SSC CGL 2025 PYQs

Concept: Convert time to hours before dividing.

1. 38 min = $\frac{38}{60}$ h
2. Speed = $5.2\times \frac{60}{38}=\frac{312}{38}$
3. = 8.21 km/h (approx.)

Trick: In 38 min ≈ 5.2 km, so in 60 min a bit over 8 km → only 8.21 fits.

Answer: (A) 8.21 km/h

Concept: Profit share ∝ capital × months.

1. $A=40\text{k}\times 12=480\text{k}$; $B=90\text{k}\times 12=1080\text{k}$; $C=140\text{k}\times 6=840\text{k}$
2. Ratio = 4 : 9 : 7 (sum 20) → 1 part = $\frac{80,000}{20}=$ ₹4,000
3. C − A = $(7-4)\times 4,000=$ ₹12,000

Trick: Find only the difference: 3 parts of ₹4,000 — no need to compute each share.

Answer: (B) ₹12,000

Concept: Value of 1 part = total ÷ sum of ratio terms.

1. Sum = $3+4+5=12$ → 1 part = $\frac{15,000}{12}=$ ₹1,250
2. Highest − lowest = $5-3=2$ parts
3. $2\times 1,250=$ ₹2,500

Trick: Difference = $\frac{2}{12}\times 15,000=\frac{15,000}{6}=2,500$.

Answer: (C) ₹2,500

Concept: Area of triangle = $\frac{1}{2}\times$ base $\times$ height.

1. Base = 3k, height = 4k
2. $\frac{1}{2}\times 3k\times 4k=6k^2=216\to k^2=36\to k=6$
3. Base = 18 cm, height = 24 cm

Trick: Check options: ratio 3 : 4 and $\frac{1}{2}\times b\times h=216$ → only $\frac{18\times 24}{2}=216$.

Answer: (B) 18 cm and 24 cm

Concept: Past = Present $÷{(1+\frac{r}{100})}^n$.

1. $72,600÷{\left(\frac{11}{10}\right)}^2$
2. $=72,600\times \frac{100}{121}$
3. $=600\times 100=60,000$

Trick: $121\times 6=726$, so $72,600÷121=600$ at sight.

Answer: (A) 60,000

Concept: Net gain = 8% of first − 6% of second.

1. Let profit item = x, other = $6000-x$
2. $0.08x-0.06(6000-x)=200$
3. $0.14x-360=200\to x=\frac{560}{0.14}=$ ₹4,000

Trick: Alligation: net $\frac{200}{6000}=\frac{10}{3}\%$. Distances: $8-\frac{10}{3}=\frac{14}{3}$ and $\frac{10}{3}+6=\frac{28}{3}$ → ratio 2 : 1 → 4,000 : 2,000.

Answer: (A) ₹4,000

Concept: Equate the efficiencies, then convert everyone into one unit.

1. 6 M = 15 W → 1 M = 2.5 W
2. 4 M + 10 W = 10 W + 10 W = 20 W
3. 15 W take 80 days → 20 W take $\frac{15\times 80}{20}=60$ days

Trick: Convert everything to women: $M_1{D}_1=M_2{D}_2\to 15\times 80=20\times D$.

Answer: (B) 60 days

Concept: Find output per minute, then multiply.

1. $110÷5=22$ bottles per minute
2. $22\times 12=264$ bottles

Trick: 12 min = $2\times 5$ min + 2 min → $220+44=264$.

Answer: (C) 264 bottles

Concept: New average = (old total − removed) ÷ remaining count.

1. Total = $12\times 18=216$ kg
2. Removed = $24+12=36$ kg → left 180 kg
3. $180÷10=18$ kg

Trick: 24 and 12 average to 18 (same as the overall average), so the rest also average 18.

Answer: (B) 18 kg

Concept: Removing part of a mixture removes acid in the same ratio; total volume stays 200 L.

1. Acid = $\frac{3}{8}\times 200=75$ L
2. After removing 40 L ($\frac{1}{5}$th), acid left = $75\times \frac{4}{5}=60$ L
3. Acid % = $\frac{60}{200}\times 100=30\%$

Trick: Original acid % = 37.5%; $\frac{1}{5}$th replaced → $37.5\times \frac{4}{5}=30\%$.

Answer: (D) 30%

Concept: Volume of a prism = base area × height; $1000{\text{cm}}^3=1\text{L}$.

1. Triangular part: $80{\text{cm}}^2\times 150\text{cm}=12,000{\text{cm}}^3$ = 12 L
2. Rectangular part: $50\times 35\times 90=1,57,500{\text{cm}}^3$ = 157.5 L
3. Total = $12+157.5=169.5$ L

Trick: Work fully in cm³ and divide by 1000 — avoids m²/m³ decimal errors.

Answer: (C) 169.5 L

Concept: Use ${\sec }^{2}A=1+{\tan }^{2}A$.

1. ${\tan }^{2}A={(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$
2. ${\sec }^{2}A=1+x^2+\frac{1}{x^2}+2$
3. $=x^2+\frac{1}{x^2}+3$

Trick: Put $x=1$: $\tan A=2$, ${\sec }^{2}A=5$. Only $x^2+\frac{1}{x^2}+3$ gives 5.

Answer: (A) x² + $\frac{1}{x\mathrm{²}}$ + 3

Concept: Slope $m=\frac{y_2-y_1}{x_2-x_1}$.

1. Rise = $-9-7=-16$
2. Run = $5-(-3)=8$
3. $m=\frac{-16}{8}=-2$

Trick: y falls as x rises → slope negative; only (B), (D) remain, and $\frac{16}{8}=2$.

Answer: (B) −2

Concept: Shortest distance = perpendicular distance = $\frac{|c|}{\sqrt{a^2+b^2}}$.

1. Line through (6, 0) and (0, 8): $\frac{x}{6}+\frac{y}{8}=1\to 4x+3y=24$
2. Distance = $\frac{24}{\sqrt{16+9}}=\frac{24}{5}=4.8$

Trick: Right triangle with legs 6, 8, hypotenuse 10: altitude to hypotenuse = $\frac{6\times 8}{10}=4.8$.

Answer: (A) 4.8

Concept: If two angles and the side between them are equal, triangles are congruent (ASA).

1. Given ∠A = ∠D and ∠B = ∠E.
2. AB is the side joining vertices A and B → it is the included side.
3. Hence ASA.

Trick: Side between the two angles → ASA; side not between them → AAS.

Answer: (C) ASA

Concept: $\sqrt{a+b-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}(a>b)$.

1. $13-2\sqrt{42}=7+6-2\sqrt{7\times 6}\to \sqrt{}=\sqrt{7}-\sqrt{6}$
2. $\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}$ (since $7-6=1$)
3. $(\sqrt{7}-\sqrt{6})-(\sqrt{7}+\sqrt{6})=-2\sqrt{6}$

Trick: Find two numbers with sum 13 and product 42 → 7 and 6.

Answer: (C) −2$\sqrt{6}$

Concept: The line from the centre to the chord's midpoint is perpendicular to the chord.

1. Let half chord = h. Then $h^2+8^2={17}^2$
2. $h^2=225\to h=15$ cm
3. Full chord = $2h=30$ cm

Trick: 8-15-17 is a Pythagorean triplet — half chord = 15 instantly.

Answer: (D) 30 cm

Concept: Angle at the centre = 2 × angle at any point on the circle (same segment).

1. Central angle = 80°
2. Angle on the circle = 80° ÷ 2 = 40°

Yaad rakhein: Angles in the same segment are equal; angle in a semicircle = 90°.

Answer: (B) 40°

Concept: ${\sec }^{2}A-{\tan }^{2}A=1\Rightarrow (\sec A+\tan A)(\sec A-\tan A)=1$.

1. $\sec A+\tan A=\frac{3}{2}$
2. $\sec A-\tan A=1÷\frac{3}{2}=\frac{2}{3}$

Trick: sec − tan is always the reciprocal of sec + tan.

Answer: (A) $\frac{2}{3}$

Concept: Wage per hour = total earning ÷ hours.

1. $4.2\text{h}=\frac{42}{10}=\frac{21}{5}\text{h}$
2. $84÷\frac{21}{5}=84\times \frac{5}{21}=4\times 5=20$
3. As a fraction: $\frac{20}{1}$ (₹20 per hour)

Trick: Shift decimals: $840÷42=20$.

Answer: (A) $\frac{20}{1}$