CBSE Class 10 Maths Basic 2025 Question Paper (Set 430/4/1) with Solutions

1. $\text{HCF} \times \text{LCM} = x \times 20$, so $2 \times 60 = 20x$ and $x = 6$. (Check: $\text{HCF}(6, 20) = 2$ ✓ and $\text{LCM}(6, 20) = 60$ ✓.)

Answer: (B) 6

1. $d = \sqrt{(2 + 6)^2 + (7 - 9)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$.

Answer: (A) $2\sqrt{17}$

1. $a_{n+1} - a_n = (5(n + 1) - 6) - (5n - 6) = 5$.

Answer: (C) 5

1. A zero makes the polynomial $0$: $k\left(-\dfrac{3}{2}\right)^2 - 9\left(-\dfrac{3}{2}\right) + 3 = 0$.
2. $\dfrac{9k}{4} + \dfrac{27}{2} + 3 = 0$, so $\dfrac{9k}{4} = -\dfrac{33}{2}$ and $k = -\dfrac{22}{3}$.

Answer: (D) $-\dfrac{22}{3}$

1. There are $2^3 = 8$ outcomes. Exactly one tail: $THH$, $HTH$, $HHT$, which is $3$. $P = \dfrac{3}{8}$.

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

1. $\pi r_1^2 h_1 = \pi r_2^2 h_2$, so $\dfrac{r_1^2}{r_2^2} = \dfrac{h_2}{h_1} = \dfrac{2}{1}$ and $r_1 : r_2 = \sqrt{2} : 1$.

Answer: (A) $\sqrt{2} : 1$

1. $\sin\theta = \dfrac{1}{\sqrt{2}}$, so $\theta = 45^\circ$. $\cot 45^\circ \times \operatorname{cosec} 45^\circ = 1 \times \sqrt{2} = \sqrt{2}$.

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

1. $OQ \perp PQ$, so $OP = \sqrt{3^2 + 4^2} = 5$ cm.
2. Perimeter of $\triangle OQP = OQ + QP + OP = 3 + 4 + 5 = 12$ cm.

Answer: (B) 12 cm

1. $\alpha + \beta = -\dfrac{5}{2}$ and $\alpha\beta = \dfrac{1}{2}$. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-5/2}{1/2} = -5$.

Answer: (D) −5

1. $a = 10\sqrt{2}$ and $d = 6\sqrt{2} - 10\sqrt{2} = -4\sqrt{2}$. $a_{20} = 10\sqrt{2} + 19(-4\sqrt{2}) = 10\sqrt{2} - 76\sqrt{2} = -66\sqrt{2}$.

Answer: (C) $-66\sqrt{2}$

1. $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta = 1$. So $\sec\theta + \tan\theta = \dfrac{1}{2}$.

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

1. There are $4$ nines in the deck. $P = \dfrac{4}{52} = \dfrac{1}{13}$.

Answer: (D) $\dfrac{1}{13}$

1. Arc $= \dfrac{210}{360} \times 2\pi r = \dfrac{7}{12} \times 2 \times \dfrac{22}{7} r = \dfrac{11}{3}r = \dfrac{44}{3}$, so $r = 4$ cm.

Answer: (B) 4 cm

1. Parallel lines need $\dfrac{14}{-3} = \dfrac{m}{2} \neq \dfrac{20}{16}$. So $m = -\dfrac{28}{3}$ (then $\dfrac{m}{2} = -\dfrac{14}{3} \neq \dfrac{5}{4}$ ✓).

Answer: (C) $-\dfrac{28}{3}$

1. $\pi r l = 550$: $\dfrac{22}{7} \times 7 \times l = 22l = 550$, so $l = 25$ cm.

Answer: (A) 25 cm

1. $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \dfrac{4}{9}} = \dfrac{\sqrt{5}}{3}$.

Answer: (B) $\dfrac{\sqrt{5}}{3}$

1. $OA \perp PA$, so $\angle OAP = 90^\circ$. In the quadrilateral $OAPB$: $\angle APB = 360^\circ - 90^\circ - 90^\circ - 105^\circ = 75^\circ$.
2. $\angle OAP + \angle APB = 90^\circ + 75^\circ = 165^\circ$.

Answer: (D) $165^\circ$

1. $AB = AP + PB = 2.4 + 3.6 = 6$ cm. $PQ \parallel BC$ gives $\triangle APQ \sim \triangle ABC$, so $\dfrac{PQ}{BC} = \dfrac{AP}{AB}$.
2. $PQ = 5.4 \times \dfrac{2.4}{6} = 2.16$ cm.

Answer: (D) 2.16 cm

1. (R) The median is the middle value: half of the data lie below it and half above. True.
2. (A) That is exactly what a median of $16$ means: half of the class scored less than $16$. True, and R explains it.

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

1. (R) The probabilities of an event and its complement add up to $1$. True.
2. (A) $P(\bar{E}) = 1 - \dfrac{1}{999} = \dfrac{998}{999} \approx 0.999$, not $0.001$ (that is close to $P(E)$ itself). False.

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

1. $45^n = (3 \times 3 \times 5)^n = 3^{2n} \times 5^n$. A number ends in $0$ only if it is divisible by both $2$ and $5$. $45^n$ has no factor $2$, so it cannot end with $0$.
2. Multiplying by the prime $a = 2$ gives the factors $2$ and $5$ together, so $2 \times 45^n$ ends with $0$.

Answer: Shown; $a = 2$

1. Let $P$ divide the segment in the ratio $k : 1$. The $y$-coordinate of $P$ is $0$: $\dfrac{-5k + 8}{k + 1} = 0$, so $k = \dfrac{8}{5}$. The ratio is $8 : 5$.
2. $x = \dfrac{8(-3) + 5(2)}{13} = \dfrac{-24 + 10}{13} = -\dfrac{14}{13}$.

Answer: Ratio $8 : 5$; $x = -\dfrac{14}{13}$

1. (a) Area of the circle $= \dfrac{22}{7} \times 0.7 \times 0.7 = 1.54$ m². Area of the rectangle $= 3 \times 2 = 6$ m². $P = \dfrac{1.54}{6} = \dfrac{77}{300}$.
2. (b) There are $36$ equally likely outcomes. (i) Difference $3$: $(1,4), (4,1), (2,5), (5,2), (3,6), (6,3)$, which is $6$: $P = \dfrac{6}{36} = \dfrac{1}{6}$.
3. (ii) Sum $8$: $(2,6), (6,2), (3,5), (5,3), (4,4)$, which is $5$: $P = \dfrac{5}{36}$.

Answer: (a) $\dfrac{77}{300}$   (b) (i) $\dfrac{1}{6}$ (ii) $\dfrac{5}{36}$

1. (a) Let $PA$ and $PB$ be tangents from $P$ to a circle with centre $O$. $OA \perp PA$ and $OB \perp PB$, so $\angle OAP = \angle OBP = 90^\circ$.
2. In the quadrilateral $OAPB$: $\angle APB + \angle PBO + \angle BOA + \angle OAP = 360^\circ$, so $\angle APB + \angle AOB = 180^\circ$. The two angles are supplementary.
3. (b) Let $AB$ be a diameter with the tangents $PQ$ at $A$ and $RS$ at $B$. $\angle OAP = \angle OBR = 90^\circ$ (radius and tangent). These are alternate (co-interior) angles that add to $180^\circ$ with the transversal $AB$, so $PQ \parallel RS$.

Answer: Proved (both parts)

1. $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, so $DE \parallel BC$ (converse of the basic proportionality theorem).
2. Then $\angle ADE = \angle ABC$ (corresponding angles). It is given that $\angle ADE = \angle ACB$, so $\angle ABC = \angle ACB$.
3. The sides opposite equal angles are equal: $AC = AB$, so $\triangle ABC$ is isosceles.

Answer: Proved

1. $6x^2 + 13x - 5 = (3x - 1)(2x + 5)$, so the zeroes are $\dfrac{1}{3}$ and $-\dfrac{5}{2}$.
2. Sum $= \dfrac{1}{3} - \dfrac{5}{2} = -\dfrac{13}{6}$ and $-\dfrac{13}{6}$ ✓. Product $= \dfrac{1}{3} \times \left(-\dfrac{5}{2}\right) = -\dfrac{5}{6}$ and $\dfrac{-5}{6}$ ✓.

Answer: Zeroes $\dfrac{1}{3}$ and $-\dfrac{5}{2}$; relations verified

1. (a) $a = 7$ and $d = 10\dfrac{1}{2} - 7 = \dfrac{7}{2}$. $84 = 7 + (n - 1) \times \dfrac{7}{2}$ gives $n - 1 = 22$ and $n = 23$.
2. $S_{23} = \dfrac{23}{2}(7 + 84) = \dfrac{23 \times 91}{2} = \dfrac{2093}{2} = 1046.5$.
3. (b) $S_n = n(n + 4) = n^2 + 4n$. $a = S_1 = 5$ and $S_2 = 12$, so $a_2 = 7$ and $d = 2$.
4. $a_{15} = 5 + 14 \times 2 = 33$.

Answer: (a) $\dfrac{2093}{2}$   (b) $a = 5$, $d = 2$, $a_{15} = 33$

1. Suppose $\sqrt{3} = \dfrac{p}{q}$ with $p, q$ co-prime and $q \neq 0$. $3q^2 = p^2$, so $3$ divides $p^2$ and so $3$ divides $p$. Write $p = 3m$.
2. $3q^2 = 9m^2$ gives $q^2 = 3m^2$, so $3$ divides $q$. Then $3$ divides both $p$ and $q$, contradicting that they are co-prime. So $\sqrt{3}$ is irrational.

Answer: Proved

1. (a) The diagonals of a parallelogram bisect each other: the mid-point of $BD$ equals the mid-point of $AC$. $\left(\dfrac{2 + x}{2}, \dfrac{-2 + 5}{2}\right) = \left(\dfrac{-5 + 8}{2}, \dfrac{y + 4}{2}\right)$.
2. $\dfrac{2 + x}{2} = \dfrac{3}{2}$ gives $x = 1$, and $\dfrac{3}{2} = \dfrac{y + 4}{2}$ gives $y = -1$.
3. $AB = \sqrt{(2 + 5)^2 + (-2 + 1)^2} = \sqrt{50} = 5\sqrt{2}$ and $BC = \sqrt{(8 - 2)^2 + (4 + 2)^2} = \sqrt{72} = 6\sqrt{2}$. The sides are $5\sqrt{2}$ and $6\sqrt{2}$.
4. (b) $AP = \sqrt{(3 - 6)^2 + (1 + 3)^2} = 5$ and $PB = \sqrt{(3 - 0)^2 + (1 - 5)^2} = 5$, so $\dfrac{AP}{PB} = 1$.
5. $AQ = \sqrt{(2 - 6)^2 + (-1 + 3)^2} = 2\sqrt{5}$ and $QC = \sqrt{(2 + 2)^2 + (-1 - 1)^2} = 2\sqrt{5}$, so $\dfrac{AQ}{QC} = 1$. Hence $\dfrac{AP}{PB} = \dfrac{AQ}{QC}$ ✓.

Answer: (a) $x = 1$, $y = -1$; sides $5\sqrt{2}$ and $6\sqrt{2}$   (b) yes, both ratios are $1$

1. Area of the sector $OAB = \dfrac{60}{360} \times 3.14 \times 10^2 = \dfrac{314}{6} \approx 52.33$ cm².
2. $OA = OB$ and $\angle AOB = 60^\circ$, so $\triangle OAB$ is equilateral with side $10$: area $= \dfrac{\sqrt{3}}{4} \times 10^2 = \dfrac{173}{4} = 43.25$ cm².
3. Shaded segment $= 52.33 - 43.25 = 9.08$ cm².

Answer: $\dfrac{314}{6} - \dfrac{173}{4} \approx 9.08$ cm²

1. Let $ABCD$ circumscribe a circle with centre $O$, touching $AB$, $BC$, $CD$, $DA$ at $P, Q, R, S$. Join $O$ to $A, B, C, D$ and to $P, Q, R, S$.
2. Tangents from a point are equal and radii are equal, so $\triangle OAP \cong \triangle OAS$ and $\angle AOP = \angle AOS$. In the same way $\angle BOP = \angle BOQ$, $\angle COQ = \angle COR$, $\angle DOR = \angle DOS$.
3. The eight angles round $O$ add up to $360^\circ$: $2(\angle AOP + \angle BOP + \angle COQ + \angle DOR) = 360^\circ$. So $\angle AOB + \angle COD = 180^\circ$, and likewise $\angle BOC + \angle AOD = 180^\circ$.

Answer: Proved

1. (a)(i) $D = (p^2 - q^2)^2 - 4(p^2)(-q^2) = (p^2 - q^2)^2 + 4p^2q^2 = (p^2 + q^2)^2$, a perfect square.
2. (ii) $x = \dfrac{-(p^2 - q^2) \pm (p^2 + q^2)}{2p^2}$. With $+$: $x = \dfrac{2q^2}{2p^2} = \dfrac{q^2}{p^2}$. With $-$: $x = \dfrac{-2p^2}{2p^2} = -1$.
3. (b) Let the integers be $x$, $x + 1$, $x + 2$. $x^2 + (x + 1)(x + 2) = 67$ gives $2x^2 + 3x - 65 = 0$, that is $(2x + 13)(x - 5) = 0$.
4. $x$ is positive, so $x = 5$ (reject $-\dfrac{13}{2}$). The numbers are $5, 6, 7$. (Check: $25 + 42 = 67$ ✓.)

Answer: (a) roots $\dfrac{q^2}{p^2}$ and $-1$   (b) $5, 6, 7$

1. $N = 6 + 16 + 17 + 4 + 5 + 2 = 50$. 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. $\sum f_iu_i = -12 - 16 + 0 + 4 + 10 + 6 = -8$. Mean $= 27.5 + \dfrac{-8}{50} \times 5 = 27.5 - 0.8 = 26.7$.
3. The modal class is $25 - 30$ (highest frequency $17$): $f_1 = 17$, $f_0 = 16$, $f_2 = 4$. Mode $= 25 + \dfrac{17 - 16}{34 - 16 - 4} \times 5 = 25 + \dfrac{5}{14} \approx 25.36$.

Answer: Mean $= 26.7$; mode $\approx 25.36$

1. (a) **Given:** in $\triangle ABC$, $DE \parallel BC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** join $BE$ and $CD$; draw $DG \perp AC$ and $EF \perp AB$.
2. $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EF}{\frac{1}{2} \cdot DB \cdot EF} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CED)} = \dfrac{\frac{1}{2} \cdot AE \cdot DG}{\frac{1}{2} \cdot EC \cdot DG} = \dfrac{AE}{EC}$.
3. $\triangle BDE$ and $\triangle CED$ have the same base $DE$ and lie between the same parallels $DE$ and $BC$, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (b) Let $AD$ meet $PQ$ at $R$. Since $PQ \parallel BC$, $PR \parallel BD$, so $\triangle ARP \sim \triangle ADB$ and $\dfrac{AR}{AD} = \dfrac{PR}{BD}$. (i)
5. Similarly $\triangle ARQ \sim \triangle ADC$ and $\dfrac{AR}{AD} = \dfrac{RQ}{DC}$. (ii)
6. From (i) and (ii): $\dfrac{PR}{BD} = \dfrac{RQ}{DC}$. $AD$ is a median, so $BD = DC$ and hence $PR = RQ$: $AD$ bisects $PQ$.

Answer: Proved (both parts)

1. Let the pedestal be $AB = h$ and the distance from the point of observation $C$ be $AC = x$. The flagstaff goes from $B$ to $D$ with $BD = 5$.
2. In $\triangle ABC$: $\tan 30^\circ = \dfrac{h}{x}$, so $x = h\sqrt{3}$. (i)
3. In $\triangle ADC$: $\tan 60^\circ = \dfrac{h + 5}{x}$, so $h + 5 = x\sqrt{3}$. (ii)
4. Substituting (i) in (ii): $h + 5 = 3h$, so $h = 2.5$ m. Then $x = 2.5 \times 1.73 = 4.325 \approx 4.33$ m.

Answer: Pedestal height $= 2.5$ m; distance $\approx 4.33$ m

1. (i) Initial amount means $t = 0$: $3C = 300$, so $C = ₹100$.
2. (ii) $3 \times 250 = 20t$, so $t = \dfrac{750}{20} = 37.5$ minutes.
3. (iii)(a) Equal charges: $20t = 10t + 300$ gives $t = 30$ minutes. $C = \dfrac{20 \times 30}{3} = ₹200$.
4. (iii)(b) At $t = 60$: plan A costs $C = \dfrac{20 \times 60}{3} = ₹400$ and plan B costs $C = \dfrac{600 + 300}{3} = ₹300$. Plan B is better because it saves ₹ $100$.

Answer: (i) ₹ $100$   (ii) $37.5$ min   (iii)(a) $30$ min, ₹ $200$   (iii)(b) plan B (₹ $300$ against ₹ $400$)

1. (i) Radius $= \dfrac{7}{2}$ cm. Volume $= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times \dfrac{7}{2} = \dfrac{539}{3} \approx 179.67$ cm³.
2. (ii) Surface of one ball $= 4\pi r^2 = 4 \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} = 154$ cm². For $10$ balls: $1540$ cm².
3. (iii)(a) Volume of the box $= 42 \times 91 \times 50 = 191100$ cm³. Volume of $600$ balls $= 600 \times \dfrac{539}{3} = 107800$ cm³. Empty space $= 191100 - 107800 = 83300$ cm³.
4. (iii)(b) Each ball sits in a square of side $7$ cm: area covered $= 7 \times 7 = 49$ cm². In the base $42 \times 91$ cm: $\dfrac{42}{7} \times \dfrac{91}{7} = 6 \times 13 = 78$ balls.