SSC CHSL 2025 PYQs

Concept: $V=\pi r^{2}h,\text{CSA}=2\pi rh\Rightarrow \text{CSA}=\frac{2V}{r}$.

1. $h=1540÷(\frac{22}{7}\times 49)=1540÷154=10\text{cm}$
2. $\text{CSA}=2\times \frac{22}{7}\times 7\times 10=440{\text{cm}}^2$

Trick: $\text{CSA}=\frac{2V}{r}=\frac{2\times 1540}{7}=440$ — no need to find h.

Answer: (A) $440{\mathrm{cm}}^2$

Concept: $(\sqrt{x}+\sqrt{y}{)}^2=x+y+2\sqrt{xy}$.

1. $a=\sqrt{5}+\sqrt{2}$
2. $a^2=5+2+2\sqrt{10}=7+2\sqrt{10}$

Trick: Rational part = sum of numbers under roots (5 + 2); surd part = $2\sqrt{\text{product}}$.

Answer: (B) $7+2\sqrt{10}$

Concept: $1-{\sin }^{2}x={\cos }^{2}x$.

1. $\frac{1-{\sin }^{2}x}{\cos x}=\frac{{\cos }^{2}x}{\cos x}$
2. $=\cos x$

Trick: Put $x=0$: $\text{LHS}=\frac{1}{1}=1$; among options only $\cos 0=1$.

Answer: (A) $\cos x$

Concept: $\frac{B}{D}=\frac{B}{C}\times \frac{C}{D}$; A : B is not needed.

1. $\frac{B}{D}=\frac{6}{7}\times \frac{8}{9}=\frac{48}{63}$
2. $=\frac{16}{21}$

Trick: Cancel before multiplying: $\frac{6}{9}=\frac{2}{3}$ → $2\times 8:7\times 3=16:21$.

Answer: (D) 16 : 21

Concept: Build the right triangle from $\cos \theta$, then find ${\sec }^2$ and ${\csc }^2$.

1. $\cos \theta =\frac{3}{5}$ → 3-4-5 triangle → $\sin \theta =\frac{4}{5}$
2. ${\sec }^2\theta =\frac{25}{9}$; ${\csc }^2\theta =\frac{25}{16}$
3. $\frac{25}{9}-\frac{25}{16}=25\times \frac{7}{144}=\frac{175}{144}$

Trick: Take 25 common: $25(\frac{1}{9}-\frac{1}{16})=25\times \frac{7}{144}$.

Answer: (C) $\frac{175}{144}$

Concept: $(a+b{)}^2=a^2+b^2+2ab$, with $\sqrt{x}\times \sqrt{x}=x$.

1. $(\sqrt{5}+\sqrt{3}{)}^2=(\sqrt{5}{)}^2+(\sqrt{3}{)}^2+2·\sqrt{5}·\sqrt{3}$.
2. $=5+3+2\sqrt{15}$.
3. = $8+2\sqrt{15}$ ($\sqrt{15}$ cannot be simplified).

Trick: Rational part = $5+3=8$ instantly; middle term is always $2\sqrt{\text{product}}=2\sqrt{15}$.

Answer: (B) $8+2\sqrt{15}$

Concept: $\text{New value}=\text{Old}\times \frac{100+r}{100}$.

1. 20% of 800 = 160.
2. New price = $800+160=$ ₹960.
3. Or: $800\times \frac{6}{5}=960$.

Trick: 20% up = multiply by $\frac{6}{5}$ → $800÷5\times 6=960$.

Answer: (B) Rs. 960

Concept: Difference in profit % corresponds to the extra money received.

1. Profit rises from 20% to 50% → extra 30% of CP.
2. 30% of CP = ₹600.
3. $\text{CP}=\frac{600\times 100}{30}=$ ₹2000.

Trick: Extra ₹ ÷ extra % × 100 = CP → $600÷30\times 100$.

Answer: (D) Rs. 2000

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

1. $\text{SI}=\frac{1500\times 12\times 4}{100}$.
2. $=15\times 48=$ ₹720.

Trick: Cancel 100 with 1500 first: $15\times 12\times 4=720$.

Answer: (B) Rs. 720

Concept: $\text{Efficiency}\propto \frac{1}{\text{time}}$ for the full work.

1. P: $\frac{1}{4}$ work in 10 days → full work in 40 days.
2. Q: $\frac{1}{5}$ work in 12 days → full work in 60 days.
3. Efficiency P : Q = 60 : 40 = 3 : 2 → P more by $\frac{3-2}{2}\times 100=$ 50%.

Trick: Time ratio 40 : 60 = 2 : 3, flip it for efficiency 3 : 2 – extra 1 on base 2 = 50%.

Answer: (A) P is $50\%$ more efficient

Concept: Central angle of a regular n-gon = 360°/n.

1. Pentagon: n = 5.
2. Central angle = 360° ÷ 5 = 72°.

Yaad rakhein: Central angle = exterior angle of a regular polygon (both 360°/n).

Answer: (A) 72°

Concept: ${\text{Slant height}}^2={\text{height}}^2+(\text{half base side}{)}^2$.

1. Half side = $\frac{16}{2}=8$ m.
2. $h^2={17}^2-8^2=289-64=225$.
3. h = 15 m.

Trick: 8-15-17 is a Pythagorean triplet – no calculation needed.

Answer: (C) 15 m

Concept: A cut parallel to the base at half height gives a similar cone with half the radius and height.

1. Small cone: h = 9 cm, r = 7 cm.
2. $V=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 9$.
3. $=22\times 7\times 3=$ 462 cm³.

Trick: Half linear size → $\frac{1}{8}$ volume: full cone = $\frac{1}{3}\times \frac{22}{7}\times 196\times 18=3696$; $3696÷8=462$.

Answer: (B) $462{\mathrm{cm}}^3$

Concept: $\text{Surface area}\propto r^2$.

1. Area ratio = $9^2:8^2=81:64$.
2. Increase = 17 parts on 64.
3. % increase = $\frac{17}{64}\times 100=$ 26.5625% ≈ 26.56%.

Trick: Use ($2x+\frac{x^2}{100}$) with $x=12.5\%$: $25+1.5625=26.5625\%$.

Answer: (B) $26.56\%$

Concept: Outer radius = r + width; compare areas.

1. $\pi (r+2{)}^2=2\pi r^2$ → $(r+2)=\sqrt{2}·r$.
2. $r(\sqrt{2}-1)=2$ → $r=\frac{2}{\sqrt{2}-1}=2(\sqrt{2}+1)$.
3. $r\approx 2\times 2.414=$ 4.83 ≈ 4.82 m.

Trick: Area doubles ⇒ radius $\times \sqrt{2}$; rationalise $\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$.

Answer: (C) 4.82

Concept: $a^3+b^3+c^3=3abc+(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

1. $ab+bc+ca=\frac{{30}^2-350}{2}=\frac{550}{2}=275$.
2. $a^2+b^2+c^2-(ab+bc+ca)=350-275=75$.
3. Sum of cubes = $30\times 75+3\times 1000=2250+3000=$ 5250.

Trick: Get $ab+bc+ca$ from $(a+b+c{)}^2$ first – the cube formula is then pure substitution.

Answer: (B) 5,250

Concept: $\tan \theta =\frac{\sin \theta }{\cos \theta }$.

1. $7\sin \theta =9\cos \theta$.
2. Divide by $7\cos \theta$: $\tan \theta =$ $\frac{9}{7}$.

Trick: $a\sin \theta =b\cos \theta \Rightarrow \tan \theta =\frac{b}{a}$.

Answer: (B) $\frac{9}{7}$

Concept: $\text{TSA}=\text{base area}+4\times (\frac{1}{2}\times \text{side}\times \text{slant height})$.

1. Base = $12\times 12=144$ m².
2. Lateral = $\frac{1}{2}\times \text{perimeter}\times \text{slant}=\frac{1}{2}\times 48\times 8=192$ m².
3. Total = $144+192=$ 336 m².

Trick: $\text{Lateral area}=\frac{1}{2}\times \text{perimeter}\times \text{slant height}$ – one step for all four faces.

Answer: (B) $336m^2$

Concept: $0.ababab\dots =\frac{ab}{99}$.

1. $0.5454\dots =\frac{54}{99}=$ $\frac{6}{11}$.
2. $\frac{6}{11}-\frac{1}{4}=\frac{24-11}{44}=$ $\frac{13}{44}$.

Trick: Two repeating digits → two 9s in the denominator.

Answer: (C) $\frac{6}{11},\frac{13}{44}$

Concept: $\sec \theta =\frac{\text{hypotenuse}}{\text{base}}$; $\tan \theta =\frac{\text{perpendicular}}{\text{base}}$.

1. Hypotenuse 5, base 4 → perpendicular = $\sqrt{25-16}=3$.
2. $\tan \theta =$ $\frac{3}{4}$ (positive in Q1).

Trick: Or use ${\tan }^2\theta ={\sec }^2\theta -1=\frac{25}{16}-1=\frac{9}{16}$ → $\frac{3}{4}$.

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