CBSE Class 10 Maths Basic 2025 Question Paper (Set 430/3/1) with Solutions

1. LCM takes each prime at its highest power: $2^3$, $3^3$ and $5$. LCM $= 2^3 \times 3^3 \times 5$.

Answer: (D) $2^3 \times 3^3 \times 5$

1. A number ending in $5$ is divisible by $5$, but $3^n$ has only the prime factor $3$. So $3^n$ can never end in $5$. (Its last digits go $3, 9, 7, 1$.)

Answer: (B) 5

1. A prime $p$ has the factors $1$ and $p$ only, and the only prime among them is $p$ itself: exactly one prime factor.

Answer: (B) exactly one prime factor

1. A unique solution needs $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$: $\dfrac{k}{2} \neq \dfrac{2}{1}$, so $k \neq 4$.

Answer: (C) $k \neq 4$

1. Squaring both sides, $x^2 = x$, that is $x^2 - x + 0 = 0$. So $a = 1$, $b = -1$, $c = 0$ and $a + b + c = 0$.

Answer: (A) 0

1. $\dfrac{3 + 2a}{2} = 1$ gives $2a = -1$ and $a = -\dfrac{1}{2}$.
2. $\dfrac{5 + b}{2} = 2$ gives $b = -1$. So $(a, b) = \left(-\dfrac{1}{2}, -1\right)$.

Answer: (C) $\left(-\dfrac{1}{2}, -1\right)$

1. The distance from the $y$-axis is the absolute value of the $x$-coordinate: $|-3| = 3$. (A distance cannot be negative.)

Answer: (B) 3

1. $\sec^2 30^\circ = \dfrac{4}{3}$, $\tan^2 30^\circ = \dfrac{1}{3}$, and $\sin^2 45^\circ + \cos^2 45^\circ = 1$. The value is $\dfrac{4/3 + 1/3}{1} = \dfrac{5}{3}$.

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

1. All equilateral triangles have angles $60^\circ, 60^\circ, 60^\circ$, so any two are similar by AAA. The other types can have different angles.

Answer: (D) Equilateral triangles

1. $\triangle ABC \sim \triangle QRP$ matches $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$. So $\dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{AC}{QP}$.
2. $\dfrac{2}{4} = \dfrac{2.5}{x} = \dfrac{3}{y} = \dfrac{1}{2}$, so $x = 5$ and $y = 6$.

Answer: (B) $x = 5,\ y = 6$

1. A tangent to the inner circle can be drawn from a point on it or outside it. $Y$ lies inside the circle of radius $r$, and no tangent can be drawn from an interior point.

Answer: (D) Y

1. In the right triangle $ABC$ at $B$: $\sin C = \dfrac{AB}{AC} = \dfrac{2}{4} = \dfrac{1}{2}$, so $\angle C = 30^\circ$. (Check: $\tan C = \dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}}$ ✓.)

Answer: (A) $30^\circ$

1. $\dfrac{\theta}{360} = \dfrac{1}{8}$, so $\theta = 45^\circ$.

Answer: (B) $45^\circ$

1. $2\pi r = 22$, so $r = \dfrac{22 \times 7}{2 \times 22} = 3.5$ cm.
2. Perimeter of a quadrant $= \dfrac{1}{4}(22) + 2r = 5.5 + 7 = 12.5$ cm.

Answer: (C) 12.5 cm

1. $\dfrac{V_{\text{cone}}}{V_{\text{cylinder}}} = \dfrac{\frac{1}{3}\pi r^2 h}{\pi r^2 h} = \dfrac{1}{3}$. The ratio is $1 : 3$.

Answer: (B) 1 : 3

1. The modal class has the highest frequency, which is $8$, for the class $30 - 40$. Its lower limit is $30$.

Answer: (C) 30

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$: $12 = 3M - 30$, so $3M = 42$ and $M = 14$.

Answer: (C) 14

1. Sum $9$: $(3,6), (4,5), (5,4), (6,3)$ — $4$. Sum $10$: $(4,6), (5,5), (6,4)$ — $3$. Sum $11$: $(5,6), (6,5)$ — $2$. Sum $12$: $(6,6)$ — $1$.
2. Total $= 4 + 3 + 2 + 1 = 10$.

Answer: (C) 10

1. (A) $\sin\theta = 1$ gives $\cos^2\theta = 1 - 1 = 0$, so $\cos\theta = 0$. True.
2. (R) The identity is $\sin^2\theta + \cos^2\theta = 1$, not with a minus sign. False.

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

1. (A) $x^2 + 1 = 0$ has no real roots, so a quadratic equation need not have two real roots. False.
2. (R) A polynomial of degree $2$ has at most $2$ zeroes. True.

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

1. $x = 3$ is a vertical line. $2x - 3y = -6$: points $(0, 2)$, $(-3, 0)$ and $(3, 4)$.
2. The lines meet at $(3, 4)$: $x = 3$ and $y = 4$. (Check: $6 - 12 = -6$ ✓.)

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

1. (a) At the same time the sun’s rays make the same angle, so the two right triangles (pole and shadow, tower and shadow) are similar: $\dfrac{10}{15} = \dfrac{h}{45}$.
2. $h = \dfrac{10 \times 45}{15} = 30$ m.
3. (b) In $\triangle PQR$, $\angle 1 = \angle 2$, so the sides opposite them are equal: $PQ = PR$.
4. The given $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ becomes $\dfrac{QR}{QS} = \dfrac{QT}{PQ}$, that is $\dfrac{QR}{QT} = \dfrac{QS}{QP}$. Also $\angle Q$ is common to $\triangle PQS$ and $\triangle TQR$.
5. So $\triangle PQS \sim \triangle TQR$ by SAS similarity.

Answer: (a) $30$ m   (b) proved

1. $TP = TQ$ (tangents from an external point), so $\angle TPQ = \angle TQP$. In $\triangle TPQ$: $50^\circ + 2\angle TPQ = 180^\circ$, so $\angle TPQ = 65^\circ$.
2. $OP \perp TP$, so $\angle OPT = 90^\circ$ and $\angle OPQ = 90^\circ - 65^\circ = 25^\circ$.

Answer: $\angle OPQ = 25^\circ$

1. (a) $\sin 3A = 1$ gives $3A = 90^\circ$, so $A = 30^\circ$. $\cos 2A - \tan^2 45^\circ = \cos 60^\circ - 1 = \dfrac{1}{2} - 1 = -\dfrac{1}{2}$.
2. (b) $\left(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\right)(1 - \sin A) = \dfrac{(1 + \sin A)(1 - \sin A)}{\cos A} = \dfrac{1 - \sin^2 A}{\cos A} = \dfrac{\cos^2 A}{\cos A}$.
3. So the left side is $\cos A = 1 \times \cos A$, and $k = 1$.

Answer: (a) $-\dfrac{1}{2}$   (b) $k = 1$

1. Area of the quadrant $= \dfrac{1}{4} \times 3.14 \times 10 \times 10 = 78.5$ cm².
2. The unshaded part is the right triangle $POS$ with legs $OP = 10$ and $OS = 6$: area $= \dfrac{1}{2} \times 10 \times 6 = 30$ cm².
3. Shaded area $= 78.5 - 30 = 48.5$ cm².

Answer: $48.5$ cm²

1. (a) Suppose $\sqrt{2} = \dfrac{p}{q}$ with $p, q$ co-prime and $q \neq 0$. $2q^2 = p^2$, so $2$ divides $p^2$ and so $2$ divides $p$. Write $p = 2m$.
2. $2q^2 = 4m^2$ gives $q^2 = 2m^2$, so $2$ divides $q$. Then $2$ divides both $p$ and $q$, contradicting that they are co-prime. So $\sqrt{2}$ is irrational.
3. (b) $a = 13(77 + 1) = 13 \times 78$: composite. $b = 4(30 + 1) = 4 \times 31$: composite. $c = 91 + 6 = 97$, which is prime.

Answer: (a) proved   (b) $a$ and $b$ are composite

1. $9s^2 - 6s + 1 = (3s - 1)^2$, so both zeroes are $s = \dfrac{1}{3}$.
2. Sum $= \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}$ and $-\dfrac{(-6)}{9} = \dfrac{2}{3}$ ✓. Product $= \dfrac{1}{9}$ and $\dfrac{\text{constant}}{\text{coefficient of } s^2} = \dfrac{1}{9}$ ✓.

Answer: Zeroes $\dfrac{1}{3}, \dfrac{1}{3}$; relations verified

1. (a) Let the fraction be $\dfrac{x}{y}$. $\dfrac{x - 1}{y} = \dfrac{1}{3}$ gives $3x - y = 3$, and $\dfrac{x}{y + 8} = \dfrac{1}{4}$ gives $4x - y = 8$.
2. Subtracting: $x = 5$ and $y = 12$. The fraction is $\dfrac{5}{12}$.
3. (b) Infinitely many solutions need $\dfrac{k}{12} = \dfrac{3}{k} = \dfrac{k - 3}{k}$. $k^2 = 36$ and $k^2 - 6k = 0$ hold only for $k = 6$.
4. For $k = 6$ both equations become $2x + y = 1$. Two solutions: $(0, 1)$ and $(1, -1)$.

Answer: (a) $\dfrac{5}{12}$   (b) $k = 6$; solutions $(0, 1)$ and $(1, -1)$

1. LHS $= \dfrac{1 + \frac{1}{\sin A}}{\frac{1}{\sin A}} = \dfrac{\sin A + 1}{\sin A} \times \sin A = 1 + \sin A$.
2. Multiply and divide by $(1 - \sin A)$: $\dfrac{(1 + \sin A)(1 - \sin A)}{1 - \sin A} = \dfrac{1 - \sin^2 A}{1 - \sin A} = \dfrac{\cos^2 A}{1 - \sin A} =$ RHS.

Answer: Proved

1. Total marbles $= 6 + 4 + 8 = 18$.
2. (i) $P(\text{white}) = \dfrac{4}{18} = \dfrac{2}{9}$. (ii) $P(\text{white or red}) = \dfrac{12}{18} = \dfrac{2}{3}$. (iii) $P(\text{not red}) = \dfrac{10}{18} = \dfrac{5}{9}$.

Answer: (i) $\dfrac{2}{9}$   (ii) $\dfrac{2}{3}$   (iii) $\dfrac{5}{9}$

1. Join $O$ to $A, B, C, D$ and to the points of contact $P, Q, R, S$. Tangents from a point are equal and radii are equal, so $\triangle OAP \cong \triangle OAS$ (SSS: $AP = AS$, $OP = OS$, $OA$ common). Hence $\angle AOP = \angle AOS$.
2. In the same way $\angle BOP = \angle BOQ$, $\angle COQ = \angle COR$ and $\angle DOR = \angle DOS$.
3. The eight angles round $O$ add up to $360^\circ$: $2(\angle AOP + \angle BOP + \angle COQ + \angle DOR) = 360^\circ$. So $(\angle AOP + \angle BOP) + (\angle COQ + \angle DOR) = 180^\circ$, that is $\angle AOB + \angle COD = 180^\circ$.

Answer: Proved

1. (a) Let the integers be $x$ and $x + 2$. $x^2 + (x + 2)^2 = 290$ gives $2x^2 + 4x - 286 = 0$, that is $x^2 + 2x - 143 = 0$ and $(x - 11)(x + 13) = 0$.
2. $x = 11$ gives $11$ and $13$. (The root $x = -13$ gives the pair $-13$ and $-11$, which also fit the words of the question; the key lists $11$ and $13$.)
3. (b) Let the breadth be $x$ m, so the length is $(2x + 1)$ m. $(2x + 1)x = 300$ gives $2x^2 + x - 300 = 0$ and $(x - 12)(2x + 25) = 0$.
4. A width cannot be negative, so $x = 12$. Length $= 25$ m and breadth $= 12$ m. (Check: $25 \times 12 = 300$ ✓.)

Answer: (a) $11$ and $13$ (also $-13$ and $-11$)   (b) length $25$ m, breadth $12$ 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, the other two sides are divided in the same ratio.
2. **Given:** $DE \parallel BC$ in $\triangle ABC$. **To prove:** $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. **Construction:** draw $DM \perp AC$ and $EN \perp AB$; join $BE$ and $CD$.
3. $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DBE)} = \dfrac{\frac{1}{2} \cdot AD \cdot EN}{\frac{1}{2} \cdot DB \cdot EN} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle ECD)} = \dfrac{\frac{1}{2} \cdot AE \cdot DM}{\frac{1}{2} \cdot EC \cdot DM} = \dfrac{AE}{EC}$.
4. $\triangle DBE$ and $\triangle ECD$ are on the same base $DE$ and between the same parallels $BC$ and $DE$, so their areas are equal. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
5. (b)(i) $\triangle ABC \sim \triangle PQR$ gives $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ and $\angle A = \angle P$. Since $CM$, $RN$ are medians, $AB = 2AM$ and $PQ = 2PN$, so $\dfrac{AC}{PR} = \dfrac{AM}{PN}$. So $\triangle AMC \sim \triangle PNR$ (SAS).
6. (ii) From (i), $\angle ACM = \angle PRN$. Also $\angle ACB = \angle PRQ$. Subtracting, $\angle ACB - \angle ACM = \angle PRQ - \angle PRN$, so $\angle BCM = \angle QRN$.
7. (iii) $\dfrac{BC}{QR} = \dfrac{AB}{PQ} = \dfrac{2BM}{2QN} = \dfrac{BM}{QN}$ and $\angle B = \angle Q$. So $\triangle BMC \sim \triangle QNR$ (SAS).

Answer: Proved (both parts)

1. Radius $r = \dfrac{5}{2}$ cm and height $h = 10$ cm. The capacity is the cylinder minus the raised hemisphere.
2. Cylinder $= \pi r^2 h = 3.14 \times \dfrac{5}{2} \times \dfrac{5}{2} \times 10 = 196.25$ cm³.
3. Hemisphere $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times 3.14 \times \dfrac{5}{2} \times \dfrac{5}{2} \times \dfrac{5}{2} \approx 32.71$ cm³.
4. Capacity $= 196.25 - 32.71 = 163.54 \approx 163.5$ cm³, that is $\dfrac{3925}{24} \approx 163.5$ mL.

Answer: About $163.5$ mL

1. Cumulative frequencies: $2, 7, 15, 21, 27, 30$. $\dfrac{N}{2} = 15$.
2. The cumulative frequency $15$ is reached at the class $50 - 55$, which is the median class: $l = 50$, $cf = 7$, $f = 8$, $h = 5$.
3. Median $= 50 + \dfrac{15 - 7}{8} \times 5 = 50 + 5 = 55$ kg.

Answer: $55$ kg

1. The numbers joining on days $1, 2, 3, \dots$ are $5, 8, 11, \dots$, an A.P. with $a = 5$ and $d = 3$.
2. (i) $5 + (n - 1) \times 3 = 59$ gives $n - 1 = 18$ and $n = 19$.
3. (ii) $a_{31} = 5 + 30 \times 3 = 95$.
4. (iii)(a) ₹ $5000$ at ₹ $100$ a person means $50$ people: $5 + (n - 1) \times 3 = 50$ gives $n = 16$.
5. (iii)(b) $S_{16} = \dfrac{16}{2}[2 \times 5 + 15 \times 3] = 8 \times 55 = 440$ people in all, so he earned $440 \times 100 = ₹44{,}000$.

Answer: (i) day $19$   (ii) $95$ people   (iii)(a) day $16$   (iii)(b) ₹ $44{,}000$

1. (i) In the right triangle $PDA$: $\tan 30^\circ = \dfrac{PD}{AD} = \dfrac{20}{AD}$, so $AD = 20\sqrt{3} = 20 \times 1.73 = 34.6$ m.
2. (ii) In $\triangle PDB$: $\tan 45^\circ = \dfrac{20}{DB} = 1$, so $DB = 20$ m. Width $= AD + DB = 34.6 + 20 = 54.6$ m.
3. (iii)(a) Boat $A$ takes $\dfrac{34.6}{10} = 3.46$ s and boat $B$ takes $\dfrac{20}{5} = 4$ s. Boat $A$ reaches $D$ first, by $4 - 3.46 = 0.54$ s.
4. (iii)(b) In $3$ s boat $A$ covers $30$ m and boat $B$ covers $15$ m. They move towards each other, so the distance between them is $54.6 - (30 + 15) = 9.6$ m.

Answer: (i) $34.6$ m   (ii) $54.6$ m   (iii)(a) boat A, $0.54$ s earlier   (iii)(b) $9.6$ m

1. (i) $O$ is the centre, the mid-point of the diameter $AB$: $O = \left(\dfrac{2 + 22}{2}, \dfrac{3 + 3}{2}\right) = (12, 3)$.
2. (ii) Radius $= OA = 12 - 2 = 10$ (the distance from $(12, 3)$ to $(2, 3)$).
3. (iii)(a) $C$ is on the boundary, so $OC = 10$: $\sqrt{(12 - 12)^2 + (y - 3)^2} = 10$ gives $y - 3 = \pm 10$, that is $y = 13$ or $y = -7$. The park is the semicircle below $AB$, so $C = (12, -7)$ (the key also lists $(12, 13)$ as a solution of the equation).
4. (iii)(b) $PA = \dfrac{1}{3}PB$ means $AP : PB = 1 : 3$. $P = \left(\dfrac{1 \times 22 + 3 \times 2}{4}, \dfrac{1 \times 3 + 3 \times 3}{4}\right) = \left(\dfrac{28}{4}, \dfrac{12}{4}\right) = (7, 3)$.

Answer: (i) $(12, 3)$   (ii) $10$   (iii)(a) $(12, -7)$   (iii)(b) $(7, 3)$