SSC CGL 2025 PYQs

Concept: Radian = degree $\times \frac{\pi }{180}$.

1. $160\times \frac{\pi }{180}$.
2. $=\frac{16\pi }{18}=\frac{8\pi }{9}$.

Trick: ${20}^{\circ }=\frac{\pi }{9}$, and ${160}^{\circ }=8\times {20}^{\circ }\Rightarrow \frac{8\pi }{9}$.

Answer: (A) $\frac{8\pi }{9}$

Concept: tan θ = cot(90° − θ).

1. tan x = cot y = tan(90° − y).
2. So x = 90° − y.
3. x + y = 90°.

Trick: Whenever tan of one = cot of the other (acute angles), they add to 90°.

Answer: (C) 90°

Concept: Area = $r\times s$, where s = semi-perimeter.

1. s = $\frac{40}{2}=20$ cm.
2. Area = $2\times 20=40$ cm².

Trick: Area = $r\times \frac{\text{perimeter}}{2}$ – works for every triangle.

Answer: (D) 40 cm²

Concept: Perpendicular from centre bisects the chord: $r^2=(\text{half chord}{)}^2+d^2$.

1. Half chord = 9 cm.
2. $r^2=9^2+{13}^2=81+169=250$.
3. $r=\sqrt{250}\approx 15.8\approx 16$ cm.

Trick: r must exceed 13 (distance) and 9 (half chord) but be less than $13+9$ – only 16 and 18 fit; ${16}^2=256\approx 250$.

Answer: (C) 16 cm

Concept: Tangent at any point is perpendicular to the radius at that point.

1. Radius to the point of contact ⟂ tangent.
2. Angle = 90°, whatever the radius (8 cm is extra data).

Yaad rakhein: Two tangents from an external point are equal in length.

Answer: (C) 90°

Concept: $V=\pi r^{2}h$, so with fixed height $V\propto r^2$.

1. New radius = 2r → new $V=\pi (2r{)}^{2}h=4\pi r^{2}h$.
2. Ratio new : old = 4 : 1.

Trick: Volume follows the square of the radius: ×2 radius → ×4 volume; ×3 radius → ×9 volume.

Answer: (C) Volume becomes four times

Concept: Take the LCM, then multiply by primes having odd powers.

1. LCM of 6, 9, 15 = $90=2^1\times 3^2\times 5^1$
2. 2 and 5 have odd powers → multiply by $2\times 5=10$
3. Least square = $90\times 10=900={30}^2$

Trick: Check options from smallest: 400 is not divisible by 9; $900÷90=10$ ✓ and 900 is a square.

Answer: (B) 900

Concept: For one year compounded yearly, CI = SI, so $R=\frac{I}{P}\times 100$.

1. Interest = ₹700 − ₹500 = ₹200
2. Rate = $\frac{200}{500}\times 100=40\%$

Trick: Amount/Principal = $\frac{700}{500}=1.4$ → growth of 0.4 = 40% in one year.

Answer: (A) 40%

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. Rate for 2.5 years = $11.5\times 2.5=28.75\%$
2. SI = 28.75% of ₹2,500 = $2500\times 0.2875=$ ₹718.75

Trick: Combine $R\times T$ first (total %), then take that % of P – 1% of 2500 is 25, so $28.75\times 25=718.75$.

Answer: (C) ₹718.75

Concept: Speed in m/s $\times \frac{18}{5}$ = speed in km/h.

1. Time = 8 min = 480 s
2. Speed = $600÷480=1.25$ m/s
3. $1.25\times \frac{18}{5}=4.5$ km/h

Trick: Metres per minute $\times 0.06$ = km/h: $\frac{600}{8}=75$ m/min → $75\times 0.06=4.5$.

Answer: (B) 4.5 km/h

Concept: Material volume = outer volume − inner volume; each inner side = outer side − 2 × thickness.

1. Outer: $8\times 6\times 5=240$ cm³
2. Inner sides: $8-1,6-1,5-1=7,5,4$ → $7\times 5\times 4=140$ cm³
3. Material = $240-140=100$ cm³

Trick: Thickness is on both walls, so subtract 2t (here 1 cm) from every dimension – not just t.

Answer: (A) 100 cm³

Concept: Sector area = $\frac{\theta }{{360}^{\circ }}\times \pi r^2$.

1. Full circle area = $\frac{22}{7}\times 7\times 7=154$ cm²
2. Fraction = $\frac{38.5}{154}=\frac{1}{4}$
3. $\theta ={360}^{\circ }\times \frac{1}{4}={90}^{\circ }$

Trick: $38.5\times 4=154$ – the sector is a quarter-circle, so the angle is a right angle.

Answer: (A) 90°

Concept: Difference of percentages of the same number = (difference %) of that number.

1. $30\%-15\%$ = 15% of the number = 90
2. $1\%=90÷15=6$
3. $100\%=600$

Trick: Find 1% first, then scale to 100% – avoids fractions.

Answer: (A) 600

Concept: New member's age = old average + (new count × increase).

1. Let old average = a. Teacher = $a+6\times 2=a+12$
2. $a+12=40\Rightarrow a=28$ years

Trick: Teacher gives 2 extra years to each of 6 people (12 years) beyond the old average: $40-12=28$.

Answer: (C) 28

Concept: Two discounts a% and b% equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. Net discount = $20+10-\frac{20\times 10}{100}=28\%$
2. Price paid = 72% of ₹800 = ₹576

Trick: Multiply the 'pay' parts: $0.8\times 0.9=0.72$ → $800\times 0.72=576$.

Answer: (D) ₹576

Concept: After n replacements, remaining acid = initial acid $\times {(1-\frac{x}{V})}^n$.

1. Initial acid = $\frac{3}{5}$ of 100 L = 60 L
2. Each operation keeps 90% of the acid: $60\times {\left(\frac{9}{10}\right)}^2=60\times 0.81=48.6$ L
3. Volume stays 100 L → acid = 48.6%

Trick: Acid % directly: $60\%\times 0.81=48.6\%$ – no need to track litres.

Answer: (D) 48.6%

Concept: Total work = efficiency × days; use efficiencies as units per day.

1. Ashish: 2 units/day for 40 days → total = 80 units
2. All three: $2+3+5=10$ units/day → 5 days = 50 units; left = 30
3. Ashish + Navin: 5 units/day → $30÷5=6$ days

Trick: Take the efficiency ratio itself as daily units – the whole problem becomes simple division.

Answer: (D) 6 days

Concept: Lateral surface area of a prism = base perimeter × height.

1. Base perimeter = $3+4+5=12$ cm
2. LSA = $12\times 8=96$ cm²

Trick: Lateral faces are three rectangles: $3\times 8+4\times 8+5\times 8=24+32+40=96$. Base area (6 cm² each) is not included.

Answer: (C) 96 cm²

Concept: Frustum volume = big cone − small (similar) cone cut from the top.

1. Similar cones: r/h same → small cone radius = $6\times \frac{6}{9}=4$ cm, height 6 cm
2. Big cone = $\frac{1}{3}\pi \times 36\times 9=108\pi$; small cone = $\frac{1}{3}\pi \times 16\times 6=32\pi$
3. Frustum = $108\pi -32\pi =76\pi$ cm³

Trick: Volumes of similar cones scale as $(\text{height ratio}{)}^3$: ${\left(\frac{2}{3}\right)}^3=\frac{8}{27}$ → frustum = $108\pi \times \frac{19}{27}=76\pi$.

Answer: (B) 76π cm³

Concept: $\tan A$ $=\frac{\text{perpendicular}}{\text{base}}$; complete the triangle with Pythagoras.

1. $\tan A=\frac{3}{4}$ → sides 3, 4, hypotenuse 5
2. $\sin A=\frac{3}{5}$, $\cos A=\frac{4}{5}$
3. $3\times \frac{3}{5}-4\times \frac{4}{5}=\frac{9-16}{5}=-\frac{7}{5}$

Trick: Divide by $\cos A$: $3\sin A-4\cos A=\cos A(3\tan A-4)=\frac{4}{5}\times (\frac{9}{4}-4)=-\frac{7}{5}$.

Answer: (A) −$\frac{7}{5}$