Chapter-wise previous year questions (2023–2026) with official marking scheme solutions. Free.
Answer: $8500$
$85 \times 17 + 85 \times 83 = 85 \times (17 + 83) = 85 \times 100 = 8500$
Answer: $\dfrac{5}{4} = 1\dfrac{1}{4}$
LCM of 4, 6, 3 = 12. $\frac{9}{12} + \frac{10}{12} - \frac{4}{12} = \frac{15}{12} = \frac{5}{4}$
Answer: $x = 6$
$3x = 25 - 7 = 18$, so $x = 18 \div 3 = 6$
Answer: $70^\circ$
Sum of angles in a triangle = $180^\circ$. Third angle = $180^\circ - (45^\circ + 65^\circ) = 180^\circ - 110^\circ = 70^\circ$
Answer: $12$ girls
$3:2 = 18:x$. So $3x = 36$, $x = 12$
Answer: Area = $400\text{ m}^2$, Perimeter = $82\text{ m}$
Area = $25 \times 16 = 400\text{ m}^2$. Perimeter = $2(25 + 16) = 2 \times 41 = 82\text{ m}$
Answer: $0$
$(-15) + (-8) - (-20) + 3 = -15 - 8 + 20 + 3 = -23 + 23 = 0$
Answer: $24$
Sum = $12+18+24+30+36 = 120$. Mean = $120 \div 5 = 24$
Answer: $\dfrac{2}{3}$
Let $x = 0.666...$ Then $10x = 6.666...$, so $10x - x = 6$, $9x = 6$, $x = \frac{6}{9} = \frac{2}{3}$
Answer: $17$
$p(2) = 3(8) - 4(4) + 7(2) - 5 = 24 - 16 + 14 - 5 = 17$
Answer: $4\sqrt{2}$ units
$d = \sqrt{(6-2)^2 + (7-3)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Answer: $x = \frac{11}{5},\; y = \frac{6}{5}$
From second eq: $x = y + 1$. Substitute: $2(y+1) + 3y = 8$, $5y + 2 = 8$, $5y = 6$, $y = \frac{6}{5}$, $x = \frac{11}{5}$
Answer: $70^\circ$
Vertically opposite angles are equal. $\angle AOC = \angle BOD = 70^\circ$
Answer: $\angle C = 50^\circ,\; \angle A = 80^\circ$
Since $AB = AC$, angles opposite equal sides are equal. So $\angle B = \angle C = 50^\circ$. $\angle A = 180^\circ - 50^\circ - 50^\circ = 80^\circ$
Answer: $\angle B = 110^\circ,\; \angle C = 70^\circ,\; \angle D = 110^\circ$
Adjacent angles are supplementary: $\angle B = 180^\circ - 70^\circ = 110^\circ$. Opposite angles are equal: $\angle C = \angle A = 70^\circ$, $\angle D = \angle B = 110^\circ$
Answer: $15$
Arrange: $8, 10, 12, \mathbf{15}, 18, 20, 25$. The 4th (middle) value is $15$.
Answer: HCF = $4$, LCM = $9696$
$96 = 2^5 \times 3$, $404 = 2^2 \times 101$. HCF = $2^2 = 4$. LCM = $\frac{96 \times 404}{4} = \frac{38784}{4} = 9696$
Answer: Zeroes: $5$ and $-2$
$x^2 - 3x - 10 = (x-5)(x+2)$. Zeroes: $5, -2$. Sum $= 5 + (-2) = 3 = -\frac{-3}{1}$ ✓. Product $= 5 \times (-2) = -10 = \frac{-10}{1}$ ✓
Answer: $x = 3,\; y = 2$
Add both eqs: $3x = 9$, so $x = 3$. Substitute: $3 + y = 5$, so $y = 2$. Intersection point: $(3, 2)$
Answer: $x = 3$ or $x = \dfrac{1}{2}$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}$. $x = \frac{12}{4} = 3$ or $x = \frac{2}{4} = \frac{1}{2}$
Answer: $820$
$a = 3$, $d = 4$, $n = 20$. $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{20}{2}[6 + 19 \times 4] = 10[6 + 76] = 10 \times 82 = 820$
Answer: $EC = 7.5\text{ cm}$
By Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$, so $\frac{4}{6} = \frac{5}{EC}$, $EC = \frac{5 \times 6}{4} = 7.5\text{ cm}$
Answer: $\cos A = \dfrac{4}{5},\; \tan A = \dfrac{3}{4}$
$\sin^2 A + \cos^2 A = 1$, so $\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25}$, $\cos A = \frac{4}{5}$. $\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4}$
Answer: $\dfrac{50}{\sqrt{3}} \approx 28.87\text{ m}$
$\tan 60^\circ = \frac{50}{d}$, so $\sqrt{3} = \frac{50}{d}$, $d = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3} \approx 28.87\text{ m}$
Answer: $\dfrac{77}{3} \approx 25.67\text{ cm}^2$
Area = $\frac{\theta}{360^\circ} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 49 = \frac{1}{6} \times 154 = \frac{77}{3}\text{ cm}^2$
Answer: $3$
Frequency: $3 \rightarrow 4$ times, $7 \rightarrow 3$ times, $8 \rightarrow 2$ times. Mode = $3$ (most frequent)
Answer: $\{3, 5, 7, 11, 13, 17, 19\}$
$A = \{2, 3, 5, 7, 11, 13, 17, 19\}$, $B = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}$. Intersection: primes that are also odd.
Answer: Verified ✓
$\sin\frac{\pi}{6} = \frac{1}{2}$, so $\sin^2 = \frac{1}{4}$. $\cos\frac{\pi}{3} = \frac{1}{2}$, so $\cos^2 = \frac{1}{4}$. $\tan\frac{\pi}{4} = 1$, so $\tan^2 = 1$. LHS = $\frac{1}{4} + \frac{1}{4} - 1 = \frac{2}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$ ✓
Answer: $8 - i$
$(2)(1) + (2)(-2i) + (3i)(1) + (3i)(-2i) = 2 - 4i + 3i - 6i^2 = 2 - i + 6 = 8 - i$ (since $i^2 = -1$)
Answer: $120$
5 letters, choose and arrange 4: $P(5,4) = \frac{5!}{(5-4)!} = \frac{120}{1} = 120$
Answer: $6$
$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3$. $\lim_{x \to 3} (x+3) = 6$
Answer: $y = 3x - 1$ or $3x - y - 1 = 0$
Point-slope form: $y - 5 = 3(x - 2)$, so $y - 5 = 3x - 6$, $y = 3x - 1$
Answer: Both one-one and onto (bijective)
One-one: $f(a) = f(b) \implies a^3 = b^3 \implies a = b$. Onto: For any $y \in \mathbb{R}$, $x = \sqrt[3]{y}$ satisfies $f(x) = y$.
Answer: $\dfrac{\pi}{2}$
$\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$, $\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$. Sum = $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. Note: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1,1]$
Answer: $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
$\det(A) = 2(2) - 3(1) = 1$. $A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
Answer: $0$
$= 1(45-48) - 2(36-42) + 3(32-35) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0$
Answer: $f'(x) = 3x^2 \sin x + x^3 \cos x$
Product rule: $f'(x) = (x^3)' \sin x + x^3 (\sin x)' = 3x^2 \sin x + x^3 \cos x$
Answer: $x > 2$ i.e., $(2, \infty)$
$f'(x) = 2x - 4$. For increasing: $f'(x) > 0$, so $2x - 4 > 0$, $x > 2$.
Answer: $x^3 + x^2 + x + C$
$\int 3x^2\,dx = x^3$, $\int 2x\,dx = x^2$, $\int 1\,dx = x$. So answer = $x^3 + x^2 + x + C$
Answer: $6$
$\int_0^2 (2x+1)dx = [x^2 + x]_0^2 = (4 + 2) - (0 + 0) = 6$
Answer: $\dfrac{3}{13}\hat{i} - \dfrac{4}{13}\hat{j} + \dfrac{12}{13}\hat{k}$
$|\vec{a}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Unit vector = $\frac{\vec{a}}{|\vec{a}|}$
Answer: $\dfrac{2}{15}$
$P(\text{both red}) = P(R_1) \times P(R_2|R_1) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$