SSC CGL 2025 PYQs

Concept: Sphere $\text{TSA}=4\pi r^2$; cylinder $\text{TSA}=2\pi r(r+h)$.

1. $4\pi r^2=2\pi r(r+h)$.
2. $2r=r+h\to h=r$.
3. $h:r=1:1$.

Trick: Cancel $2\pi r$ first: $2r=r+h$, done in one line.

Answer: (A) 1 : 1

Concept: To get n pieces along an edge you need $n-1$ cuts in that direction.

1. $27=3\times 3\times 3$ → 3 pieces along each edge.
2. Cuts = $2+2+2=6$.

Trick: For $n^3$ equal cubes, minimum cuts = $3(n-1)$.

Answer: (B) 6

Concept: $V\propto r^{2}h$, so multiply the change factors.

1. Factor = $(1.25{)}^2\times 0.9=1.5625\times 0.9=1.40625$.
2. Change = $+40.625\%\approx 40.6\%$ increase.

Trick: Fractions: $(\frac{5}{4}{)}^2\times \frac{9}{10}=\frac{225}{160}=\frac{45}{32}\to \frac{13}{32}\approx 40.6\%$.

Answer: (B) 40.6% increase

Concept: In quadrant II, sin is positive and cos is negative.

1. $\sin A=\frac{5}{13}$ → triplet 5-12-13 → $\cos A=-\frac{12}{13}$.
2. $\sin A+\cos A=\frac{5}{13}-\frac{12}{13}=-\frac{7}{13}$.
3. Square: $\frac{49}{169}$.

Trick: Add first, then square: $(-\frac{7}{13}{)}^2$ – no need to expand $1+2\sin A\cos A$.

Answer: (C) $\frac{49}{169}$

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

1. $\sin A=\frac{p}{q}$ → ${\cos }^{2}A=1-\frac{p^2}{q^2}=\frac{q^2-p^2}{q^2}$.
2. ${\sec }^{2}A=\frac{q^2}{q^2-p^2}$.

Trick: Use ${\cos }^2=1-{\sin }^2$ directly; no need to find the third side.

Answer: (A) $\frac{q\mathrm{²}}{q\mathrm{²}-p\mathrm{²}}$

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

1. $m=\frac{8-4}{5-3}=\frac{4}{2}=2$.

Trick: Rise over run: y goes up 4 while x goes up 2.

Answer: (A) 2

Concept: Same radius → sector area is proportional to central angle.

1. $\frac{\pi }{2}\text{rad}={90}^{\circ }$.
2. Ratio = ${180}^{\circ }:{90}^{\circ }=2:1$.

Trick: Radius (24 cm) is a distractor when radii are equal.

Answer: (B) 2 : 1

Concept: Sum = $(n-2)\times {180}^{\circ }$.

1. $n=8\to 6\times {180}^{\circ }={1080}^{\circ }$.

Trick: Answer must be a multiple of 180° – 1080 is the only one fitting n = 8 (900 is for 7 sides).

Answer: (B) 1080°

Concept: It passes through the midpoint and has slope = $-\frac{1}{\text{segment slope}}$.

1. Midpoint = $(\frac{4+2}{2},\frac{8+6}{2})=(3,7)$.
2. Segment slope = $\frac{6-8}{2-4}=1$ → perpendicular slope = −1.
3. $y-7=-(x-3)\to y=-x+10$.

Trick: Slope −1 leaves (A) and (C); plug (3, 7): only $y=-x+10$ works.

Answer: (C) y = −x + 10

Concept: Area ratio of similar triangles = $(\text{ratio of perimeters/sides}{)}^2$.

1. Area ratio = 25 : 49 → total 74 parts.
2. 74 parts = 74 cm² → 1 part = 1 cm².
3. Larger = 49 cm².

Trick: Total $74=25+49$ is a giveaway – the squares themselves are the areas.

Answer: (C) 49 cm²

Concept: $18+20-29\ne 0$, so no identity shortcut – compute directly.

1. Positive part: ${18}^3+{20}^3=5832+8000=13832$; add 116 → 13948.
2. Negative part: ${29}^3=24389$.
3. Result = $-(24389-13948)=-10441$.

Trick: Unit digits: $2+0-9+6=-1$ → a negative answer ending in 1; −10441 fits, −16789 does not.

Answer: (A) −10441

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

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

Trick: $(x+\frac{1}{x}{)}^2$ always gives +2; then add 1 → +3.

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

Concept: Radius ⟂ tangent, so $P{A}^2=O{P}^2-r^2$.

1. $P{A}^2={25}^2-7^2=625-49=576$.
2. PA = 24 cm.

Trick: 7-24-25 is a Pythagorean triplet.

Answer: (C) 24 cm

Concept: $OA⟂PA$, $OB⟂PB$ and $\angle APB={90}^{\circ }$ → OAPB is a square.

1. $\angle AOB={360}^{\circ }-{90}^{\circ }-{90}^{\circ }-{90}^{\circ }={90}^{\circ }$; $OA=OB=6$ and $PA=PB$.
2. So OAPB is a square of side 6.
3. $OP=\text{diagonal}=6\sqrt{2}$ cm.

Trick: $\angle APB={90}^{\circ }\Rightarrow OP=r\sqrt{2}$ (in general $OP=\frac{r}{\sin (\frac{1}{2}\angle APB)}$).

Answer: (B) 6$\sqrt{2}$ cm

Concept: Multiply as whole numbers, then place decimal = total decimal digits.

1. $48\times 7=336$.
2. Decimal places: $2+1=3\to 0.336$.

Trick: Estimate: about $0.5\times 0.7=0.35$ – slightly less → 0.336.

Answer: (B) 0.336

Concept: Solve brackets first, then × / ÷, then −.

1. First bracket: $5.125+2.8=7.925$
2. $0.45\times 7.925=3.56625$
3. Second bracket: $0.9÷0.3=3$
4. Result: $0.56625\approx 0.566$

Trick: $7.925\times 0.45$ is a bit above 3.5, minus 3 gives about 0.5–0.6 — only 0.566 fits.

Answer: (A) 0.566

Concept: Convert both percentages and collect P and Q terms.

1. 0.4(P + Q) = 0.6(P − Q) → 2(P + Q) = 3(P − Q)
2. 2P + 2Q = 3P − 3Q
3. P = 5Q → P : Q = 5 : 1

Trick: Divide the % by 20 first: 40 : 60 = 2 : 3, then solve 2(P+Q) = 3(P−Q).

Answer: (A) 5 : 1

Concept: Cost of mixture = weighted average of the two costs.

1. CP of 4 kg = $3\times 50+1\times 90=240$ → ₹60/kg
2. SP = ₹96/kg → profit = ₹36/kg
3. Profit % = $\frac{36}{60}\times 100=60\%$ gain

Trick: SP 96 is $1.6\times 60$, so gain = 60% directly.

Answer: (A) 60% Gain

Concept: Profit is shared in the ratio capital × time.

1. A → 8 months × ₹60,000; B → 9 months × ₹80,000
2. In thousands: 480 : 720 = 2 : 3
3. A gets 2 parts out of 5 → $40,000\times \frac{2}{5}=$ ₹16,000

Trick: Drop zeros: $(6\times 8):(8\times 9)=48:72=2:3$.

Answer: (A) ₹16,000

Concept: Removed value = old total − new total.

1. Old total = $40\times 70=2800$ kg
2. New total = $39\times 69.5=2710.5$ kg
3. Removed = $2800-2710.5=89.5$ kg

Trick: Removed = new avg + (old count × fall) = $69.5+40\times 0.5=89.5$.

Answer: (B) 89.5 kg