SSC CPO Paper I 2020, 2022, 2023, 2024 PYQs

Concept: Percentage $=\frac{\text{class frequency}}{\text{total}}\times 100$.

1. Total $=50+70+85+90+60+45=400$.
2. 90 and above (90–100) $=45$.
3. $\frac{45}{400}\times 100=11.25\%$.

Trick: Total 400 ⇒ divide count by 4 to get %.

Answer: (B) $11.25\%$

Concept: For equal time, profit is shared in the ratio of investments.

1. $92500:112500=37:45$.
2. A’s profit $=9000\times \frac{37}{45}=7400$.

Trick: B’s 45 parts $=9000$ ⇒ 1 part $=200$ ⇒ A $=37\times 200$.

Answer: (C) ₹7,400

Concept: Find B’s full time, then A’s rate $=\frac{1}{25}-\frac{1}{B}$.

1. $66\frac{2}{3}\%=\frac{2}{3}$ in 30 days ⇒ B alone $=45$ days.
2. A’s rate $=\frac{1}{25}-\frac{1}{45}=\frac{9-5}{225}=\frac{4}{225}$.
3. Time for $\frac{4}{15}$ work $=\frac{4}{15}÷\frac{4}{225}=15$ days.

Trick: $66\frac{2}{3}\%=\frac{2}{3}$.

Answer: (B) 15 days

Concept: Employees in a department $=\%\times$ total; females $=100\%-75\%=25\%$ of them.

1. D $=8\%$ of 450 $=36$.
2. Females $=25\%$ of 36 $=9$.

Trick: 36 is the whole department, not the females.

Answer: (A) 9

Concept: Replace each factor by its remainder and multiply step by step.

1. Remainders by 35: $72\to 2$, $73\to 3$, $78\to 8$, $76\to 6$.
2. $2\times 3=6$; $6\times 8=48\to 13$.
3. $13\times 6=78\to 8$.

Trick: Reduce after every multiplication to keep numbers small.

Answer: (D) 8

Concept: Employees $=\frac{\%}{100}\times$ total.

1. F $=22\%$.
2. $\frac{22}{100}\times 450=99$.

Trick: $1\%$ of 450 $=4.5$ ⇒ $22\times 4.5=99$.

Answer: (C) 99

Concept: Least number divisible by all = LCM.

1. $4=2^2$, $5$, $8=2^3$, $10=2\times 5$, $12=2^2\times 3$.
2. LCM $=2^3\times 3\times 5=120$.

Trick: Take the highest power of each prime. 240 is divisible too, but not least.

Answer: (B) 120

Concept: Savings = Income − Expenditure; take income 100.

1. Before: income 100, expenditure 75, savings 25.
2. After: income 128, expenditure 90, savings 38.
3. Change $=\frac{38-25}{25}\times 100=52\%$ increase.

Trick: Income 100 makes every figure a direct percentage.

Answer: (B) 52% increase

Concept: Convert each average into a total.

1. $A+B+C=210$; $A+B+C+D=240\Rightarrow D=30$, so $E=35$.
2. $B+C+D+E=236\Rightarrow B+C=236-65=171$.
3. $A=210-171=39$.
4. Average of A, D, E $=\frac{39+30+35}{3}=\frac{104}{3}\approx 34.67\approx 35$.

Trick: Replacing A by E dropped the total by $240-236=4$ ⇒ $A=E+4=39$.

Answer: (B) 35 kg

Concept: $x^3+y^3+z^3-3xyz=(x+y+z)[(x+y+z{)}^2-3(xy+yz+zx)]$.

1. $x^3+y^3+z^3-3xyz=17\times (289-333)=17\times (-44)=-748$.
2. $x^3+y^3+z^3=-748+513=-235$.
3. $x^3+y^3+z^3+xyz=-235+171=-64$.
4. $\sqrt[3]{-64}=-4$.

Trick: Cube root of a negative number is negative; −64 is the value under the root, not the answer.

Answer: (D) −4

Concept: Area of a square $=\frac{d^2}{2}$, so area $\propto d^2$ and $d\propto \sqrt{\text{Area}}$.

1. New area $=100\%-19\%=81\%$ of old area.
2. New diagonal $=\sqrt{0.81}=0.9$ times old diagonal.
3. Decrease $=1-0.9=0.1=10\%$.

Trick: Area factor $0.81=(0.9{)}^2$, so every length (side, diagonal) falls by $10\%$.

Answer: (A) $10\%$

Concept: If a sum of squares is zero, each square is zero. Split the expression into two perfect squares.

1. $x^2-4xy+4y^2+4y^2-12y+9=0$
2. $(x-2y{)}^2+(2y-3{)}^2=0$
3. $2y-3=0\Rightarrow y=\frac{3}{2}$; $x=2y=3$
4. $7x-8y=21-12=9$

Trick: Write $8y^2=4y^2+4y^2$: one part completes $(x-2y{)}^2$, the other $(2y-3{)}^2$.

Answer: (D) $9$

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $433.50=P\times {\left(\frac{17}{100}\right)}^2=P\times \frac{289}{10000}$
2. $P=\frac{433.50\times 10000}{289}=1.5\times 10000=15000$

Trick: $289\times 1.5=433.5$, so $P=1.5\times 10000$ directly.

Answer: (A) ₹$15000$

Concept: Add the three pair-rates: it gives twice the rate of A, B and C together.

1. Take work $=\text{LCM}(36,60,45)=180$ units. $A+B=5$, $B+C=3$, $A+C=4$ units/day.
2. $2(A+B+C)=12\Rightarrow A+B+C=6$
3. $A=6-(B+C)=6-3=3$ units/day
4. Days for A $=\frac{180}{3}=60$

Trick: One alone $=$ (all three) $-$ (the other two).

Answer: (B) $60$ days

Concept: Radius ⟂ tangent, so in quadrilateral OAPB, $\angle AOB={180}^{\circ }-\angle APB$; triangle OAB is isosceles ($OA=OB$).

1. $\angle AOB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$
2. $\angle OAB=\angle OBA=\frac{{180}^{\circ }-{80}^{\circ }}{2}={50}^{\circ }$

Trick: $\angle OAB=\frac{\angle APB}{2}={50}^{\circ }$ (always half the angle between tangents).

Answer: (D) ${50}^{\circ }$

Concept: Department strength $=\text{percentage}\times \text{total}$; males $=100\%-60\%=40\%$ of it.

1. Department E $=20\%$ of $450=90$
2. Males $=40\%$ of $90=36$

Trick: Directly: $450\times 0.2\times 0.4=36$.

Answer: (B) $36$

Concept: For the same time period, profit is shared in the ratio of investments.

1. $92500:112500=37:45$
2. B's share $=\frac{45}{82}$ of profit $=9000$
3. Total profit $=\frac{9000\times 82}{45}=200\times 82=16400$

Trick: 1 ratio unit $=\frac{9000}{45}=200$; total $=82\times 200$.

Answer: (B) ₹$16400$

Concept: Area of a trapezium $=\frac{1}{2}\times (\text{sum of parallel sides})\times \text{height}$; $1{\text{m}}^2=10000{\text{cm}}^2$.

1. Area $=\frac{1}{2}\times (27+13)\times 7=140{\text{cm}}^2$
2. In ${\text{m}}^2$: $\frac{140}{10000}=0.014$

Yaad rakhein: ${\text{cm}}^2\to {\text{m}}^2$: divide by ${10}^4$ (not $100$).

Answer: (B) $0.014$

Concept: Numbers in ratio $a:b$ (co-prime) with HCF $h$ are $ha$ and $hb$; LCM $=h\times a\times b$.

1. Numbers $=8\times 7=56$ and $8\times 13=104$
2. LCM $=8\times 7\times 13=728$

Trick: Check: $\text{HCF}\times \text{LCM}=8\times 728=5824=56\times 104$.

Answer: (D) $728$

Concept: Girls $\%=\frac{\text{girls}}{\text{boys}+\text{girls}}\times 100$ for each school.

1. A: $\frac{500}{1100}\approx 45.5\%$, B: $\frac{550}{1000}=55\%$
2. C: $\frac{650}{1400}\approx 46.4\%$, D: $\frac{600}{1300}\approx 46.2\%$
3. E: $\frac{650}{1450}\approx 44.8\%$ – the lowest

Trick: Compare boys : girls; the highest boys-to-girls ratio ($\frac{800}{650}\approx 1.23$ in E) means the lowest girls share.

Answer: (C) E