SSC GD 2025 PYQs

Concept: Do ÷ and × first (left to right), then + and −.

1. Division: $12÷3$ → 4
2. Multiplication: $3\times 4$ → 12
3. Now $16+4=20$; $20-12=$ 8

Trick: Never go left-to-right blindly: $(16+12)÷3$ … gives a wrong option – that is the trap.

Answer: (D) 8

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

1. $R\times T=6\times 3=18\%$ total interest in 3 years
2. 18% of P = ₹576 → 1% = $576÷18$ = ₹32
3. $P=32\times 100$ = ₹3200

Trick: SI for T years = R×T % of P. Find 1% first, then ×100.

Answer: (C) 3200

Concept: $\text{Final price}=\text{CP}\times (1+p\%)\times (1-l\%)$.

1. Multipliers: profit 5% → $\frac{21}{20}$; loss 24% → $\frac{19}{25}$
2. $\text{CP}\times \frac{21}{20}\times \frac{19}{25}=3990$
3. $\text{CP}=\frac{3990\times 20\times 25}{21\times 19}=\frac{3990\times 500}{399}=10\times 500$ = ₹5000

Trick: Note $21\times 19=399$ and $3990=399\times 10$ – spotting this makes it a 10-second sum.

Answer: (C) 5000

Concept: For r% yearly growth, $\text{previous year's population}=\text{present}\times \frac{100}{100+r}$.

1. 25% growth → multiply by $\frac{4}{5}$ for each year back.
2. 2 years ago: $110000\times \frac{16}{25}=70400$
3. 3 years ago: $70400\times \frac{4}{5}=56320$
4. Difference = $70400-56320=14080$

Trick: Difference = 2-years-ago value × $\frac{1}{5}=70400÷5=14080$ (no need to find the 3rd value).

Answer: (C) 14080

Concept: If $x\propto \frac{1}{y}$, the product of x and y stays constant.

1. $x\propto \frac{1}{y}\Rightarrow x=\frac{k}{y}$
2. Multiply both sides by y $\Rightarrow xy=k$ (constant)
3. $\frac{x}{y}=k$ is direct proportion, so (A) is wrong.

Trick: Direct → ratio constant (÷); Inverse → product constant (×).

Answer: (D) $\mathrm{xy}=k$

Concept: Same distance: $\text{time}\propto \frac{1}{\text{speed}}$.

1. Speed ratio new : normal = 3 : 4 ⇒ time ratio = 4 : 3
2. Difference of 1 part = 24 min
3. Normal time = 3 parts = $3\times 24=72$ min

Trick: At $\frac{3}{4}$ speed the delay equals $\frac{1}{3}$ of usual time, so usual time = $24\times 3=72$.

Answer: (C) 72

Concept: $\text{Expenditure}=\text{Price}\times \text{Consumption}$.

1. New price factor = 1.5; new expenditure factor = 1.11
2. New consumption factor = $1.11÷1.5=0.74$
3. Reduction = $1-0.74=0.26$ → 26%

Trick: Careful: 74% is what remains; the cut is 26% – (D) is the trap.

Answer: (B) 26%

Concept: Convert averages to totals.

1. S + R + M = 138; S + R = 78; R + M = 96
2. Add last two: S + 2R + M = 174
3. Subtract first: R = 174 − 138 = 36 kg

Trick: Common member = (sum of pair totals) − (group total).

Answer: (A) 36

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

1. $a+b=18.5+11.5=30$
2. $\frac{ab}{100}=\frac{18.5\times 11.5}{100}=\frac{212.75}{100}=2.1275$
3. Single discount = $30-2.1275=27.8725\%$

Trick: Answer must be a little less than $a+b$ (30%) – only 27.8725% fits.

Answer: (C) 27.8725%

Concept: $\text{CP}=\text{SP}\times \frac{100}{100+\text{gain}\%}$.

1. $\text{CP}=217\times \frac{100}{108.5}$
2. $\frac{217}{108.5}=2$
3. $\text{CP}=2\times 100=$ ₹200

Trick: Check options: 8.5% of 200 = 17, and $200+17=217$ ✔.

Answer: (D) 200

Concept: Solve inner brackets first (BODMAS).

1. $\frac{16-12}{4}=\frac{4}{4}=1$
2. $\{7÷7\times 1\}=1$
3. $(19\times 8)\times 1=152$

Trick: When the curly bracket becomes 1, the answer is just $19\times 8$.

Answer: (C) 152

Concept: If p% of x = N, then $x=N\times \frac{100}{p}$.

1. 4% → 48
2. 1% → 12
3. 100% → 1200

Trick: Unitary method: divide by the % then multiply by 100.

Answer: (D) 1200

Concept: Open brackets inside-out: ( ) → { } → [ ].

1. $(6-5)=1\to \{16-2\times 1\times 8\}=16-16=0$
2. $[2+5\times 0÷31]=2$
3. $50-7\times 2=36$

Trick: Once a bracket becomes 0, everything multiplied with it vanishes – skip the ÷ 31.

Answer: (A) 36

Concept: ÷ and × before + and −.

1. $12÷4\to 3$; $3\times 4\to 12$
2. $16+3=19$
3. $19-12=7$

Trick: Compare with the similar sum $16+12÷3-3\times 4=8$; only the divisor changed.

Answer: (C) 7

Concept: For an odd count of consecutive numbers, average = middle term.

1. 19 numbers → middle = 10th term = 41
2. Smallest = $41-(10-1)=41-9=32$

Trick: $\text{Smallest}=\text{average}-\frac{n-1}{2}=41-\frac{18}{2}=32$.

Answer: (B) 32

Concept: For 2 years: CI − SI = interest on one year's SI; also $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. 1 year SI = $42÷2=$ ₹21; $\text{CI}-\text{SI}=$ ₹1
2. Rate $=\frac{1}{21}\times 100\Rightarrow \frac{R}{100}=\frac{1}{21}$
3. $P=21÷\frac{1}{21}=21\times 21=$ ₹441

Trick: $P=(\text{SI per year}{)}^2÷(\text{CI}-\text{SI})={21}^2÷1=441$.

Answer: (A) 441

Concept: 1-day work of C = (1-day work of all) − (A + B).

1. C's rate = $\frac{1}{4}-\frac{1}{16}-\frac{1}{12}$
2. LCM 48: $\frac{12}{48}-\frac{3}{48}-\frac{4}{48}=\frac{5}{48}$
3. C alone takes $\frac{48}{5}$ days

Trick: Using total work = LCM (48 units) avoids fractions: $C=12-3-4=5$ units/day.

Answer: (B) $\frac{48}{5}$

Concept: Saving = Income − Expenditure.

1. Expenditure 5 parts = ₹4,50,000 → 1 part = 90,000 → income = ₹5,40,000
2. Income +40%: $5,40,000\times 1.4=7,56,000$; expenditure −20%: $4,50,000\times 0.8=3,60,000$
3. Final saving = $7,56,000-3,60,000=$ ₹3,96,000

Trick: In parts: new income = $6\times 1.4=8.4$, new expenditure = $5\times 0.8=4$; saving = 4.4 parts $\times 90,000=3,96,000$.

Answer: (A) $3,96,000$

Concept: Inner surface of a hemispherical bowl = $2\pi r^2$.

1. $2\pi r^2=2\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=77{\text{cm}}^2$
2. Cost = 77 × ₹4 = ₹308

Trick: With $r=3.5$, $\pi r^2=38.5$ – remember it; $2\pi r^2=77$.

Answer: (A) ₹308

Concept: Same distance both ways: average speed = $\frac{2xy}{x+y}$.

1. $x=4,y=12$
2. $2\times 48÷16=96÷16=6$ km/h

Trick: Assume distance = LCM 12 km: time 3 h + 1 h = 4 h; average = $24÷4=6$.

Answer: (B) 6