at $C$ and $R$, the two sides that include them must be in proportion: the sides of $\angle C$ are $CA$ and $CB$, and of $\angle R$ are $RP$ and $RQ$.
2. $\dfrac{CA}{RP} = \dfrac{CB}{RQ}$, that is $\dfrac{AC}{BC} = \dfrac{PR}{QR}$.

Answer: (D) $\dfrac{AC}{BC} = \dfrac{PR}{QR}$

1. $\bar{x} = a + h\bar{u}$, so $\bar{u} = \dfrac{\bar{x} - a}{h} = \dfrac{64 - 62.5}{5} = \dfrac{1.5}{5} = 0.3$.

Answer: (C) 0.3

1. The largest cone has the base of the hemisphere as its base and its tip on the pole: radius $r = 10$ and height $h = 10$.
2. Slant height $l = \sqrt{10^2 + 10^2} = 10\sqrt{2}$. Curved surface area $= \pi r l = 3.14 \times 10 \times 10\sqrt{2} = 314\sqrt{2}$ cm².

Answer: (A) $314\sqrt{2}$ cm²

1. $d = -\dfrac{10}{4} + \dfrac{15}{4} = \dfrac{5}{4}$. $a_{16} - a_{12} = (16 - 12)d = 4 \times \dfrac{5}{4} = 5$.

Answer: (C) 5

1. The roots are rational when $D = p^2 - 24$ is a perfect square.
2. Check: $p = 1$ gives $-23$ (no real roots), $p = 25$ gives $601$ (not a square), $p = \sqrt{5}$ gives $-19$. $p = -5$ gives $25 - 24 = 1$, a perfect square: $x^2 + 5x + 6 = (x + 2)(x + 3)$ with roots $-2$ and $-3$.

Answer: (B) −5

1. A probability must lie between $0$ and $1$.
2. $\dfrac{10}{0.2} = 50 > 1$, so it cannot be a probability. The others are $0.39$, $0.00005$ and $0.1$, all between $0$ and $1$.

Answer: (C) $\dfrac{10}{0.2}$

1. There are $4$ tens and $4$ kings, $8$ cards in all. $P = \dfrac{8}{52} = \dfrac{2}{13}$.

Answer: (B) $\dfrac{2}{13}$

1. Curved surface of the hemisphere $= 2\pi r^2 = 2 \times \dfrac{22}{7} \times 1.4 \times 1.4 = 12.32$ m².
2. The door is an opening, so subtract it: $12.32 - 0.50 = 11.82$ m².

Answer: (C) 11.82 m²

1. Arc $= \dfrac{\theta}{360} \times 2\pi r$: $2.2 = \dfrac{\theta}{360} \times 2 \times \dfrac{22}{7} \times 2.8 = \dfrac{\theta}{360} \times 17.6$, so $\theta = \dfrac{2.2 \times 360}{17.6} = 45^\circ$.

Answer: (C) $45^\circ$

1. $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta} = 9$, so $9\operatorname{cosec}\theta = 81$. The expression is $\dfrac{81 + 1}{81 - 1} = \dfrac{82}{80}$.

Answer: (D) $\dfrac{82}{80}$

1. $PQ \parallel YZ$, so $\triangle XPQ \sim \triangle XYZ$ and $\dfrac{PQ}{YZ} = \dfrac{XP}{XY}$.
2. $XP : PY = 2 : 3$ gives $XP : XY = 2 : 5$. $\dfrac{5}{YZ} = \dfrac{2}{5}$, so $YZ = 12.5$ cm.

Answer: (A) 12.5 cm

1. $P = \dfrac{\text{her tickets}}{800} = 0.08$, so her tickets $= 0.08 \times 800 = 64$.

Answer: (A) 64

1. In the right triangle $ABC$: $\cos 60^\circ = \dfrac{AB}{AC}$, so $AC = \dfrac{5\sqrt{3}}{1/2} = 10\sqrt{3}$ m.

Answer: (B) $10\sqrt{3}$ m

1. $\sec^2 A - 1 = \tan^2 A$, so $\sqrt{\sec^2 A - 1} = \tan A$. $\dfrac{\sec A}{\tan A} = \dfrac{1/\cos A}{\sin A/\cos A} = \dfrac{1}{\sin A} = \operatorname{cosec} A$.

Answer: (C) $\operatorname{cosec} A$

1. (A) $\dfrac{a_1}{a_2} = \dfrac{3}{-6} = -\dfrac{1}{2}$, $\dfrac{b_1}{b_2} = \dfrac{-5}{10} = -\dfrac{1}{2}$ and $\dfrac{c_1}{c_2} = \dfrac{7}{14} = \dfrac{1}{2}$. The first two are equal and the third is different: parallel lines, inconsistent. True.
2. (R) Without a unique solution the lines may also be coincident (infinitely many solutions), not only parallel. False.

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

1. (A) $36m^2 = 2 \times 18 \times m \times m$ and $18m$ divides $36m^2$ (as $36m^2 = 18m \times 2m$). So the HCF is $18m$. True.
2. (R) The HCF divides both numbers, so it cannot exceed the smaller one. True. But it does not tell us that the HCF is $18m$ here, so it is not the explanation.

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

1. Suppose $2 - 5\sqrt{3} = \dfrac{a}{b}$ with integers $a, b$ and $b \neq 0$.
2. Then $\sqrt{3} = \dfrac{2b - a}{5b}$, a ratio of integers, so $\sqrt{3}$ would be rational. That contradicts the given fact, so $2 - 5\sqrt{3}$ is irrational.

Answer: Proved

1. (a) $BC = \sqrt{(-8 - 1)^2 + (7 - 1)^2} = \sqrt{117} = 3\sqrt{13}$ and $AB = \sqrt{(3 - 1)^2 + (4 - 1)^2} = \sqrt{13}$.
2. $\tan A = \dfrac{BC}{AB} = \dfrac{3\sqrt{13}}{\sqrt{13}} = 3$ (opposite over adjacent, as the right angle is at $B$).
3. (b) $AB = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}$, $BC = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$ and $AC = \sqrt{9 + 1} = \sqrt{10}$.
4. $AC + BC = \sqrt{10} + 2\sqrt{10} = 3\sqrt{10} = AB$, so the points are collinear.

Answer: (a) $\tan A = 3$   (b) collinear

1. $AB \parallel DE$, so $\angle DEF = \angle ABC$ (corresponding angles). $AC \parallel DF$, so $\angle DFE = \angle ACB$. By AA, $\triangle ABC \sim \triangle DEF$.
2. $EF = EB + BC + CF = 5 + 10 + 5 = 20$ cm. $\dfrac{BC}{EF} = \dfrac{AB}{DE}$: $\dfrac{10}{20} = \dfrac{7}{DE}$, so $DE = 14$ cm.

Answer: $DE = 14$ cm

1. (a) $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, so $\sin^3 60^\circ = \dfrac{3\sqrt{3}}{8}$. $\tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$. $\cos^2 45^\circ = \dfrac{1}{2}$.
2. $\dfrac{\frac{3\sqrt{3}}{8} - \frac{\sqrt{3}}{3}}{\frac{1}{2}} = 2 \times \sqrt{3}\left(\dfrac{9 - 8}{24}\right) = \dfrac{\sqrt{3}}{12}$ (which is the same as $\dfrac{1}{4\sqrt{3}}$).
3. (b) $A + 2B = 60^\circ$ and $2A + B = 45^\circ$. Adding: $3(A + B) = 105^\circ$, so $A + B = 35^\circ$. Then $B = 60^\circ - 35^\circ = 25^\circ$ and $A = 10^\circ$.

Answer: (a) $\dfrac{\sqrt{3}}{12}$   (b) $A = 10^\circ$, $B = 25^\circ$

1. $P(\text{yellow}) = 1 - P(\text{green}) = 1 - \dfrac{3}{5} = \dfrac{2}{5}$.
2. Yellow balls $= \dfrac{2}{5} \times 25 = 10$.

Answer: $10$ yellow balls

1. Radius $= 10$ cm. Each sector has central angle $\dfrac{360^\circ}{5} = 72^\circ$.
2. Area $= \dfrac{72}{360} \times \dfrac{22}{7} \times 10 \times 10 = \dfrac{440}{7} \approx 62.86$ cm².
3. Perimeter $=$ arc $+ 2r = \dfrac{72}{360} \times 2 \times \dfrac{22}{7} \times 10 + 10 + 10 = \dfrac{88}{7} + 20 = \dfrac{228}{7} \approx 32.57$ cm.

Answer: Area $= \dfrac{440}{7} \approx 62.86$ cm²; perimeter $= \dfrac{228}{7} \approx 32.57$ cm

1. (a) $a_{15} - a_8 = 7d = 21$, so $d = 3$. $S_{10} = \dfrac{10}{2}[2a + 9 \times 3] = 55$ gives $2a + 27 = 11$ and $a = -8$.
2. The A.P. is $-8, -5, -2, 1, \dots$.
3. (b) $a_1 = S_1 = 2 + 13 = 15$. $S_2 = 8 + 26 = 34$, so $a_2 = 34 - 15 = 19$ and $d = 4$.
4. $a_n = 15 + (n - 1) \times 4 = 4n + 11$. $a_{10} = 40 + 11 = 51$.

Answer: (a) $-8, -5, -2, \dots$   (b) $a_n = 4n + 11$, $a_{10} = 51$

1. LHS $= \dfrac{\frac{\sin\theta}{\cos\theta}}{1 - \frac{\cos\theta}{\sin\theta}} + \dfrac{\frac{\cos\theta}{\sin\theta}}{1 - \frac{\sin\theta}{\cos\theta}} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} + \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)}$.
2. $= \dfrac{1}{\sin\theta - \cos\theta}\left[\dfrac{\sin^2\theta}{\cos\theta} - \dfrac{\cos^2\theta}{\sin\theta}\right] = \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}$.
3. $\sin^3\theta - \cos^3\theta = (\sin\theta - \cos\theta)(\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta) = (\sin\theta - \cos\theta)(1 + \sin\theta\cos\theta)$.
4. LHS $= \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta} + 1 = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} + 1 = \tan\theta + \cot\theta + 1 =$ RHS.

Answer: Proved

1. (a) The centre $O(2, 1)$ is equidistant from $A$ and $B$: $OA = OB$, so $(5 - 2)^2 + (6 - 1)^2 = (-3 - 2)^2 + (K - 1)^2$.
2. $34 = 25 + (K - 1)^2$, so $(K - 1)^2 = 9$ and $K = 4$ or $K = -2$.
3. For $K = -2$: $AB = \sqrt{(5 + 3)^2 + (6 + 2)^2} = \sqrt{128} = 8\sqrt{2}$. For $K = 4$: $AB = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$.
4. (b) $P = \left(\dfrac{3(4) + 2(-1)}{5}, \dfrac{3(-3) + 2(7)}{5}\right) = (2, 1)$. On the line: $x - 3y = 2 - 3 = -1$ ✓.
5. $PA = \sqrt{(2 + 1)^2 + (1 - 7)^2} = \sqrt{45} = 3\sqrt{5}$ and $PB = \sqrt{(2 - 4)^2 + (1 + 3)^2} = \sqrt{20} = 2\sqrt{5}$ (ratio $3 : 2$ ✓).

Answer: (a) $K = 4$ ($AB = 2\sqrt{17}$) or $K = -2$ ($AB = 8\sqrt{2}$)   (b) $P(2, 1)$ on the line, $PA = 3\sqrt{5}$, $PB = 2\sqrt{5}$

1. $x = 4$ is a vertical line through $(4, 0)$.
2. $3x - 2y = 6$: when $x = 2$, $y = 0$; when $x = 4$, $y = 3$; when $x = 0$, $y = -3$.
3. The lines meet at $(4, 3)$, so $x = 4$ and $y = 3$.

Answer: $x = 4$, $y = 3$

1. $156 = 2^2 \times 3 \times 13$ and $216 = 2^3 \times 3^3$. The largest square side that fits an exact number of times along both is the HCF $= 2^2 \times 3 = 12$ cm.
2. Number of squares $= \dfrac{156}{12} \times \dfrac{216}{12} = 13 \times 18 = 234$.

Answer: Side $12$ cm; $234$ squares

1. Let the stream go at $x$ km/h. Upstream speed $= 12 - x$ and downstream speed $= 12 + x$. $\dfrac{15}{12 - x} + \dfrac{18}{12 + x} = 3$.
2. $15(12 + x) + 18(12 - x) = 3(144 - x^2)$, so $396 - 3x = 432 - 3x^2$ and $3x^2 - 3x - 36 = 0$, that is $x^2 - x - 12 = 0$.
3. $(x - 4)(x + 3) = 0$. A speed cannot be negative, so $x = 4$. (Check: $\dfrac{15}{8} + \dfrac{18}{16} = 1.875 + 1.125 = 3$ ✓.)

Answer: $4$ km/h

1. (a)(i) $D$ is the mid-point of $BC$ and $OD = DG$, so the diagonals $OG$ and $BC$ of $OBGC$ bisect each other. So $OBGC$ is a parallelogram.
2. (ii) $CO \parallel GB$, so $CE \parallel GB$. In $\triangle AGB$, $OE \parallel GB$, so $\dfrac{AO}{OG} = \dfrac{AE}{EB}$. In the same way in $\triangle AGC$, $\dfrac{AO}{OG} = \dfrac{AF}{FC}$.
3. Hence $\dfrac{AE}{EB} = \dfrac{AF}{FC}$, so $EF \parallel BC$ (converse of the basic proportionality theorem).
4. (iii) In $\triangle AEF$ and $\triangle ABC$: $\angle AEF = \angle ABC$ (corresponding angles, $EF \parallel BC$) and $\angle A$ is common. So $\triangle AEF \sim \triangle ABC$.
5. (b)(i) $QC \parallel AB$ (as $ABCD$ is a parallelogram), so $\triangle RQC \sim \triangle RAB$ and $\dfrac{QR}{AR} = \dfrac{QC}{AB} = \dfrac{1}{2}$ ($Q$ is the mid-point of $CD$ and $CD = AB$). So $AR = 2QR$ and $AQ = QR$.
6. (ii) $DQ \parallel AB$, so $\triangle PQD \sim \triangle PAB$ and $\dfrac{PQ}{AP} = \dfrac{DQ}{AB} = \dfrac{1}{2}$. So $AP = 2PQ$.
7. (iii) $AQ = QR$ gives $AP + PQ = PR - PQ$. With $PQ = \dfrac{1}{2}AP$: $AP + \dfrac{1}{2}AP = PR - \dfrac{1}{2}AP$, so $PR = 2AP$.

Answer: Proved (both parts)

1. (a) The sum of frequencies: $12 + p + 27 + 20 + q + 6 = 65 + p + q = 100$, so $p + q = 35$.
2. Class marks: $5, 15, 25, 35, 45, 55$. $\sum f_ix_i = 60 + 15p + 675 + 700 + 45q + 330 = 1765 + 15p + 45q$. Mean $= 28$: $1765 + 15p + 45q = 2800$, so $p + 3q = 69$.
3. Subtracting: $2q = 34$, so $q = 17$ and $p = 18$.
4. (b) Cumulative frequencies: $15, 25, 37, 46, 54, 64, 70$. $\dfrac{N}{2} = 35$ lies in $30 - 45$: $l = 30$, $cf = 25$, $f = 12$, $h = 15$. Median $= 30 + \dfrac{35 - 25}{12} \times 15 = 30 + 12.5 = 42.5$.
5. Modal class $0 - 15$ (the highest frequency $15$): $f_1 = 15$, $f_0 = 0$, $f_2 = 10$. Mode $= 0 + \dfrac{15 - 0}{30 - 0 - 10} \times 15 = \dfrac{225}{20} = 11.25$.