SSC GD 2025 PYQs

Concept: SP = CP + Profit; then convert to the required unit.

1. CP of 400 apples = 4 × ₹1210 = ₹4840
2. $\text{SP}=4840+860=$ ₹5700
3. SP per apple = $5700÷400=$ ₹14.25
4. Per dozen = $14.25\times 12=$ ₹171

Trick: Per dozen = $\text{total SP}\times \frac{12}{400}=5700\times \frac{3}{100}=171$.

Answer: (C) 171

Concept: If two shares grow to the same amount, the share invested for one year less must be bigger by the factor $(1+\frac{r}{100})$.

1. Ketan's money stays $23-15$ = 8 years; Mahesh's stays $23-16$ = 7 years.
2. $K\times {1.04}^8=M\times {1.04}^7\Rightarrow M=1.04K\Rightarrow K:M=25:26$
3. Total 51 parts = ₹25500 ⇒ 1 part = ₹500
4. Mahesh = $26\times 500=$ ₹13000

Trick: $4\%=\frac{1}{25}$, so the younger-time share is $\frac{26}{25}$ of the other – ratio 25 : 26 directly.

Answer: (D) 13000

Concept: Successive changes multiply: new = old × product of $(1\pm \frac{r}{100})$.

1. Factors: 0.75, 0.65, 0.60, 1.60
2. $0.75\times 0.65=0.4875$; $\times 0.6=0.2925$; $\times 1.6=0.468$
3. Price = 46.8% of original ⇒ fall of $100-46.8=$ 53.2%

Trick: Pair smartly: $0.6\times 1.6=0.96$ and $0.75\times 0.65=0.4875$; $0.4875\times 0.96\approx 0.468$.

Answer: (B) Decreases by 53.2%

Concept: If a part of a group has the same average as the whole, the rest also has that average.

1. Total of 5 = $5\times 55=275$; total of 3 = $3\times 55=165$
2. Remaining two = $275-165=110$
3. Average = $110÷2=$ 55

Trick: Same average for whole and part ⇒ remaining part has the same average – no calculation needed.

Answer: (A) 55

Concept: Subtract each remainder from its number; the answer is the HCF of the results.

1. $2248-3=2245$; $1761-1=1760$
2. $2245=5\times 449$ (449 is prime); $1760=5\times 352$
3. HCF = 5 (greater than both remainders ✔)

Trick: Test options on 2245 and 1760: 2245 is odd (rules out 14, 24) and not a multiple of 19 → only 5 works.

Answer: (B) 5

Concept: Distance = speed × time, with time converted to hours.

1. 35 min 15 s = $35\frac{1}{4}\text{min}=\frac{141}{4}\text{min}=\frac{141}{240}\text{h}$
2. Distance = $65\times \frac{141}{240}=\frac{9165}{240}=38.1875$ km
3. $0.1875\text{km}\times 1000=187.5$ m ⇒ 38 km 187.5 m

Trick: 65 km/h ≈ 1.0833 km per minute; 35 min ≈ 37.9 km, so the km part must be 38 – options with 36 drop out.

Answer: (B) 38; 187.5

Concept: Opposite directions: relative speed = sum; distance = sum of both lengths.

1. Relative speed = $50+80=130$ km/h = $130\times \frac{5}{18}$ m/s
2. Total length = $130\times \frac{5}{18}\times 28=\frac{18200}{18}=1011.11$ m
3. Other train = $1011.11-480=$ 531.11 m

Trick: Check the decimals: $\frac{18200}{18}$ ends in .11, and 480 is whole – so the answer must end in .11.

Answer: (D) 531.11

Concept: For 2 years, $\text{CI}=\text{SI}\times (1+\frac{r}{200})$.

1. Here $1+\frac{4}{200}=1.02$
2. $\text{SI}=127.5÷1.02=$ ₹125
3. (Check: $P=125÷0.08=1562.5$; $\text{CI}=1562.5\times 0.0816=127.5$ ✔)

Trick: 2-year $\text{CI}-\text{SI}=\text{SI}\times \frac{r}{200}$; so divide CI by $(1+\frac{r}{200})$ – one step.

Answer: (C) 125

Concept: Take total work = LCM of days; find each person's daily units.

1. Work = $\text{LCM}(28,18,16)=1008$ units
2. $A+B=36,B+C=56,A+B+C=63$ per day
3. $C=63-36=27;A=63-56=7\Rightarrow A+C=34$
4. Days = $\frac{1008}{34}=\frac{504}{17}=$ $29\frac{11}{17}$

Trick: B alone = $(A+B)+(B+C)-(A+B+C)=36+56-63=29$, so $A+C=63-29=34$.

Answer: (A) $29\frac{11}{17}$

Concept: Third proportional to a and b = $\frac{b^2}{a}$ (since $a:b=b:x$).

1. $\frac{b^2}{a}=ab$ (given)
2. $b^2=a^{2}b\Rightarrow b=a^2$ ($b\ne 0$)

Trick: Put $a=2,b=4$: third proportional = $\frac{16}{2}=8=ab$ ✔ – matches $a^2=b$.

Answer: (C) $a^2=b$

Concept: Number leaving the same remainder r with several divisors = $k\times \text{LCM}+r$.

1. $\text{LCM}(12,15,18,20,25)=2^2\times 3^2\times 5^2=900$
2. $x=900k+3$
3. $k=1\Rightarrow 903=7\times 129$ ✔ divisible by 7

Trick: Options must end in 3 after subtracting 3 → multiple of 900; only 903 fits.

Answer: (C) 903

Concept: Net discount for a% and b% = $a+b-\frac{ab}{100}$.

1. 10, 3: $13-0.3=12.7\%$
2. 10, 10: $20-1=$ 19%
3. 7, 10: $17-0.7=16.3\%$; 5, 10: $15-0.5=14.5\%$

Trick: The pair with the largest sum ($10+10=20$) wins here, since all the $\frac{ab}{100}$ terms are small.

Answer: (B) Two successive discounts of 10% each

Concept: Profit = $\text{SP}-\text{CP}$.

1. SP = 5 × ₹220 = ₹1100
2. Profit = $1100-875=$ ₹225

Trick: Per kg: $\text{CP}=\frac{875}{5}=175$, profit/kg = 45, × 5 = 225.

Answer: (C) ₹225

Concept: Fourth proportional of p, q, r = $\frac{qr}{p}$.

1. From $a:c=bc:d\Rightarrow d=\frac{b{c}^2}{a}$
2. Required = $(\frac{ad}{c}\times c)÷\left(\frac{b{c}^2}{a}\right)=ad\times \frac{a}{b{c}^2}=\frac{a^{2}d}{b{c}^2}$
3. Put $d=\frac{b{c}^2}{a}$ ⇒ result = a

Trick: Note $ad=b{c}^2$; so the first term $\frac{b{c}^2}{a}=d$. Then $x=\frac{ad}{c}·c÷d=\frac{ad}{d}=a$.

Answer: (B) a

Concept: Solve innermost brackets first; ÷ and × left to right.

1. $19-12=7$
2. $\{4÷4\times 7÷7\}=1\times 7÷7=1$
3. $22\times 6=132\Rightarrow 132\times 1=$ 132

Trick: The curly bracket is 'something ÷ itself' style = 1, so the answer is just $22\times 6$.

Answer: (C) 132

Concept: $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}\Rightarrow \text{base}\propto \text{area}÷\text{height}$.

1. $b_1:b_2=(4÷3):(3÷4)$
2. = $\frac{4}{3}\times \frac{4}{3}=$ 16 : 9

Trick: Divide by a ratio = multiply by its inverse: 4:3 × 4:3 = 16:9.

Answer: (A) 16 : 9

Concept: Numbers = HCF × co-primes a, b, and $\text{LCM}=\text{HCF}\times a\times b$.

1. $\text{LCM}=126\times 7\Rightarrow a\times b=126$
2. Difference = $5\times 7\Rightarrow a-b=5$
3. Co-prime pair: 14 and 9
4. Sum = $7\times (14+9)=7\times 23=$ 161

Trick: The sum must be a multiple of HCF 7: among options only 161 $(=7\times 23)$ qualifies.

Answer: (C) 161

Concept: Common member = (sum of the two pair totals) − (total of all three).

1. Total of three = $3\times 45=135$
2. Pair totals: $2\times 43=86$ each
3. Akshay = $86+86-135=$ 37 kg

Trick: Akshay is counted twice in the two pairs – remove the triple total once.

Answer: (A) 37

Concept: At the same rate and time, the interest difference comes only from the principal difference.

1. Rupesh has ₹4000 more ($9400-5400$).
2. In 4 years, interest on that ₹4000 = ₹600 ⇒ yearly ₹150
3. $R=\frac{150}{4000}\times 100=$ 3.75%

Trick: $R=600\times 100÷(4000\times 4)=\frac{60000}{16000}=3.75$.

Answer: (C) 3.75

Concept: $25\%=\frac{1}{4}$.

1. Promoted = $1224÷4=$ 306

Trick: Halve twice: 1224 → 612 → 306.

Answer: (B) 306