CBSE Class 10 Maths Basic 2025 Question Paper (Set 430/6/1) with Solutions

1. The zeroes are the points where the graph meets the $x$-axis. The curve crosses it at $3$ points.

Answer: (A) 3

1. $a = \dfrac{3}{2}$ and $d = \dfrac{1}{2} - \dfrac{3}{2} = -1$. $a_{22} = \dfrac{3}{2} + 21(-1) = -\dfrac{39}{2}$.

Answer: (C) $-\dfrac{39}{2}$

1. On the $x$-axis $y = 0$, so $2x = 6$ and $x = 3$. The point is $(3, 0)$.

Answer: (D) (3, 0)

1. Height of one cone $= \sqrt{6^2 - 4^2} = \sqrt{20} = 2\sqrt{5}$ cm.
2. The solid is two such cones tip to tip: height $= 2 \times 2\sqrt{5} = 4\sqrt{5}$ cm.

Answer: (B) $4\sqrt{5}$ cm

1. Inconsistent needs $\dfrac{3}{k} = \dfrac{-7}{14} \neq \dfrac{1}{6}$. $\dfrac{3}{k} = -\dfrac{1}{2}$ gives $k = -6$, and then $\dfrac{3}{k} = -\dfrac{1}{2} \neq \dfrac{1}{6}$ ✓.

Answer: (A) −6

1. Sum $10$: $(4,6), (5,5), (6,4)$ — $3$. Sum $11$: $(5,6), (6,5)$ — $2$. Sum $12$: $(6,6)$ — $1$. Total $6$ of $36$: $P = \dfrac{1}{6}$.

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

1. A diagonal joins the opposite vertices $(2, -2)$ and $(4, 8)$: $\sqrt{(4 - 2)^2 + (8 + 2)^2} = \sqrt{4 + 100} = \sqrt{104} = 2\sqrt{26}$. (The other diagonal, $(8, 4)$ to $(-2, 2)$, gives $\sqrt{100 + 4}$, the same.)

Answer: (D) $2\sqrt{26}$

1. $OA \perp PA$, so in the right triangle $OAP$: $\sin 30^\circ = \dfrac{OA}{OP}$, giving $OP = \dfrac{2.5}{1/2} = 5$ cm.

Answer: (B) 5 cm

1. $P(\text{not happening}) = 1 - 0.57 = 0.43$.

Answer: (A) 0.43

1. $\dfrac{22}{3} = \dfrac{\theta}{360} \times 2 \times \dfrac{22}{7} \times 7 = \dfrac{\theta}{360} \times 44$, so $\theta = \dfrac{22 \times 360}{3 \times 44} = 60^\circ$.

Answer: (C) $60^\circ$

1. $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{x + 1}{3} = \dfrac{2x + 1}{4}$. $4x + 4 = 6x + 3$, so $x = \dfrac{1}{2}$.

Answer: (B) $\dfrac{1}{2}$

1. There are $8$ outcomes. Exactly one head: $HTT$, $THT$, $TTH$, which is $3$. $P = \dfrac{3}{8}$.

Answer: (D) $\dfrac{3}{8}$

1. $OC \perp AB$ (radius and tangent). $AC = \sqrt{OA^2 - OC^2} = \sqrt{12.25 - 4.41} = \sqrt{7.84} = 2.8$ cm.
2. The perpendicular from the centre bisects the chord, so $AB = 2 \times 2.8 = 5.6$ cm.

Answer: (A) 5.6 cm

1. $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{1}{\sqrt{3}}$, so $\theta = 30^\circ$.

Answer: (D) $30^\circ$

1. With $d = x - A$, $x = A + d$, so the mean is $\bar{x} = A + \bar{d}$. (The factor $h$ appears only in the step-deviation method, where $u = \dfrac{x - A}{h}$.)

Answer: (A) $A + \bar{d}$

1. The first term is $S_1 = \dfrac{1}{2}(3 + 1) = 2$.

Answer: (A) 2

1. $\sin C = \dfrac{AB}{AC} = \dfrac{1}{2}$, so $C = 30^\circ$ and $\cos C = \cos 30^\circ = \dfrac{\sqrt{3}}{2}$.

Answer: (C) $\dfrac{\sqrt{3}}{2}$

1. The cone has one third of the volume of the cylinder: $\dfrac{1}{3} \times 450 = 150$ cm³. The empty space is $450 - 150 = 300$ cm³.

Answer: (D) $300$ cm³

1. (A) $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$, an integer, so it is rational. True.
2. (R) $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ is irrational, so the product of two irrationals need not be rational. False.

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

1. (R) The angle sum of a triangle is $180^\circ$. True.
2. (A) $\angle B = 180^\circ - 65^\circ - 60^\circ = 55^\circ$, and in similar triangles $\angle Q = \angle B = 55^\circ$. True. But the equality $\angle Q = \angle B$ comes from similarity, which R does not state, so R alone is not the full explanation.

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

1. (a) $D = (-9)^2 - 4(4)(3) = 81 - 48 = 33$. $x = \dfrac{9 \pm \sqrt{33}}{8}$.
2. (b) $D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$, so the roots are real and equal.

Answer: (a) $\dfrac{9 \pm \sqrt{33}}{8}$   (b) real and equal

1. In $\triangle OAB$ and $\triangle OCD$: $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (alternate angles, $AB \parallel DC$).
2. So $\triangle OAB \sim \triangle OCD$ by AA, and the corresponding sides are proportional: $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

Answer: Proved

1. (i) $2$-digit numbers from $10$ to $99$: $90$ of the $120$ discs. $P = \dfrac{90}{120} = \dfrac{3}{4}$.
2. (ii) Perfect squares up to $120$: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$, which is $10$. $P = \dfrac{10}{120} = \dfrac{1}{12}$.

Answer: (i) $\dfrac{3}{4}$   (ii) $\dfrac{1}{12}$

1. (a) $\tan 30^\circ + \sin 60^\circ = \dfrac{1}{\sqrt{3}} + \dfrac{\sqrt{3}}{2} = \dfrac{2 + 3}{2\sqrt{3}} = \dfrac{5}{2\sqrt{3}}$. The value is $\dfrac{1/\sqrt{2}}{5/(2\sqrt{3})} = \dfrac{2\sqrt{3}}{5\sqrt{2}} = \dfrac{\sqrt{6}}{5}$.
2. (b) LHS $= \sin 60^\circ = \dfrac{\sqrt{3}}{2}$. RHS $= \dfrac{2/\sqrt{3}}{1 + 1/3} = \dfrac{2/\sqrt{3}}{4/3} = \dfrac{\sqrt{3}}{2}$. LHS $=$ RHS ✓.

Answer: (a) $\dfrac{\sqrt{6}}{5}$   (b) verified

1. $180 = 2^2 \times 3^2 \times 5$, $140 = 2^2 \times 5 \times 7$, $210 = 2 \times 3 \times 5 \times 7$.
2. The common primes at their smallest powers are $2$ and $5$: HCF $= 2 \times 5 = 10$.

Answer: $10$

1. (a) $\alpha + \beta = \dfrac{5}{8}$ and $\alpha\beta = -\dfrac{1}{8}$. New sum $= \dfrac{2}{\alpha} + \dfrac{2}{\beta} = \dfrac{2(\alpha + \beta)}{\alpha\beta} = \dfrac{2 \times 5/8}{-1/8} = -10$. New product $= \dfrac{4}{\alpha\beta} = -32$.
2. The polynomial is $x^2 + 10x - 32$ (or $k(x^2 + 10x - 32)$ for any non-zero $k$).
3. (b) $3x^2 + x - 10 = (x + 2)(3x - 5)$, so the zeroes are $-2$ and $\dfrac{5}{3}$.
4. Sum $= -2 + \dfrac{5}{3} = -\dfrac{1}{3} = -\dfrac{1}{3}$ ✓. Product $= -2 \times \dfrac{5}{3} = -\dfrac{10}{3} = \dfrac{-10}{3}$ ✓.

Answer: (a) $x^2 + 10x - 32$   (b) zeroes $-2, \dfrac{5}{3}$; relations verified

1. Let the length be $a$ and the breadth $b$. $2(a + b) = 100$ gives $a + b = 50$ and $ab = 600$.
2. $a(50 - a) = 600$ gives $a^2 - 50a + 600 = 0$, so $(a - 30)(a - 20) = 0$ and $a = 30$ or $20$.
3. The length is $30$ m and the breadth $20$ m (the other root just swaps the two).

Answer: Length $30$ m, breadth $20$ m

1. $120 = 2^3 \times 3 \times 5$, $100 = 2^2 \times 5^2$, $150 = 2 \times 3 \times 5^2$. LCM $= 2^3 \times 3 \times 5^2 = 600$ cm.
2. The rods are used $\dfrac{600}{120} = 5$, $\dfrac{600}{100} = 6$ and $\dfrac{600}{150} = 4$ times respectively.

Answer: $600$ cm; used $5$, $6$ and $4$ times

1. The shaded region is two sectors of $30^\circ$ (the vertically opposite angles $\angle BOD$ and $\angle AOC$), each of radius $7$ cm.
2. Area $= 2 \times \dfrac{30}{360} \times \dfrac{22}{7} \times 7 \times 7 = \dfrac{77}{3} \approx 25.67$ cm².
3. Perimeter $= 2 \times (\text{arc} + 2 \text{ radii}) = 2\left(\dfrac{30}{360} \times 2 \times \dfrac{22}{7} \times 7 + 14\right) = 2\left(\dfrac{11}{3} + 14\right) = \dfrac{106}{3} \approx 35.33$ cm.

Answer: Area $\approx 25.67$ cm²; perimeter $\approx 35.33$ cm

1. LHS $= \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} + \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)} = \dfrac{1}{\sin\theta - \cos\theta}\left[\dfrac{\sin^2\theta}{\cos\theta} - \dfrac{\cos^2\theta}{\sin\theta}\right]$.
2. $= \dfrac{\sin^3\theta - \cos^3\theta}{(\sin\theta - \cos\theta)\sin\theta\cos\theta} = \dfrac{(\sin\theta - \cos\theta)(1 + \sin\theta\cos\theta)}{(\sin\theta - \cos\theta)\sin\theta\cos\theta}$.
3. $= \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta} + 1 = \sec\theta\operatorname{cosec}\theta + 1 =$ RHS.

Answer: Proved

1. (a) $a + 2d = 16$ and $a_7 - a_5 = 2d = 12$, so $d = 6$ and $a = 4$. The A.P. is $4, 10, 16, \dots$.
2. $S_{29} = \dfrac{29}{2}[2(4) + 28 \times 6] = \dfrac{29}{2} \times 176 = 2552$.
3. (b) $a_1 = 7$ and $d = 2$. $S_{20} = \dfrac{20}{2}[14 + 19 \times 2] = 10 \times 52 = 520$.
4. If $52$ is a term: $7 + (n - 1) \times 2 = 52$ gives $n = \dfrac{47}{2}$, which is not a natural number. So $52$ is not a term.

Answer: (a) $4, 10, 16, \dots$; $S_{29} = 2552$   (b) $S_{20} = 520$; $52$ is not a term

1. (a) $2x + y = 9$: points $(4, 1)$, $(3, 3)$, $(0, 9)$. $x - 2y = 2$: points $(2, 0)$, $(4, 1)$, $(0, -1)$. The lines meet at $(4, 1)$, so $x = 4$, $y = 1$. (Check: $8 + 1 = 9$ ✓, $4 - 2 = 2$ ✓.)
2. (b) $\dfrac{6x}{100} + \dfrac{5y}{100} = 1200$ gives $6x + 5y = 120000$ (i). $\dfrac{3x}{100} + \dfrac{8y}{100} = 1260$ gives $3x + 8y = 126000$ (ii).
3. (ii) $\times 2$: $6x + 16y = 252000$. Subtracting (i): $11y = 132000$, so $y = 12000$. Then $6x = 120000 - 60000$ and $x = 10000$.

Answer: (a) $x = 4$, $y = 1$   (b) $x = ₹10{,}000$, $y = ₹12{,}000$

1. (a) Tangents from a point are equal: $BF = BD = 6$, $CE = CD = 10$ and $AE = AF = x$. The sides are $AB = 6 + x$, $AC = 10 + x$, $BC = 16$, so $s = 16 + x$.
2. By Heron's formula, area $= \sqrt{(16 + x)(x)(6)(10)}$. Also, area $= \dfrac{1}{2} \times r \times \text{perimeter} = \dfrac{1}{2} \times 4 \times 2(16 + x) = 4(16 + x)$.
3. Equating the squares: $60x(16 + x) = 16(16 + x)^2$, so $60x = 16(16 + x)$ and $44x = 256$, $x = \dfrac{64}{11} \approx 5.8$ cm. $AE = \dfrac{64}{11}$ cm.
4. (b)(i) $OA \perp AP$ and $OP$ bisects $\angle AOB$, so $\angle AOP = 60^\circ$ and $\angle OPA = 30^\circ$.
5. (ii) $\sin 30^\circ = \dfrac{OA}{OP}$ gives $OP = 20$ cm. $\tan 30^\circ = \dfrac{OA}{PA}$ gives $PA = 10\sqrt{3}$ cm. Perimeter $= 10 + 20 + 10\sqrt{3} = 30 + 10\sqrt{3}$ cm.
6. (iii) $PA = PB$ and $\angle APB = 60^\circ$, so $\triangle APB$ is equilateral and $AB = PA = 10\sqrt{3}$ cm.

Answer: (a) $\dfrac{64}{11}$ cm   (b) (i) $30^\circ$ (ii) $(30 + 10\sqrt{3})$ cm (iii) $10\sqrt{3}$ cm

1. Let the multi-storeyed building be $CD = h$, the small one $AB = 9$ m, and the distance between them $x$.
2. The foot $A$ is at $60^\circ$ from the top $D$: $\tan 60^\circ = \dfrac{h}{x}$, so $h = x\sqrt{3}$. (i)
3. The top $B$ is at $30^\circ$: it is $h - 9$ below $D$, so $\tan 30^\circ = \dfrac{h - 9}{x}$ and $x = (h - 9)\sqrt{3}$. (ii)
4. Substituting (i) in (ii): $x = (x\sqrt{3} - 9)\sqrt{3} = 3x - 9\sqrt{3}$, so $x = \dfrac{9\sqrt{3}}{2} = 7.79$ m. Then $h = 7.79 \times 1.73 = 13.5$ m.

Answer: Height $= 13.5$ m; distance $\approx 7.79$ m

1. $N = 12 + 10 + 15 + 11 + 7 + 5 = 60$. Class marks: $17.5, 22.5, 27.5, 32.5, 37.5, 42.5$. Take $a = 27.5$ and $h = 5$: $u = -2, -1, 0, 1, 2, 3$.
2. $\sum f_iu_i = -24 - 10 + 0 + 11 + 14 + 15 = 6$. Mean $= 27.5 + \dfrac{6}{60} \times 5 = 28$.
3. The modal class is $25 - 30$ ($f_1 = 15$, $f_0 = 10$, $f_2 = 11$): Mode $= 25 + \dfrac{15 - 10}{30 - 10 - 11} \times 5 = 25 + \dfrac{25}{9} \approx 27.78$.

Answer: Mean $= 28$; mode $\approx 27.78$

1. (i) $PQ \parallel AC$, so $\angle BQP = \angle BCA$ (corresponding angles) and $\angle B$ is common. By AA, $\triangle BPQ \sim \triangle BAC$.
2. (ii) $P$ and $R$ trisect $AB$, so $BP = \dfrac{1}{3}BA$. From the similarity, $\dfrac{PQ}{AC} = \dfrac{BP}{BA} = \dfrac{1}{3}$, that is $PQ = \dfrac{1}{3}AC$.
3. (iii)(a) $AB = AC = 3$ m, so $PQ = 1$ m and $BP = 1$ m. In the right triangle $BPQ$: $BQ = \sqrt{1^2 + 1^2} = \sqrt{2}$ m.
4. $BR = \dfrac{2}{3} \times 3 = 2$ m and in the same way $RS = 2$ m, so $BS = \sqrt{2^2 + 2^2} = 2\sqrt{2}$ m. $\dfrac{1}{2}BS = \sqrt{2} = BQ$ ✓.
5. (iii)(b) $\triangle BRS \sim \triangle BAC$ with ratio $\dfrac{2}{3}$, so $BR = \dfrac{2}{3}AB$ and $RS = \dfrac{2}{3}AC$. $BR^2 + RS^2 = \dfrac{4}{9}(AB^2 + AC^2) = \dfrac{4}{9}BC^2$ (Pythagoras in $\triangle ABC$).

Answer: (iii)(a) $BQ = \sqrt{2}$ m, $BS = 2\sqrt{2}$ m; $BQ = \dfrac{1}{2}BS$   (iii)(b) proved

1. (i) The bowl just fits, so the box is as wide as the outer diameter, $2 \times 10.5 = 21$ cm, and as tall as the outer radius, $10.5$ cm. The box is $21 \times 21 \times 10.5$ cm.
2. (ii) Total outer surface area $= 2(lb + bh + hl) = 2(441 + 220.5 + 220.5) = 2 \times 882 = 1764$ cm².
3. (iii)(a) Capacity of the bowl (inner radius $10$) $= \dfrac{2}{3} \times 3.14 \times 10^3 = \dfrac{6280}{3} \approx 2093.33$ cm³. Volume of the box $= 21 \times 21 \times 10.5 = 4630.5$ cm³.
4. Difference $= 4630.5 - 2093.33 \approx 2537.17$ cm³ (here the capacity is taken as the volume, as in the key’s note).
5. (iii)(b) Painted area = inner curved surface $+$ the ring of the rim $= 2\pi r^2 + \pi(R^2 - r^2) = 2 \times \dfrac{22}{7} \times 100 + \dfrac{22}{7}(110.25 - 100) = \dfrac{4400}{7} + \dfrac{451}{14} = \dfrac{9251}{14} \approx 660.79$ cm².