CBSE Class 10 Maths Basic 2025 Question Paper (Set 430/5/1) with Solutions

1. $(2 - 5\sqrt{3})^2 = 4 - 20\sqrt{3} + 75 = 79 - 20\sqrt{3}$. Since $\sqrt{3}$ is irrational, this is irrational.

Answer: (B) an irrational number

1. Equal roots need $D = (4k)^2 - 4(6)(k) = 16k^2 - 24k = 8k(2k - 3) = 0$, so $k = 0$ or $k = \dfrac{3}{2}$.
2. Only $k = 0$ is among the options (then $6x^2 = 0$ has the equal roots $0, 0$).

Answer: (A) 0

1. $d = \sqrt{(2 + 2)^2 + (3 + 3)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

Answer: (C) $2\sqrt{13}$

1. The zeroes are the points where the graph meets the $x$-axis. The parabola crosses it twice, so $p(x)$ has $2$ zeroes.

Answer: (D) 2

1. The angle between the tangent $CD$ and the chord $CA$ equals the angle in the alternate segment: $\angle CBA = \angle ACD = 40^\circ$.
2. (Check: $\angle ACB = 90^\circ$ (angle in a semicircle), so $\angle CAB = 50^\circ$ and $\angle OCA = 50^\circ$, giving $\angle OCD = 90^\circ = 50^\circ + 40^\circ$ ✓.)

Answer: (B) $40^\circ$

1. $a = -12$ and $d = -7$. $a_{10} = -12 + 9(-7) = -75$.

Answer: (A) −75

1. $x^2 = 8$ gives $x = \pm 2\sqrt{2}$: two different irrational numbers. ($D = 32 > 0$, not a perfect square.)

Answer: (B) irrational and distinct

1. Let the ratio be $k : 1$. The $y$-coordinate is $0$: $\dfrac{-10k + 5}{k + 1} = 0$, so $k = \dfrac{1}{2}$. The ratio is $1 : 2$.

Answer: (D) 1 : 2

1. $51$ cards remain. The king of hearts is red, so it is still there: $P = \dfrac{1}{51}$.

Answer: (C) $\dfrac{1}{51}$

1. The biggest cone has its base circle inscribed in a face of the cube, so its diameter is $25$ cm and the radius is $12.5$ cm.

Answer: (B) 12.5 cm

1. $P(\bar{E}) = 1 - 0.1 = 0.9$.

Answer: (A) 0.9

1. $\tan A = 1$ gives $A = 45^\circ$, so $\sin A = \cos A = \dfrac{1}{\sqrt{2}}$. $3\sin A + \cos A = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}$.

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

1. $(x + 2)^2 = (x + 2)(x + 2)$ has the repeated zero $-2$ and no other. ($x^2 + 2x$ has the zeroes $0$ and $-2$.)

Answer: (D) $(x + 2)^2$

1. $OQ \perp PQ$, so $\triangle OQP$ is right-angled at $Q$ with $OQ = QP$. The two base angles are equal and add to $90^\circ$: $\angle OPQ = 45^\circ$.

Answer: (A) $45^\circ$

1. The mean uses every value (sum divided by the number of values). The median and mode depend on the position and frequency, and the range on only the largest and smallest values.

Answer: (B) Mean

1. $OQ \perp PQ$, so $OP = \sqrt{OQ^2 + PQ^2} = \sqrt{9 + 25} = \sqrt{34}$ cm.

Answer: (D) $\sqrt{34}$ cm

1. Exactly one $4$: the first die shows $4$ and the second does not ($5$ ways) or the other way round ($5$ ways), $10$ outcomes in $36$. $P = \dfrac{10}{36} = \dfrac{5}{18}$.

Answer: (C) $\dfrac{5}{18}$

1. $22 = \dfrac{x}{360} \times 2 \times \dfrac{22}{7} \times 36$, so $x = \dfrac{22 \times 360 \times 7}{2 \times 22 \times 36} = 35$.

Answer: (A) 35

1. (R) The curved surface area of a hemisphere is $2\pi r^2$. True.
2. (A) Carving out the hemisphere removes the flat circle $\pi r^2$ of the cylinder’s top and adds the hollow of area $2\pi r^2$. The net increase is $2\pi r^2 - \pi r^2 = \pi r^2$, not $2\pi r^2$. False.

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

1. (R) The angle sum is $180^\circ$ and $\cos 90^\circ = 0$. True.
2. (A) $A + C = 180^\circ - 90^\circ = 90^\circ$, so $\cos(A + C) = \cos 90^\circ = 0$. True, and R is exactly this reasoning.

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

1. $15^n \times 2^n = (3 \times 5)^n \times 2^n = 3^n \times 5^n \times 2^n$. Both $2$ and $5$ are prime factors.
2. A number divisible by both $2$ and $5$ is divisible by $10$, so it ends with the digit $0$. (Indeed, $15^n \times 2^n = 30^n$.)

Answer: Yes, it ends with 0

1. (a) $2(1)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 - (1)^2 = 2 + \dfrac{3}{4} - 1 = \dfrac{7}{4}$.
2. (b) LHS $= \cos 60^\circ = \dfrac{1}{2}$. RHS $= \dfrac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \dfrac{2/3}{4/3} = \dfrac{1}{2}$. LHS $=$ RHS ✓.

Answer: (a) $\dfrac{7}{4}$   (b) verified

1. Let $A(-4, 1)$ and $B(6, 5)$. The point $P$ with $AP : PB = 1 : 2$ is $\left(\dfrac{1(6) + 2(-4)}{3}, \dfrac{1(5) + 2(1)}{3}\right) = \left(-\dfrac{2}{3}, \dfrac{7}{3}\right)$.
2. The point $Q$ with $AQ : QB = 2 : 1$ is $\left(\dfrac{2(6) + 1(-4)}{3}, \dfrac{2(5) + 1(1)}{3}\right) = \left(\dfrac{8}{3}, \dfrac{11}{3}\right)$.

Answer: $\left(-\dfrac{2}{3}, \dfrac{7}{3}\right)$ and $\left(\dfrac{8}{3}, \dfrac{11}{3}\right)$

1. (a) Black marbles $= \dfrac{3}{5} \times 40 = 24$, so white marbles $= 40 - 24 = 16$.
2. (b) The cards $20$ to $59$ number $59 - 20 + 1 = 40$. (i) Primes: $23, 29, 31, 37, 41, 43, 47, 53, 59$, which is $9$: $P = \dfrac{9}{40}$.
3. (ii) Perfect squares: $25, 36, 49$, which is $3$: $P = \dfrac{3}{40}$.

Answer: (a) $16$   (b) (i) $\dfrac{9}{40}$ (ii) $\dfrac{3}{40}$

1. Let $A(1, 5)$, $B(2, 3)$, $C(3, 1)$. $AB = \sqrt{1^2 + (-2)^2} = \sqrt{5}$, $BC = \sqrt{1^2 + (-2)^2} = \sqrt{5}$ and $AC = \sqrt{2^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5}$.
2. $AB + BC = \sqrt{5} + \sqrt{5} = 2\sqrt{5} = AC$, so the three points lie on one straight line.

Answer: Proved

1. (a) $\alpha + \beta = \dfrac{8}{3}$ and $\alpha\beta = \dfrac{4}{3}$. The new sum $= \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{8/3}{4/3} = 2$ and the new product $= \dfrac{1}{\alpha\beta} = \dfrac{3}{4}$.
2. The polynomial is $x^2 - 2x + \dfrac{3}{4}$, or $k(4x^2 - 8x + 3)$ for any non-zero $k$.
3. (b) $6x^2 - 7x - 3 = (2x - 3)(3x + 1)$, so the zeroes are $\dfrac{3}{2}$ and $-\dfrac{1}{3}$.
4. Sum $= \dfrac{3}{2} - \dfrac{1}{3} = \dfrac{7}{6} = -\dfrac{-7}{6}$ ✓. Product $= \dfrac{3}{2} \times \left(-\dfrac{1}{3}\right) = -\dfrac{1}{2} = \dfrac{-3}{6}$ ✓.

Answer: (a) $x^2 - 2x + \dfrac{3}{4}$   (b) zeroes $\dfrac{3}{2}, -\dfrac{1}{3}$; relations verified

1. Let the numbers be $x$ and $x + 2$. $x^2 + (x + 2)^2 = 452$ gives $2x^2 + 4x - 448 = 0$, so $x^2 + 2x - 224 = 0$ and $(x + 16)(x - 14) = 0$.
2. $x = 14$ gives $14$ and $16$. (Check: $196 + 256 = 452$ ✓. The root $x = -16$ gives the pair $-16, -14$, which also fits the words; the key lists $14$ and $16$.)

Answer: $14$ and $16$

1. LHS $= \operatorname{cosec}^2 A + 2\operatorname{cosec} A\sin A + \sin^2 A + \sec^2 A + 2\sec A\cos A + \cos^2 A$.
2. Since $\operatorname{cosec} A\sin A = 1$ and $\sec A\cos A = 1$: LHS $= (\sin^2 A + \cos^2 A) + (1 + \cot^2 A) + (1 + \tan^2 A) + 2 + 2$.
3. $= 1 + 1 + \cot^2 A + 1 + \tan^2 A + 4 = 7 + \tan^2 A + \cot^2 A =$ RHS.

Answer: Proved

1. They change together again after the LCM of $45$, $75$ and $60$ seconds. $45 = 3^2 \times 5$, $75 = 3 \times 5^2$, $60 = 2^2 \times 3 \times 5$.
2. LCM $= 2^2 \times 3^2 \times 5^2 = 900$ seconds $= 15$ minutes. They next change together at $5{:}15$ a.m.

Answer: At $5{:}15$ a.m.

1. (a) Sphere: $\dfrac{4}{3} \times \dfrac{22}{7} \times 10^3 \approx 4190.48$ cm³. Neck (a cylinder of radius $4$ and length $7$): $\dfrac{22}{7} \times 4 \times 4 \times 7 = 352$ cm³.
2. Volume $= 4190.48 + 352 = 4542.48$ cm³.
3. (b) Cylinder: $\dfrac{22}{7} \times 7 \times 7 \times 20 = 3080$ cm³. The two cones: $2 \times \dfrac{1}{3} \times \dfrac{22}{7} \times 2.1 \times 2.1 \times 5 = 46.2$ cm³.
4. Remaining volume $= 3080 - 46.2 = 3033.8$ cm³.

Answer: (a) $\approx 4542.48$ cm³   (b) $3033.8$ cm³

1. (i) In $\triangle DFE$ and $\triangle CFB$: $\angle DFE = \angle CFB$ (vertically opposite angles) and $\angle EDF = \angle BCF$ (alternate angles, since $AD \parallel BC$ and $DC$ is a transversal... with $E$ on $AD$ produced and $DF \subset DC$). So $\triangle DFE \sim \triangle CFB$ by AA.
2. (ii) In $\triangle AEB$ and $\triangle CBF$: $\angle AEB = \angle CBF$ (alternate angles, $AE \parallel BC$ with transversal $BE$) and $\angle EAB = \angle BCF$ (opposite angles of the parallelogram $ABCD$).
3. So $\triangle AEB \sim \triangle CBF$ by AA.

Answer: Shown

1. $2x + 3y = 12$: points $(0, 4)$, $(6, 0)$, $(3, 2)$. $x - y = 1$: points $(0, -1)$, $(1, 0)$, $(3, 2)$.
2. The two lines are drawn on the same axes and intersect at exactly one point, $(3, 2)$. (Check: $6 + 6 = 12$ ✓ and $3 - 2 = 1$ ✓.)
3. Since the lines intersect at a single point, the pair has a unique solution, $x = 3$, $y = 2$.

Answer: Unique solution: $x = 3$, $y = 2$

1. (a) **Given:** $DE \parallel BC$ in $\triangle ABC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** join $BE$ and $DC$; 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 $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{\frac{1}{2} \cdot AE \cdot DG}{\frac{1}{2} \cdot EC \cdot DG} = \dfrac{AE}{EC}$.
3. $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
4. (b) Given $\dfrac{AB}{PQ} = \dfrac{AD}{PM} = \dfrac{AC}{PR}$. Extend $AD$ to $E$ with $DE = AD$ and $PM$ to $N$ with $MN = PM$; join $CE$ and $RN$.
5. $\triangle DAB \cong \triangle DEC$ (SAS: $BD = DC$, $AD = DE$, vertically opposite angles), so $CE = AB$. Likewise $\triangle MPQ \cong \triangle MNR$ and $RN = PQ$.
6. Then $\dfrac{CE}{NR} = \dfrac{AB}{PQ} = \dfrac{AD}{PM} = \dfrac{AE}{PN}$ (as $AE = 2AD$, $PN = 2PM$) $= \dfrac{AC}{PR}$. So $\triangle CAE \sim \triangle RPN$ (SSS) and $\angle 1 = \angle 2$. In the same way $\angle 3 = \angle 4$.
7. Adding, $\angle 1 + \angle 3 = \angle 2 + \angle 4$, that is $\angle BAC = \angle QPR$. With $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$, we get $\triangle ABC \sim \triangle PQR$ by SAS.

Answer: Proved (both parts)

1. Let the building be $AB = h$ and the tower $CD = 30$ m, with $AC = x$.
2. From the foot $A$ of the building, the top of the tower is at $60^\circ$: $\tan 60^\circ = \dfrac{30}{x}$, so $x = 10\sqrt{3}$. (i)
3. From the foot $C$ of the tower, the top of the building is at $30^\circ$: $\tan 30^\circ = \dfrac{h}{x}$, so $x = h\sqrt{3}$. (ii)
4. From (i) and (ii), $h\sqrt{3} = 10\sqrt{3}$, so $h = 10$ m. $x = 10 \times 1.73 = 17.3$ m.

Answer: Building height $= 10$ m; distance $= 17.3$ m

1. (a) $N = 70$. Take $a = 47.5$ and $h = 15$: $u = -2, -1, 0, 1, 2, 3$ for the class marks $17.5, 32.5, 47.5, 62.5, 77.5, 92.5$. $\sum f_iu_i = -24 - 10 + 0 + 13 + 16 + 36 = 31$.
2. Mean $= 47.5 + \dfrac{31}{70} \times 15 \approx 54.14$. Modal class $40 - 55$ ($f_1 = 15$, $f_0 = 10$, $f_2 = 13$): Mode $= 40 + \dfrac{15 - 10}{30 - 10 - 13} \times 15 = 40 + \dfrac{75}{7} \approx 50.71$.
3. (b) Cumulative frequencies: $7, 17, 38, 60, 75, 80$. $N = 80$ and $\dfrac{N}{2} = 40$ lies in $35 - 45$: $l = 35$, $cf = 38$, $f = 22$, $h = 10$. Median $= 35 + \dfrac{40 - 38}{22} \times 10 \approx 35.91$.
4. The modal class is $35 - 45$ (frequency $22$): $f_1 = 22$, $f_0 = 21$, $f_2 = 15$. Mode $= 35 + \dfrac{22 - 21}{44 - 21 - 15} \times 10 = 35 + 1.25 = 36.25$.

Answer: (a) mean $\approx 54.14$, mode $\approx 50.71$   (b) median $\approx 35.91$, mode $= 36.25$

1. (i) $O$ is the centre of the circle inscribed in the equilateral triangle, so $OB$ bisects $\angle B = 60^\circ$: $\angle OBD = 30^\circ$. In the right triangle $ODB$ ($OD = 4$): $\sin 30^\circ = \dfrac{4}{OB}$, so $OB = 8$ cm.
2. (ii) Yes, $DE \parallel CA$. $ABC$ is equilateral and $AD \perp BC$, so $D$ is the mid-point of $BC$. In the same way $E$ is the mid-point of $AB$, and the line joining the mid-points of two sides is parallel to the third side: $DE \parallel CA$.
3. (iii)(a) $\angle EBD = 60^\circ$. The radii to the points of contact are perpendicular to the tangents: $\angle OEB = \angle ODB = 90^\circ$. So $\angle EOD = 360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ$. $\angle OEB + \angle ODB = 180^\circ$ (opposite angles supplementary), so $OEBD$ is a cyclic quadrilateral.
4. (iii)(b) $\cos 30^\circ = \dfrac{BD}{OB}$, so $BD = 8 \times \dfrac{\sqrt{3}}{2} = 4\sqrt{3} = 6.92$ cm. $BC = 2BD = 13.84$ cm. Perimeter $= 3 \times 13.84 = 41.52$ cm.

Answer: (i) $8$ cm   (ii) yes (mid-point theorem)   (iii)(a) $60^\circ, 90^\circ, 120^\circ, 90^\circ$; cyclic   (iii)(b) $41.52$ cm

1. (i) The blades are equally placed in the quarter circle ($\angle AOR = 90^\circ$), and two blades of $30^\circ$ each leave $90^\circ - 60^\circ = 30^\circ$ for the gaps, shared equally: $\angle BOP = \dfrac{30^\circ}{2} = 15^\circ$.
2. (ii) Arc $CD = \dfrac{30}{360} \times 2 \times \dfrac{22}{7} \times 21 = 11$ cm.
3. (iii)(a) Area $CABD = \dfrac{30}{360} \times \dfrac{22}{7} \times (28^2 - 21^2) = \dfrac{1}{12} \times \dfrac{22}{7} \times 343 = \dfrac{539}{6} \approx 89.8$ cm².
4. (iii)(b) Arc $AB = \dfrac{30}{360} \times 2 \times \dfrac{22}{7} \times 28 = \dfrac{44}{3} \approx 14.67$ cm. Perimeter $= AB\text{ arc} + CD\text{ arc} + CA + DB = 14.67 + 11 + 2 \times (28 - 21) = 39.67$ cm.

Answer: (i) $15^\circ$   (ii) $11$ cm   (iii)(a) $\approx 89.8$ cm²   (iii)(b) $\approx 39.67$ cm

1. (i) $h_1 = 44$, $h_2 = 44 + 10 = 54$, $h_3 = 64$, $h_4 = 74$, $h_5 = 84$.
2. (ii) The difference between consecutive heights is always $10$ ($54 - 44 = 64 - 54 = 74 - 64 = 84 - 74 = 10$), so they form an A.P. with first term $a = 44$ and common difference $d = 10$.
3. (iii)(a) $h_n = 44 + (n - 1) \times 10 \leq 160$ gives $(n - 1) \times 10 \leq 116$, so $n \leq 12.6$. The maximum number of chairs is $12$.
4. (iii)(b) $h_{15} = 44 + 14 \times 10 = 184$ cm, and $184 > 180$. So $15$ chairs cannot be stacked within $180$ cm.

Answer: (i) $44, 54, 64, 74, 84$   (ii) A.P., $a = 44$, $d = 10$   (iii)(a) $12$ chairs   (iii)(b) no, $h_{15} = 184 > 180$