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

1. $960 = 2^6 \times 3 \times 5$ and $432 = 2^4 \times 3^3$.
2. HCF $=$ common primes at the smaller powers $= 2^4 \times 3 = 48$.

Answer: (A) 48

1. $2$ has exactly two factors, $1$ and $2$, so it is prime (the only even prime).

Answer: (A) a prime number

1. $6 \times 6 = 36$, $36 \times 6 = 216$, and any number ending in $6$ times $6$ ends in $6$. So $6^n$ always ends in $6$.

Answer: (B) 6

1. The zeroes of $f(x)$ are the $x$-values where the graph meets the $x$-axis. This curve stays above the $x$-axis everywhere and never touches it, so there are no zeroes.

Answer: (A) 0

1. Coincident lines lie on top of each other, so every point of the line satisfies both equations: infinitely many solutions.

Answer: (D) an infinite number of solutions

1. $d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$.

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

1. $\dfrac{DE}{AB} = 2$, so the scale factor is $2$ and $EF = 2 \times BC = 2 \times 8 = 16$ cm.

Answer: (D) 16 cm

1. Mid-point $= \left(\dfrac{5 + 6}{2}, \dfrac{-4 + 4}{2}\right) = \left(\dfrac{11}{2}, 0\right)$. Its $y$-coordinate is $0$, so it lies on the $x$-axis.

Answer: (A) $x$-axis

1. $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \dfrac{a^2}{b^2}} = \dfrac{\sqrt{b^2 - a^2}}{b}$.

Answer: (C) $\dfrac{\sqrt{b^2 - a^2}}{b}$

1. $\sin^2 A + 2\cos^2 A = (\sin^2 A + \cos^2 A) + \cos^2 A = 1 + \cos^2 A = 1 + \dfrac{1}{4} = \dfrac{5}{4}$.

Answer: (B) $\dfrac{5}{4}$

1. $\tan\theta = \dfrac{\text{height}}{\text{distance}} = \dfrac{30}{10\sqrt{3}} = \sqrt{3}$, so $\theta = 60^\circ$.

Answer: (D) $60^\circ$

1. $OP \perp TP$ and $OQ \perp TQ$, so in the quadrilateral $OPTQ$ the angles at $P$ and $Q$ are $90^\circ$.
2. $\angle PTQ = 360^\circ - 90^\circ - 90^\circ - 120^\circ = 60^\circ$. (The angle between the tangents and the angle at the centre add to $180^\circ$.)

Answer: (A) $60^\circ$

1. $A$, $O$, $B$ are on one straight line (a diameter), so $\angle POA = 180^\circ - 125^\circ = 55^\circ$.
2. $OA \perp PA$ (radius and tangent), so in $\triangle OAP$: $\angle APO = 180^\circ - 90^\circ - 55^\circ = 35^\circ$.

Answer: (D) $35^\circ$

1. Arc $= \dfrac{60}{360} \times 2\pi r = \dfrac{1}{6} \times 2 \times \dfrac{22}{7} \times 21 = 22$ cm.

Answer: (A) 22 cm

1. The hour hand turns $360^\circ$ in $12$ hours, which is $30^\circ$ per hour, or $0.5^\circ$ per minute.
2. From $7{:}00$ to $8{:}10$ is $70$ minutes, so the angle is $70 \times 0.5 = 35^\circ$. (The $7$ cm length is not needed for the angle.)

Answer: (C) $35^\circ$

1. Diameter $2d$ means the radius is $d$. Total surface area of a solid hemisphere $= 3\pi r^2 = 3\pi d^2$.

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

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$, so $21 = 3M - 24$ and $M = 15$.

Answer: (C) 15

1. $P(3) = \dfrac{1}{6}$, so $P(\text{not } 3) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$.

Answer: (C) $\dfrac{5}{6}$

1. (A) A leap year has $366 = 52$ weeks $+ 2$ days. The two extra days are one of $7$ equally likely pairs, and $2$ of them contain a Monday (Sun–Mon and Mon–Tue). So $P = \dfrac{2}{7}$. True.
2. (R) A non-leap year has $52$ weeks $+ 1$ day, so $P(53 \text{ Mondays}) = \dfrac{1}{7}$, not $\dfrac{5}{7}$. False.

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

1. (A) $y^2 + 4y + 3 = (y + 1)(y + 3)$, so the zeroes are $-1$ and $-3$: two zeroes. True.
2. (R) A polynomial of degree $2$ has at most $2$ zeroes. True. But R is a general limit, it does not show that this particular polynomial has two (we needed to factorise), so it is not the explanation.

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

1. $\alpha + \beta = -\dfrac{-3}{1} = 3$ and $\alpha\beta = \dfrac{-1}{1} = -1$.
2. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{3}{-1} = -3$.

Answer: $-3$

1. (a) $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{x}{x - 2} = \dfrac{x + 2}{x - 1}$.
2. $x(x - 1) = (x + 2)(x - 2)$, so $x^2 - x = x^2 - 4$ and $x = 4$.
3. (b) Similar triangles have proportional sides: $\dfrac{AB}{XY} = \dfrac{BC}{YZ} = \dfrac{AC}{XZ}$, that is $\dfrac{4}{x} = \dfrac{6}{7.2} = \dfrac{y}{6}$.
4. $\dfrac{6}{7.2} = \dfrac{5}{6}$. So $x = 4 \times \dfrac{6}{5} = 4.8$ cm and $y = 6 \times \dfrac{5}{6} = 5$ cm.

Answer: (a) $x = 4$   (b) $x = 4.8$ cm, $y = 5$ cm

1. The distance from the centre to the point equals the radius: $(x - 7 + 9)^2 + (2x - 11)^2 = (5\sqrt{2})^2 = 50$.
2. $(x + 2)^2 + (2x - 11)^2 = 50$, so $x^2 + 4x + 4 + 4x^2 - 44x + 121 = 50$ and $5x^2 - 40x + 75 = 0$, that is $x^2 - 8x + 15 = 0$.
3. $(x - 3)(x - 5) = 0$, so $x = 3$ or $x = 5$.

Answer: $x = 3$ or $x = 5$

1. (a) $\tan\theta = \dfrac{24}{7}$: the hypotenuse is $\sqrt{24^2 + 7^2} = 25$, so $\sin\theta = \dfrac{24}{25}$ and $\cos\theta = \dfrac{7}{25}$.
2. $\sin\theta + \cos\theta = \dfrac{24 + 7}{25} = \dfrac{31}{25}$.
3. (b) $\dfrac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)} = \dfrac{1 - \sin^2\theta}{1 - \cos^2\theta} = \dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \dfrac{49}{64}$.

Answer: (a) $\dfrac{31}{25}$   (b) $\dfrac{49}{64}$

1. Let the chord $AB$ of the larger circle touch the smaller circle at $M$. The radius to the point of contact is perpendicular to the tangent, so $OM \perp AB$ and $OM = 4$ cm.
2. The perpendicular from the centre bisects the chord. In the right triangle $OMA$: $AM = \sqrt{5^2 - 4^2} = 3$ cm.
3. $AB = 2 \times 3 = 6$ cm.

Answer: $6$ cm

1. Suppose $\sqrt{3}$ is rational: $\sqrt{3} = \dfrac{p}{q}$ with $p, q$ co-prime integers, $q \neq 0$.
2. Squaring, $3q^2 = p^2$, so $3$ divides $p^2$ and therefore $3$ divides $p$. Write $p = 3a$.
3. Then $9a^2 = 3q^2$, so $q^2 = 3a^2$; $3$ divides $q^2$ and so $3$ divides $q$.
4. Now $3$ divides both $p$ and $q$, which contradicts that they are co-prime. So $\sqrt{3}$ is irrational.

Answer: Proved

1. A point on the $x$-axis has $y = 0$. Let it divide the segment from $A(-6, 5)$ to $B(-4, -1)$ in the ratio $k : 1$.
2. $y = \dfrac{k(-1) + 1(5)}{k + 1} = 0$, so $-k + 5 = 0$ and $k = 5$. The ratio is $5 : 1$.
3. $x = \dfrac{5(-4) + 1(-6)}{5 + 1} = \dfrac{-26}{6} = -\dfrac{13}{3}$. The point is $\left(-\dfrac{13}{3}, 0\right)$.

Answer: Ratio $5 : 1$; point $\left(-\dfrac{13}{3}, 0\right)$

1. (a) $\dfrac{x - h}{a} = \cos\theta$ and $\dfrac{y - k}{b} = \sin\theta$.
2. LHS $= \cos^2\theta + \sin^2\theta = 1 =$ RHS.
3. (b) LHS $= \tan A \cdot \dfrac{(1 - \sec A) - (1 + \sec A)}{(1 + \sec A)(1 - \sec A)} = \tan A \cdot \dfrac{-2\sec A}{1 - \sec^2 A}$.
4. $1 - \sec^2 A = -\tan^2 A$, so LHS $= \dfrac{-2\tan A\sec A}{-\tan^2 A} = \dfrac{2\sec A}{\tan A} = \dfrac{2}{\cos A} \cdot \dfrac{\cos A}{\sin A} = 2\operatorname{cosec} A =$ RHS.

Answer: Proved (both parts)

1. (a) $AC = \sqrt{4^2 + 3^2} = 5$ cm. Let the circle touch $AB$, $BC$, $CA$ at $D$, $E$, $F$ and let $BD = BE = x$.
2. Tangents from a point are equal: $AD = AF = 4 - x$ and $CE = CF = 3 - x$. $AF + CF = AC$ gives $(4 - x) + (3 - x) = 5$, so $x = 1$.
3. $\angle B = 90^\circ$ and $BD = BE$, so $BDIE$ is a square and the radius equals $x = 1$ cm. (Check with areas: $\dfrac{1}{2} \times 3 \times 4 = 6 = \dfrac{1}{2} r (3 + 4 + 5) = 6r$, so $r = 1$ ✓.)
4. (b) Tangents from a point are equal: $PM = PN$, $QM = QS$ and $RN = RS$.
5. $2PM = PM + PN = (PQ + QM) + (PR + RN) = PQ + PR + QS + RS = PQ + PR + QR$. So $PM = \dfrac{1}{2}(PQ + QR + PR)$.

Answer: (a) $1$ cm   (b) proved

1. Radius of the cylinder and of each hemisphere $= \dfrac{7}{2}$ cm. The two hemispheres take $\dfrac{7}{2} + \dfrac{7}{2} = 7$ cm of the height, so the cylinder is $20 - 7 = 13$ cm tall.
2. Volume $= \pi r^2 h + 2 \times \dfrac{2}{3}\pi r^3 = \pi r^2\left(13 + \dfrac{4}{3} \times \dfrac{7}{2}\right) = \pi r^2\left(13 + \dfrac{14}{3}\right)$.
3. $= \dfrac{22}{7} \times \dfrac{49}{4} \times \dfrac{53}{3} = \dfrac{77}{2} \times \dfrac{53}{3} = \dfrac{4081}{6} \approx 680.17$ cm³.

Answer: $\dfrac{4081}{6} \approx 680.17$ cm³

Common mistake: Taking the cylinder as 20 − 7 = 13 only after adding the two radii. Using 20 cm or 20 − 3.5 for the cylinder overcounts.

1. Writing (number on the first die, number on the second): $(1,1), (1,2), \dots, (1,6)$; $(2,1), \dots, (2,6)$; $(3,1), \dots, (3,6)$; $(4,1), \dots, (4,6)$; $(5,1), \dots, (5,6)$; $(6,1), \dots, (6,6)$. That is $36$ equally likely outcomes.
2. (i) Same number: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$, which is $6$ outcomes. $P = \dfrac{6}{36} = \dfrac{1}{6}$.
3. (ii) Different numbers: $P = 1 - \dfrac{1}{6} = \dfrac{5}{6}$.

Answer: (i) $\dfrac{1}{6}$   (ii) $\dfrac{5}{6}$

1. Line 1, $2x - 3y + 6 = 0$: points $(0, 2)$ and $(-3, 0)$; also $(3, 4)$. Line 2, $2x + 3y - 18 = 0$: points $(0, 6)$ and $(9, 0)$; also $(3, 4)$. Line 3, $x = 0$, is the $y$-axis.
2. The lines meet in pairs at $A(0, 2)$ (line 1 and the $y$-axis), $C(0, 6)$ (line 2 and the $y$-axis) and $B(3, 4)$ (lines 1 and 2).
3. Taking $AC$ on the $y$-axis as the base: $AC = 6 - 2 = 4$. The height is the distance of $B$ from the $y$-axis, which is $3$.
4. Area $= \dfrac{1}{2} \times 4 \times 3 = 6$ square units.

Answer: Vertices $(0, 2)$, $(3, 4)$, $(0, 6)$; area $= 6$ square units

1. (a) Let the faster train go at $x$ km/h; the slower goes at $(x - 10)$ km/h. $\dfrac{200}{x - 10} - \dfrac{200}{x} = 1$.
2. $200x - 200(x - 10) = x(x - 10)$, so $2000 = x^2 - 10x$ and $x^2 - 10x - 2000 = 0$.
3. $(x - 50)(x + 40) = 0$. Speed cannot be negative, so $x = 50$. The faster train goes at $50$ km/h and the slower at $40$ km/h.
4. (b) Let the sides be $x$ and $y$ ($x > y$). $x^2 + y^2 = 640$ and $4x - 4y = 64$, so $y = x - 16$.
5. $x^2 + (x - 16)^2 = 640$ gives $2x^2 - 32x + 256 = 640$, so $x^2 - 16x - 192 = 0$ and $(x - 24)(x + 8) = 0$.
6. $x = 24$ (reject $-8$) and $y = 24 - 16 = 8$. The sides are $24$ m and $8$ m. (Check: $576 + 64 = 640$ ✓.)

Answer: (a) $50$ km/h and $40$ km/h   (b) $24$ m and $8$ m

1. (a) **Statement.** If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
2. **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 $CD$; draw $DM \perp AC$ and $EN \perp AB$.
3. $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EN}{\frac{1}{2} \cdot DB \cdot EN} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{\frac{1}{2} \cdot AE \cdot DM}{\frac{1}{2} \cdot EC \cdot DM} = \dfrac{AE}{EC}$.
4. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
5. (b)(i) $\triangle ABC \sim \triangle PQR$ gives $\angle A = \angle P$ and $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$. $CM$ and $RN$ are medians, so $AB = 2AM$ and $PQ = 2PN$, and $\dfrac{2AM}{2PN} = \dfrac{AC}{PR}$, that is $\dfrac{AM}{PN} = \dfrac{AC}{PR}$.
6. So $\triangle AMC \sim \triangle PNR$ by SAS.
7. (ii) $\angle B = \angle Q$ and $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ gives $\dfrac{2MB}{2NQ} = \dfrac{BC}{QR}$, so $\dfrac{MB}{NQ} = \dfrac{BC}{QR}$. Hence $\triangle CMB \sim \triangle RNQ$ by SAS.

Answer: (a) proved   (b) both proved

1. Sum of frequencies: $1 + x + 5 + 7 + y + 3 + 1 = 17 + x + y = 25$, so $x + y = 8$. (i)
2. Class marks: $5, 15, 25, 35, 45, 55, 65$. $\sum f_ix_i = 5 + 15x + 125 + 245 + 45y + 165 + 65 = 605 + 15x + 45y$.
3. Mean $= \dfrac{605 + 15x + 45y}{25} = 35$, so $15x + 45y = 875 - 605 = 270$ and $x + 3y = 18$. (ii)
4. (ii) $-$ (i): $2y = 10$, so $y = 5$ and $x = 3$.

Answer: $x = 3$ and $y = 5$

1. (i) The first potato is $5$ m away, so the round trip is $5 + 5 = 10$ m.
2. (ii) The second potato is $5 + 3 = 8$ m away, so the round trip is $8 + 8 = 16$ m.
3. (iii)(a) The round trips are $10, 16, 22, \dots$, an A.P. with $a = 10$ and $d = 6$ for $10$ terms. $S_{10} = \dfrac{10}{2}[2(10) + 9(6)] = 5 \times 74 = 370$ m.
4. (iii)(b) Time $= \dfrac{\text{distance}}{\text{speed}} = \dfrac{370}{5} = 74$ seconds.

Answer: (i) $10$ m   (ii) $16$ m   (iii)(a) $370$ m   (iii)(b) $74$ s

1. (i) In the right triangle $OPB$: $\cos 30^\circ = \dfrac{OP}{OB}$, so $\dfrac{\sqrt{3}}{2} = \dfrac{6}{OB}$ and $OB = \dfrac{12}{\sqrt{3}} = 4\sqrt{3}$ m.
2. (ii) In the right triangle $OPA$: $\cos 60^\circ = \dfrac{OP}{OA}$, so $\dfrac{1}{2} = \dfrac{6}{OA}$ and $OA = 12$ m.
3. (iii)(a) $\tan 30^\circ = \dfrac{BP}{6}$ gives $BP = 2\sqrt{3}$ m. $\tan 60^\circ = \dfrac{AP}{6}$ gives $AP = 6\sqrt{3}$ m. $AB = AP - BP = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}$ m.
4. (iii)(b) Area of $\triangle OPB = \dfrac{1}{2} \times OP \times BP = \dfrac{1}{2} \times 6 \times 2\sqrt{3} = 6\sqrt{3}$ m².