CBSE Class 10 Maths Basic 2025 Question Paper (Set 430/2/1) with Solutions

1. $3$ and $5$ are co-prime: HCF $= 1$ and LCM $= 15$. HCF $-$ LCM $= 1 - 15 = -14$.

Answer: (D) −14

1. A number ending in $0$ is divisible by $5$, but $2^n$ has only the prime factor $2$. So $2^n$ can never end in $0$. (It can end in $2, 4, 8, 6$.)

Answer: (D) 0

1. Substituting $x = 0$, $y = 0$: $0 + 0 = c - 1$, so $c = 1$.

Answer: (B) 1

1. Equal roots need $D = b^2 - 4ac = 0$: $9 - 4a = 0$, so $a = \dfrac{9}{4}$.

Answer: (B) $\dfrac{9}{4}$

1. (A) $x^2 = x^2 + 2x + 1$ reduces to $2x + 1 = 0$: linear. (B) $x^2 + x - 2 = 2x + 1$ gives $x^2 - x - 3 = 0$: quadratic.
2. (C) $x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$ is cubic in form (the $x^3$ terms do not cancel, leaving degree $3$). (D) has $\sqrt{x}$, not a polynomial.

Answer: (B) $(x - 1)(x + 2) = 2x + 1$

1. $\sqrt{2^2 + 3^2} = \sqrt{13}$.

Answer: (D) $\sqrt{13}$

1. Mid-point $= \left(\dfrac{1 + 1}{2}, \dfrac{3 - 3}{2}\right) = (1, 0)$. Its $y$-coordinate is $0$, so it lies on the $x$-axis.

Answer: (C) on $x$-axis

1. $DE \parallel BC$, so $\dfrac{AD}{DB} = \dfrac{AE}{EC}$: $\dfrac{1.5}{3} = \dfrac{AE}{2}$, so $AE = 1$ cm.
2. $AC = AE + EC = 1 + 2 = 3$ cm.

Answer: (B) 3 cm

1. The tangent to the smaller circle lies inside the larger circle at its point of contact, so it must leave the larger circle in both directions: it cuts the larger circle at two points (it is a chord of the larger circle).

Answer: (C) two points

1. $\tan 60^\circ = \sqrt{3}$: $\dfrac{2\sqrt{3}}{1 - 3} = \dfrac{2\sqrt{3}}{-2} = -\sqrt{3}$. (This is $\tan 120^\circ$.)

Answer: (D) $-\sqrt{3}$

1. Each bracket is a known identity: $\sin^2 A + \cos^2 A = 1$, $\sec^2 A - \tan^2 A = 1$, $\cot^2 A - \operatorname{cosec}^2 A = -1$. The value is $1 + 1 - (-1) = 3$.

Answer: (C) 3

1. The shadow length is $\dfrac{h}{\tan\theta}$. A higher sun means a larger $\tan\theta$, so the shadow gets shorter: from $h\sqrt{3}$ at $30^\circ$ to $\dfrac{h}{\sqrt{3}}$ at $60^\circ$.

Answer: (A) become shorter

1. The shaded part is bounded by a chord and the shorter arc and does not contain the centre, so it is a minor segment (a sector would reach the centre).

Answer: (C) minor segment

1. A quadrant has a quarter arc, $\dfrac{1}{4} \times 2\pi r = \dfrac{\pi r}{2}$, and two radii: perimeter $= \dfrac{\pi r}{2} + 2r$.

Answer: (D) $\dfrac{\pi r}{2} + 2r$

1. $\dfrac{1}{3}\pi r^2 h_1 = \pi r^2 h_2$, so $h_1 = 3h_2$ and $h_1 : h_2 = 3 : 1$.

Answer: (B) 3 : 1

1. In the mode formula $f_1$ is the frequency of the modal class, $f_0$ the frequency of the class before it, and $f_2$ the frequency of the class after it.

Answer: (A) frequency of the modal class

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 3a - 2a = a$.

Answer: (C) a

1. Each die has $6$ outcomes, so three dice give $6 \times 6 \times 6 = 216$.

Answer: (D) 216

1. (A) $36 = 2^2 \times 3^2$ and $6 = 2 \times 3$. The primes $2$ and $3$ divide both. True.
2. (R) $8$ divides $4^2 = 16$ but $8$ does not divide $4$. False (it is true only for prime divisors).

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

1. (R) Congruent triangles have equal corresponding sides, so the ratio is $1 : 1$. True.
2. (A) Equal corresponding angles and sides in the ratio $1 : 1$ make the triangles similar (scale factor $1$). True, and R is exactly the reason.

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

1. (a) Infinitely many solutions need $\dfrac{c}{12} = \dfrac{3}{c} = \dfrac{c - 3}{c}$.
2. $\dfrac{c}{12} = \dfrac{3}{c}$ gives $c^2 = 36$, so $c = \pm 6$. $\dfrac{3}{c} = \dfrac{c - 3}{c}$ gives $c - 3 = 3$, so $c = 6$. Both hold only for $c = 6$.
3. (b) Adding: $5x + 5y = 125$, so $x + y = 25$. Subtracting: $x - y = 5$. So $x = 15$ and $y = 10$. (Check: $45 + 20 = 65$ ✓, $30 + 30 = 60$ ✓.)

Answer: (a) $c = 6$   (b) $x = 15$, $y = 10$

1. In $\triangle ADC$ and $\triangle BAC$: $\angle ADC = \angle BAC$ (given) and $\angle C$ is common.
2. So $\triangle ADC \sim \triangle BAC$ by AA, and $\dfrac{CA}{CB} = \dfrac{CD}{CA}$. Hence $(CA)^2 = CB \cdot CD$.

Answer: Proved

1. **Given:** a circle with centre $O$ and a tangent $\ell$ at the point $A$. **To prove:** $OA \perp \ell$.
2. Take any other point $B$ on $\ell$. A tangent touches the circle only at $A$, so $B$ lies outside the circle and $OB >$ radius $= OA$.
3. This is true for every point $B \neq A$ of $\ell$, so $OA$ is the shortest distance from $O$ to $\ell$. The shortest distance from a point to a line is the perpendicular, so $OA \perp \ell$.

Answer: Proved

1. $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$, $\sec 30^\circ = \dfrac{2}{\sqrt{3}}$, $\operatorname{cosec} 30^\circ = 2$.
2. $\dfrac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} = \dfrac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} = \dfrac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$.
3. Rationalise by multiplying top and bottom by $(2\sqrt{6} - 2\sqrt{2})$: the denominator becomes $24 - 8 = 16$ and the numerator $\sqrt{3}(2\sqrt{6} - 2\sqrt{2}) = 6\sqrt{2} - 2\sqrt{6}$. The value is $\dfrac{6\sqrt{2} - 2\sqrt{6}}{16} = \dfrac{3\sqrt{2} - \sqrt{6}}{8}$.

Answer: $\dfrac{3\sqrt{2} - \sqrt{6}}{8} \approx 0.224$

1. (a) $\pi r^2 = \dfrac{120}{360}\pi R^2$, so $\dfrac{r^2}{R^2} = \dfrac{1}{3}$ and $r : R = 1 : \sqrt{3}$.
2. (b) $OA = OB = 7$ and $\angle AOB = 60^\circ$, so $\triangle OAB$ is equilateral and $AB = 7$ cm.
3. Minor arc $AB = \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 7 = \dfrac{22}{3}$ cm. Perimeter of the shaded region (arc + chord) $= \dfrac{22}{3} + 7 = \dfrac{43}{3} \approx 14.33$ cm.

Answer: (a) $1 : \sqrt{3}$   (b) $\dfrac{43}{3} \approx 14.33$ cm

1. (a) Suppose $\sqrt{5} = \dfrac{p}{q}$ with $p, q$ co-prime and $q \neq 0$. $5q^2 = p^2$, so $5$ divides $p^2$ and so $5$ divides $p$. Write $p = 5m$.
2. $5q^2 = 25m^2$ gives $q^2 = 5m^2$, so $5$ divides $q$. Then $p$ and $q$ have a common factor $5$, contradicting that they are co-prime. So $\sqrt{5}$ is irrational.
3. (b) **Fundamental Theorem of Arithmetic:** every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the primes.
4. $36 = 2^2 \times 3^2$ and $54 = 2 \times 3^3$. LCM $= 2^2 \times 3^3 = 108$.

Answer: (a) proved   (b) LCM $= 108$

1. $3x^2 - 2x - 1 = (3x + 1)(x - 1)$, so the zeroes are $x = -\dfrac{1}{3}$ and $x = 1$.
2. Sum of zeroes $= -\dfrac{1}{3} + 1 = \dfrac{2}{3}$, and $-\dfrac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\dfrac{-2}{3} = \dfrac{2}{3}$ ✓.
3. Product $= -\dfrac{1}{3} \times 1 = -\dfrac{1}{3}$, and $\dfrac{\text{constant term}}{\text{coefficient of } x^2} = \dfrac{-1}{3}$ ✓.

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

1. (a) $2x - y = 2$: points $(0, -2)$, $(1, 0)$, $(2, 2)$. $4x - y = 4$: points $(0, -4)$, $(1, 0)$, $(2, 4)$.
2. The lines meet at $(1, 0)$, so $x = 1$ and $y = 0$. They cut the $y$-axis at $(0, -2)$ and $(0, -4)$.
3. (b) Let a bat cost ₹ $x$ and a ball ₹ $y$. $10x + 5y = 32500$, that is $2x + y = 6500$ (i), and $2x + 8y = 10000$ (ii).
4. (i) $-$ (ii): $-7y = -3500$, so $y = 500$. Then $2x = 6500 - 500 = 6000$ and $x = 3000$. A bat costs ₹ $3000$ and a ball ₹ $500$.

Answer: (a) $(1, 0)$; $y$-intercepts $(0, -2)$ and $(0, -4)$   (b) bat ₹ $3000$, ball ₹ $500$

1. $\angle POQ = 360^\circ - 250^\circ = 110^\circ$.
2. The radius is perpendicular to the tangent at the point of contact: $\angle OPT = \angle OQT = 90^\circ$.
3. In the quadrilateral $OPTQ$ the angles add up to $360^\circ$: $\angle PTQ = 360^\circ - 90^\circ - 90^\circ - 110^\circ = 70^\circ$.

Answer: $\angle POQ = 110^\circ$, $\angle OPT = \angle OQT = 90^\circ$, $\angle PTQ = 70^\circ$

1. LHS $= \dfrac{\cos^2\theta + (1 + \sin\theta)^2}{(1 + \sin\theta)\cos\theta}$.
2. The numerator is $\cos^2\theta + 1 + 2\sin\theta + \sin^2\theta = 2 + 2\sin\theta = 2(1 + \sin\theta)$.
3. LHS $= \dfrac{2(1 + \sin\theta)}{(1 + \sin\theta)\cos\theta} = \dfrac{2}{\cos\theta} = 2\sec\theta =$ RHS.

Answer: Proved

1. There are $10$ equally likely outcomes.
2. (i) Primes greater than $2$: $3, 5, 7$, which is $3$ numbers. $P = \dfrac{3}{10}$.
3. (ii) Odd numbers less than $9$: $1, 3, 5, 7$, which is $4$ numbers. $P = \dfrac{4}{10} = \dfrac{2}{5}$.
4. (iii) Multiples of $4$: $4, 8$, which is $2$ numbers. $P = \dfrac{2}{10} = \dfrac{1}{5}$.

Answer: (i) $\dfrac{3}{10}$   (ii) $\dfrac{2}{5}$   (iii) $\dfrac{1}{5}$

1. (a) Let the sides be $x$ and $y$. $x^2 + y^2 = 2650$ (i) and $4x + 4y = 280$, so $x + y = 70$ (ii).
2. Substituting $y = 70 - x$: $x^2 + (70 - x)^2 = 2650$ gives $2x^2 - 140x + 4900 = 2650$, so $x^2 - 70x + 1125 = 0$.
3. $(x - 25)(x - 45) = 0$, so $x = 25$ or $45$ and $y = 45$ or $25$. The sides are $25$ cm and $45$ cm (both roots are valid: they are just the two squares).
4. (b) $\dfrac{(x - 2) - x}{x(x - 2)} = 3$ gives $-2 = 3x(x - 2)$, so $3x^2 - 6x + 2 = 0$.
5. $D = 36 - 24 = 12$. $x = \dfrac{6 \pm \sqrt{12}}{6} = 1 \pm \dfrac{\sqrt{3}}{3}$.

Answer: (a) $25$ cm and $45$ cm   (b) $3x^2 - 6x + 2 = 0$; roots $1 + \dfrac{\sqrt{3}}{3}$ and $1 - \dfrac{\sqrt{3}}{3}$

1. **SAS similarity:** if one angle of a triangle equals one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
2. **Proof:** $OA \cdot OC = OB \cdot OD$ gives $\dfrac{OA}{OB} = \dfrac{OD}{OC}$.
3. In $\triangle OAD$ and $\triangle OBC$: $\angle AOD = \angle BOC$ (vertically opposite angles) and the sides including them are proportional, $\dfrac{OA}{OB} = \dfrac{OD}{OC}$.
4. So $\triangle OAD \sim \triangle OBC$ by SAS, and $\angle OAD = \angle OBC$, that is $\angle A = \angle B$.
5. These are alternate interior angles for the lines $AD$ and $CB$, so $AD \parallel CB$.

Answer: Proved

1. Radius $r = 7$ cm for both the cylinder and the hemisphere; cylinder height $h = 10$ cm.
2. The surface of the remaining article is the curved side of the cylinder, the bottom base, and the inside surface of the hemispherical hollow (the top face is completely removed because the hemisphere has the same diameter).
3. Curved side $= 2\pi r h = 2 \times \dfrac{22}{7} \times 7 \times 10 = 440$ cm². Bottom base $= \pi r^2 = \dfrac{22}{7} \times 49 = 154$ cm². Hemispherical hollow $= 2\pi r^2 = 308$ cm².
4. Total surface area $= 440 + 154 + 308 = 902$ cm².

Answer: $902$ cm²

1. (a) Cumulative frequencies: $8, 17, 29, 34, 40$. $\dfrac{N}{2} = 20$ lies in $140 - 160$: $l = 140$, $cf = 17$, $f = 12$, $h = 20$.
2. Median $= 140 + \dfrac{20 - 17}{12} \times 20 = 140 + 5 = 145$ mm.
3. (b) The table has classes $0-4, 4-8, 8-12, 12-16, 16-20, 20-24$ with number of absent students $1, 8, x, 6, 5, y$ (class marks $2, 6, 10, 14, 18, 22$).
4. The total is $30$: $1 + 8 + x + 6 + 5 + y = 20 + x + y = 30$, so $x + y = 10$. (i)
5. $\sum f_ix_i = 2 + 48 + 10x + 84 + 90 + 22y = 224 + 10x + 22y$. Mean $= 12$: $224 + 10x + 22y = 360$, so $5x + 11y = 68$. (ii)
6. From (i) $x = 10 - y$: $50 - 5y + 11y = 68$, so $6y = 18$ and $y = 3$, $x = 7$.

Answer: (a) $145$ mm   (b) $x = 7$, $y = 3$

1. (i) $\sin 30^\circ = \dfrac{60}{l}$, so $l = \dfrac{60}{1/2} = 120$ m.
2. (ii) Ravi’s string is also $120$ m. His kite is $60 - 10 = 50$ m above his rooftop: $\sin\theta = \dfrac{50}{120} = \dfrac{5}{12}$.
3. (iii)(a) At $\theta = 60^\circ$ the height above the rooftop is $120\sin 60^\circ = 60\sqrt{3} = 60 \times 1.7 = 102$ m. Height above the ground $= 102 + 10 = 112$ m.
4. (iii)(b) At $30^\circ$ the height above the rooftop is $120 \times \dfrac{1}{2} = 60$ m, so the height above the ground is $60 + 10 = 70$ m.

Answer: (i) $120$ m   (ii) $\dfrac{5}{12}$   (iii)(a) $112$ m   (iii)(b) $70$ m

1. The instalments form an A.P. with $a = 10000$ and $d = 5000$.
2. (i) $a_{10} = 10000 + 9 \times 5000 = 55000$. She paid ₹ $55{,}000$.
3. (ii) $40000 = 10000 + (n - 1) \times 5000$, so $n - 1 = 6$ and $n = 7$: the $7^{\text{th}}$ instalment.
4. (iii)(a) $S_n = \dfrac{n}{2}[2(10000) + (n - 1)5000] = 1150000$ gives $n[20000 + 5000(n - 1)] = 2300000$, so $n^2 + 3n - 460 = 0$ and $(n - 20)(n + 23) = 0$. $n = 20$ (reject $-23$).
5. (iii)(b) $\dfrac{n}{2}[20000 + 5000(n - 1)] = 325000$ gives $n^2 + 3n - 130 = 0$ and $(n - 10)(n + 13) = 0$. $n = 10$: by the $10^{\text{th}}$ instalment.

Answer: (i) ₹ $55{,}000$   (ii) $7^{\text{th}}$   (iii)(a) $20$ instalments   (iii)(b) the $10^{\text{th}}$ instalment

1. (i) The diagonals of a rectangle bisect each other, so $E$ is the mid-point of $AC$: $\left(\dfrac{10 + 40}{2}, \dfrac{10 + 50}{2}\right) = (25, 30)$.
2. (ii) $BC = \sqrt{(40 - 40)^2 + (50 - 10)^2} = 40$.
3. (iii)(a) $E$ is also the mid-point of $BD$: $\dfrac{x + 40}{2} = 25$ and $\dfrac{y + 10}{2} = 30$, so $x = 10$ and $y = 50$. $D = (10, 50)$. $BD = \sqrt{(40 - 10)^2 + (10 - 50)^2} = \sqrt{900 + 1600} = 50$.
4. (iii)(b) $AE = \sqrt{15^2 + 20^2} = 25$, $EB = \sqrt{15^2 + 20^2} = 25$, $BA = 30$. Total $= 25 + 25 + 30 = 80$ units.

Answer: (i) $(25, 30)$   (ii) $40$   (iii)(a) $D(10, 50)$, $BD = 50$   (iii)(b) $80$ units