SSC CGL 2025 PYQs

Concept: Girls are unchanged; use that to find the multiplier.

1. Boys = 5x, girls = 3x.
2. $\frac{5x+10}{3x}=2\Rightarrow 5x+10=6x\Rightarrow x=10$.
3. Initial total = $8x=80$.

Trick: Total must be a multiple of 8 (5 + 3) – only 80 among options.

Answer: (B) 80

Concept: Use LCM as total work and add work done in each 3-day cycle.

1. Work = LCM(24, 36, 72) = 72 units; P = 3, Q = 2, R = 1 per day.
2. Cycle of 3 days: $3+3+(3+2+1)=12$ units.
3. $72÷12=6$ cycles = 18 days.

Trick: When the work ends exactly at a cycle end, days = cycles × cycle length.

Answer: (B) 18 days

Concept: Pure milk is added, so water only decreases by the part removed.

1. Water initially = $\frac{2}{5}\times 200=80$ L.
2. Final water = $\frac{3}{10}\times 200=60$ L.
3. Removed S litres carry $0.4S$ water ⇒ $80-0.4S=60\Rightarrow S=50$ L.

Trick: Track the component that is NOT added back (water) – one equation solves it.

Answer: (C) 50 litres

Concept: Time is inversely proportional to efficiency.

1. Ram : Shyam efficiency = 5 : 2 ⇒ time = 2 : 5.
2. 2 parts = 20 days ⇒ 1 part = 10 ⇒ Shyam = 50 days.

Trick: Shyam's time = $20\times \frac{5}{2}=50$.

Answer: (B) 50 days

Concept: The faster vehicle must close the initial gap at the relative speed.

1. Gap = 800 m (SUV ahead); relative speed = $20-15=5$ m/s.
2. Time = $800÷5=160$ s.
3. Car's distance = $20\times 160=3,200$ m.

Trick: Car distance = gap × (car speed ÷ relative speed) = $800\times \frac{20}{5}=3,200$.

Answer: (D) 3200 m

Concept: Path area = $\pi (R^2-r^2)=\pi (R+r)(R-r)$.

1. $R=15+1.5=16.5$ m, $r=15$ m.
2. Area = $\pi \times 31.5\times 1.5=47.25\pi$.
3. ≈ $47.25\times 3.14\approx 148.5$ m².

Trick: Use $(R+r)(R-r)$ instead of squaring decimals.

Answer: (A) 148.5 m²

Concept: $\pi \text{rad}={180}^{\circ }$.

1. $\frac{7\pi }{6}=7\times \frac{{180}^{\circ }}{6}$.
2. = $7\times {30}^{\circ }={210}^{\circ }$.

Trick: $\frac{\pi }{6}={30}^{\circ }$, so just multiply ${30}^{\circ }$ by the numerator.

Answer: (B) 210°

Concept: A point lies on the line if it satisfies the equation.

1. y = −x + 6 ⇔ x + y = 6.
2. (4,2), (5,1), (6,0) all give 6.
3. (3,4): 3 + 4 = 7 ≠ 6 → not on the line.

Trick: Rewrite as x + y = 6 and add coordinates mentally.

Answer: (B) (3, 4)

Concept: Arc length $l=r\theta$, $\theta$ in radians.

1. $\theta =\frac{l}{r}=\frac{20}{10}$.
2. $\theta =2$ radians.

Trick: Arc = 2 × radius → angle = 2 rad (≈ ${114.6}^{\circ }$).

Answer: (C) 2 radians

Concept: $\tan ({90}^{\circ }-\theta )=\cot \theta$.

1. $\cot \theta =\sqrt{3}\Rightarrow \theta ={30}^{\circ }$.
2. $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$.

Trick: $\cot \theta =\sqrt{3}$ is the standard value of 30° – recall the table.

Answer: (B) $\frac{\surd 3}{2}$

Concept: Add and rationalise the denominator.

1. $x=\sqrt{3}\Rightarrow x+\frac{1}{x}=\sqrt{3}+\frac{1}{\sqrt{3}}$.
2. $=\frac{3+1}{\sqrt{3}}=\frac{4}{\sqrt{3}}$.
3. Rationalise: $\frac{4\sqrt{3}}{3}$.

Trick: $x+\frac{1}{x}=\frac{x^2+1}{x}=\frac{4}{\sqrt{3}}$ directly.

Answer: (C) $\frac{4\surd 3}{3}$

Concept: $\text{Area ratio}=(\text{side ratio}{)}^2$.

1. Sides 2 : 5.
2. Areas = $2^2:5^2=4:25$.

Trick: Squares of the given ratio – look for 4 and 25.

Answer: (C) 4 : 25

Concept: Expand $x^2$ and substitute.

1. $x^2=(2+\sqrt{3}{)}^2=4+4\sqrt{3}+3=7+4\sqrt{3}$.
2. $x^2-x=7+4\sqrt{3}-2-\sqrt{3}=5+3\sqrt{3}$.
3. $-7\Rightarrow -2+3\sqrt{3}$.

Trick: $\sqrt{3}$ coefficient: $4-1=3$; constant: $7-2-7=-2$.

Answer: (A) −2 + 3$\sqrt{3}$

Concept: Opposite angles of a cyclic quadrilateral add up to 180°.

1. Total ratio = 12 parts = 360° ⇒ 1 part = 30°.
2. Pairs (1 + 5) and (2 + 4) = 6 parts = 180° each ✓.
3. Smallest = 30°.

Trick: Any quadrilateral: 360° ÷ sum of ratio = one part.

Answer: (A) 30°

Concept: $\sec =\frac{\text{hypotenuse}}{\text{base}}$; use Pythagoras for the third side.

1. Hypotenuse = 13, base = 5.
2. Perpendicular = $\sqrt{169-25}=12$.
3. $\tan A=\frac{12}{5}$.

Trick: 5–12–13 is a Pythagorean triplet – no calculation needed.

Answer: (B) $\frac{12}{5}$

Concept: Compare fractions by converting to decimals or by cross-multiplying.

1. $\frac{6}{11}\approx 0.545$, $\frac{4}{7}\approx 0.571$, $\frac{3}{5}=0.600$, $\frac{5}{8}=0.625$.
2. Smallest to largest: 0.545, 0.571, 0.600, 0.625.
3. Order: $\frac{6}{11}$, $\frac{4}{7}$, $\frac{3}{5}$, $\frac{5}{8}$.

Trick: Compare each with 'how much short of 1': $\frac{6}{11}$ misses $\frac{5}{11}$ (biggest gap) → smallest; $\frac{5}{8}$ misses $\frac{3}{8}$ (smallest gap) → largest.

Answer: (B) $\frac{6}{11}$, $\frac{4}{7}$, $\frac{3}{5}$, $\frac{5}{8}$

Concept: Change the mixed fraction to a decimal, then add/subtract.

1. Whole-number parts: $3+4-1=6$.
2. Decimal parts: $0.5+0.5-0.9=0.1$ ($3\frac{1}{2}$ gives 0.5).
3. Total = $6+0.1=6.1$.

Trick: Pair numbers that make whole numbers first ($3.5+4.5=8$) – then only one subtraction is left.

Answer: (C) 6.1

Concept: Solve brackets from the innermost: ( ) → { } → [ ].

1. $(\frac{3}{4}-\frac{1}{4})=\frac{1}{2}$; then $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.
2. $\{\frac{1}{3}-\frac{1}{4}\}=\frac{4-3}{12}=\frac{1}{12}$.
3. $[\frac{13}{6}÷\frac{1}{12}]=\frac{13}{6}\times 12=26$.
4. $27\frac{1}{4}-26=1\frac{1}{4}$.

Trick: Dividing by $\frac{1}{12}$ = multiplying by 12 — no need for common denominators there.

Answer: (C) 1$\frac{1}{4}$

Concept: Total in a set = 'only that set' + 'both'.

1. Python users = 70 (only) + 60 (both) = 130.
2. Java users = 80 (only) + 60 (both) = 140.
3. Ratio = 130 : 140 = 13 : 14.

Trick: Check total: 60 + 70 + 80 + 20 = 230 ✔ — the 'neither' group plays no role in the ratio.

Answer: (B) 13 : 14

Concept: Profit is shared in the ratio of capital × months.

1. Sonu: $40,000\times 12$; Prem: $80,000\times 8$; Alok: $1,20,000\times 4$.
2. In thousands: 480 : 640 : 480 = 3 : 4 : 3 (total 10).
3. Alok = $\frac{3}{10}\times 1,56,000=$ ₹46,800.

Trick: Divide capitals by 40,000 first (1 : 2 : 3), then multiply by months (12, 8, 4) → 12 : 16 : 12 = 3 : 4 : 3.

Answer: (D) ₹46,800