SSC CHSL 2025 PYQs

Concept: y − y₁ = m(x − x₁).

1. y − 5 = −6(x − 4).
2. y = −6x + 24 + 5 = −6x + 29.

Trick: Slope −6 → only option with −6x is D; check (4, 5): −24 + 29 = 5 ✓.

Answer: (D) $y=-6x+29$

Concept: Read the data as $x+y=\sqrt{19}$, $x-y=1$ (with $x+y=19$ no option fits); use squares.

1. $(x+y{)}^2=19$, $(x-y{)}^2=1$.
2. Add: $2(x^2+y^2)=20$ → $x^2+y^2=10$.
3. Subtract: $4xy=18$ → $xy=\frac{9}{2}$.
4. $\frac{x^2+y^2}{xy}=10÷\frac{9}{2}=\frac{20}{9}$.

Trick: $x^2+y^2=\frac{S^2+D^2}{2}$ and $xy=\frac{S^2-D^2}{4}$.

Answer: (C) $\frac{20}{9}$

Concept: Use $(1-\sin x)(1+\sin x)={\cos }^{2}x$ as the common denominator.

1. Numerator = $(1+\cos x)(1+\sin x)+(1-\cos x)(1-\sin x)$.
2. $=2+2\sin x\cos x$ (sinx, cosx terms cancel).
3. Result = $\frac{2(1+\cos x\sin x)}{{\cos }^{2}x}$.

Trick: Put $x=0$: value = $\frac{2}{1}+\frac{0}{1}=2$; only B gives $\frac{2(1+0)}{1}=2$.

Answer: (B) $\frac{2(1+\cos (x\left)\sin \right(x\left)\right)}{{\cos }^2\left(x\right)}$

Concept: $\cot \theta =\frac{\text{base}}{\text{perpendicular}}$.

1. Base = 15, perpendicular = 8 (k = 1).
2. Hypotenuse = $\sqrt{225+64}=\sqrt{289}=17$.

Yaad rakhein: 8-15-17 is a Pythagorean triplet.

Answer: (A) 17

Concept: $\sin A=\frac{\text{perpendicular}}{\text{hypotenuse}}$; get base by Pythagoras.

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

Trick: cos can never exceed 1 → options with $\frac{13}{12}$ are out instantly.

Answer: (A) $\frac{12}{13}$ and $\frac{5}{12}$

Concept: SP at loss r% = $CP\times \frac{100-r}{100}$; SP at gain g% = $CP\times \frac{100+g}{100}$.

1. 80% of CP = ₹2000 → $CP=2000\times \frac{100}{80}=$ ₹2500.
2. Required SP = 110% of 2500 = ₹2750.

Trick: Direct ratio: new SP = $2000\times \frac{110}{80}=25\times 110=$ ₹2750.

Answer: (D) ₹2750

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

1. P: $10000\times 12=1,20,000$.
2. Q: $12000\times 6+8000\times 6=1,20,000$.
3. R: $16000\times 6+24000\times 6=2,40,000$.
4. Ratio 1 : 1 : 2 → R = $84000\times \frac{2}{4}=$ ₹42,000.

Trick: For two 6-month halves, just add the two capitals: P 20k, Q 20k, R 40k → 1 : 1 : 2 at once.

Answer: (C) ₹42,000

Concept: Convert cosec and sec into sin and cos.

1. $\csc x·\cos x=\frac{\cos x}{\sin x}=\cot x$.
2. $\sec x·\sin x=\frac{\sin x}{\cos x}=\tan x$.
3. Sum = $\cot x+\tan x$.

Trick: Put $x={45}^{\circ }$: value = $1+1=2$; only $\cot x+\tan x$ gives 2 (option C gives $\sqrt{2}$).

Answer: (D) $\cot x+\tan x$

Concept: Equal ratio parts have equal value.

1. 5 parts = 50 shoes → 1 part = 10.
2. Socks = 3 parts = 30.

Trick: Scale the ratio: 5 → 50 is ×10, so 3 → 30.

Answer: (A) 30

Concept: Yearly compounding: full years at R%, the leftover part-year at proportional rate on the new amount.

1. Year 1: 20% → $6000\times 1.2=7200$.
2. Next 6 months: 10% of 7200 = 720 → 7920.
3. CI = $7920-6000=$ ₹1920.

Trick: Effective rate for 20% then 10%: $20+10+\frac{20\times 10}{100}=32\%$ → 32% of 6000 = 1920.

Answer: (A) ₹1920

Concept: Work year by year; the second year's interest is on the balance left.

1. Year-1 interest = 20% of 2,50,000 = 50,000.
2. Withdraws 40% of 50,000 = 20,000 → balance 2,80,000.
3. Year-2: $2,80,000\times 1.2=$ ₹3,36,000.

Trick: Keep 60% of the interest: $2,50,000+30,000=2,80,000$, then $\times 1.2$.

Answer: (C) ₹3,36,000

Concept: Numbers in ratio a : b (co-prime) = ax, bx; their LCM = abx.

1. Numbers 5x, 3x → LCM = $15x=180$ → $x=12$.
2. Numbers: 60 and 36.
3. Sum = 96.

Trick: Sum = $(5+3)\times x=8\times 12=96$ – no need to find each number.

Answer: (A) 96

Concept: Total = average × number; find the totals before and after.

1. Old total = $25\times 16=400$.
2. New total = $30\times 18=540$.
3. 5 new students: 140 → average = $\frac{140}{5}=28$.

Trick: $\text{New average}+\frac{\text{old count}\times \text{rise}}{\text{new count}}$: $18+\frac{25\times 2}{5}=18+10=28$.

Answer: (C) 28

Concept: Compute each percentage value and compare.

1. 16% of 350 = 56.
2. Others: B = 100, C = 84, D = 102.
3. Least = 56.

Trick: Option A has both the smallest % and a small base – check it first.

Answer: (A) $16\%$ of 350

Concept: Same distance both ways: total time = $\frac{d}{s_1}+\frac{d}{s_2}$.

1. $\frac{d}{7}+\frac{d}{5}=6$.
2. $\frac{5d+7d}{35}=6$ → $12d=210$.
3. $d=17.5$ km.

Trick: Average speed = $\frac{2\times 7\times 5}{12}=\frac{35}{6}$ km/h; total distance = $\frac{35}{6}\times 6=35$ → one way 17.5 km.

Answer: (C) 17.5 km

Concept: Take total work = LCM of days; work/day = total ÷ days.

1. Total work = LCM(16, 24, 32) = 96 units.
2. P+Q+R = 6/day, P+Q = 4/day → R = 2/day.
3. R alone = $\frac{96}{2}=48$ days.

Trick: $\frac{1}{16}-\frac{1}{24}=\frac{1}{48}$ directly gives R's rate.

Answer: (C) 48 days

Concept: Discount applies (3 items); find MP from SP and CP from the profit %.

1. $MP=2700\times \frac{100}{90}=$ ₹3000.
2. $CP=2700\times \frac{100}{125}=$ ₹2160.
3. Profit at full price = $3000-2160=$ ₹840.

Trick: Use fractions: $10\%=\frac{1}{10},25\%=\frac{1}{4}$ → $MP=2700\times \frac{10}{9},CP=2700\times \frac{4}{5}$.

Answer: (B) ₹840

Concept: $SI=\frac{P\times R\times T}{100}$; tripling means $SI=2P$.

1. $2P=\frac{P\times R\times 4}{100}$.
2. $R=\frac{200}{4}=50\%$.

Trick: Rate = $\frac{(n-1)\times 100}{T}=\frac{2\times 100}{4}=50\%$.

Answer: (D) 50%

Concept: TSA of cuboid = $2(lb+bh+hl)$.

1. $b=3,l=6\times 3=18,h=4\times 3=12$ (cm).
2. $lb=54,bh=36,hl=216$ → sum 306.
3. TSA = $2\times 306=612{\text{cm}}^2$.

Trick: In terms of b: TSA = $2(6+4+24)b^2=68b^2=68\times 9=612$.

Answer: (B) $612{\mathrm{cm}}^2$

Concept: $n^2+n+(n+1)=n^2+2n+1=(n+1{)}^2$.

1. Take $n=1236$: ${1236}^2+1236+1237=(1237{)}^2$.
2. $\sqrt{(1237{)}^2}=1237$.

Trick: Pattern $\sqrt{n^2+n+(n+1)}$ always equals $n+1$.

Answer: (C) 1237