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

Concept: In a 100% stacked bar, each model's share $=$ upper edge $-$ lower edge of its segment.

1. 2020, model A: $0\%$ to $40\%$ ⇒ $40\%$ of $65$ lakh $=26$ lakh.
2. 2019, model E: $75\%$ to $90\%$ ⇒ $15\%$ of $40$ lakh $=6$ lakh.
3. Total $=26+6=32$ lakh $=32,00,000$.

Yaad rakhein: Read segment heights, not the top edge of the segment.

Answer: (B) $32,00,000$

Concept: Add the four cities for each year and pick the smallest total.

1. 2016: $18+25+18+20=81$; 2017: $25+23+18.5+18=84.5$.
2. 2018: $23+24+18.4+17.6=83$; 2019: $24.4+19.5+19+16=78.9$.
3. 2020: $22.5+21+20+17.5=81$. Least $=78.9$ in 2019.

Yaad rakhein: 2019 has the lowest Kolkata bar ($16$) and no year makes it up elsewhere.

Answer: (C) 2019

Concept: Net effect of $+a\%$ then $-b\%$: $a-b-\frac{ab}{100}$.

1. $26-12-\frac{26\times 12}{100}=14-3.12=10.88\%$.
2. Check: CP $100$ → MP $126$ → SP $126\times 0.88=110.88$.

Trick: Successive change formula saves time.

Answer: (A) $10.88\%$

Concept: ${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$ and ${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$.

1. $x^2+\frac{1}{x^2}=\sqrt{47+2}=7$.
2. $x+\frac{1}{x}=\sqrt{7+2}=3$ (as $x>0$).
3. $x+\frac{1}{x}-2=3-2=1$.

Trick: Add $2$ and take the root: $47\to 7\to 3$.

Answer: (A) $1$

Concept: $\frac{\sec \theta -1}{\sec \theta +1}=\frac{1-\cos \theta }{1+\cos \theta }={\left(\frac{1-\cos \theta }{\sin \theta }\right)}^2=(\csc \theta -\cot \theta {)}^2$.

1. Multiply numerator and denominator by $1-\cos \theta$: $\frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }$.
2. So ${\left(\frac{\sec \theta -1}{\sec \theta +1}\right)}^n=(\csc \theta -\cot \theta {)}^{2n}$.
3. $2n=1\Rightarrow n=\frac{1}{2}=0.5$.

Trick: $\theta ={90}^{\circ }$: LHS $=1$ for any $n$, so try $\theta ={60}^{\circ }$: ${\left(\frac{1}{3}\right)}^n=\frac{1}{\sqrt{3}}\Rightarrow n=0.5$.

Answer: (B) $0.5$

Concept: If $DE\parallel BC$, corresponding angles are equal: $\angle AED=\angle ACB$.

1. $\angle AED=\angle ACB={40}^{\circ }$.
2. In $△ADE$: $\angle DAE+\angle ADE={180}^{\circ }-{40}^{\circ }={140}^{\circ }$.

Trick: It equals $\angle A+\angle B$ of the big triangle $={180}^{\circ }-\angle C$.

Answer: (C) ${140}^{\circ }$

Concept: Surface area of a cube $=6a^2$, volume $=a^3$.

1. $6a^2=54\Rightarrow a^2=9\Rightarrow a=3\text{cm}$.
2. Volume $=3^3=27{\text{cm}}^3$.

Trick: $V={\left(\frac{S}{6}\right)}^{3/2}=9^{3/2}=27$.

Answer: (C) $27{\text{cm}}^3$

Concept: ${\sec }^{2}A-{\tan }^{2}A=1$.

1. $16{\sec }^{2}A-16{\tan }^{2}A=16({\sec }^{2}A-{\tan }^{2}A)$.
2. $=16\times 1=16$.

Yaad rakhein: Take the common factor out first.

Answer: (C) $16$

Concept: Price of mixture $=\frac{q_1{p}_1+q_2{p}_2}{q_1+q_2}$.

1. $\frac{3\times 300+2\times 375}{5}=\frac{900+750}{5}$.
2. $=\frac{1650}{5}=330$ per kg.

Trick: $300+(375-300)\times \frac{2}{5}=300+30=330$.

Answer: (B) ₹$330$

Concept: $\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=\frac{2(1+{\tan }^2\theta )}{1-{\tan }^2\theta }=\frac{2}{\cos 2\theta }$, and $2{\cos }^2\theta -1=\cos 2\theta$.

1. $\frac{(1+t{)}^2+(1-t{)}^2}{1-t^2}=\frac{2(1+t^2)}{1-t^2}$ with $t=\tan \theta$.
2. $\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }=\frac{1}{\cos 2\theta }$.
3. Value $=\cos 2\theta \times \frac{2}{\cos 2\theta }=2$.

Trick: Put $\theta =0^{\circ }$: $(2-1)(1+1)=2$.

Answer: (A) $2$

Concept: Read the class frequencies from the histogram and use $\%\text{ more}=\frac{\text{difference}}{\text{base}}\times 100$.

1. Less than 250 marks: classes 150–200 and 200–250 $\Rightarrow 30+45=75$.
2. 400 or more marks: only class 400–450 $\Rightarrow 35$.
3. $\%\text{ more}=\frac{75-35}{35}\times 100=\frac{4000}{35}\approx 114.3\%$.

Trick: Base is the number after 'than' (here 35), so divide the difference by 35.

Answer: (C) $114.3\%$

Concept: Area of four walls of a room $=2h(l+b)$.

1. Area $=2\times 10\times (12+8)=400{\text{m}}^2$.
2. Cost $=400\times 25=10000$ rupees.

Trick: Four walls = perimeter of floor × height: $40\times 10=400$.

Answer: (B) ₹10,000

Concept: New fraction $=\frac{x\times 1.6}{y\times 1.4}$; equate to the given value.

1. $\frac{x}{y}\times \frac{160}{140}=\frac{16}{63}$, i.e. $\frac{x}{y}\times \frac{8}{7}=\frac{16}{63}$.
2. $\frac{x}{y}=\frac{16}{63}\times \frac{7}{8}=\frac{2}{9}$.

Trick: Check: $\frac{2\times 1.6}{9\times 1.4}=\frac{3.2}{12.6}=\frac{16}{63}$ ✓.

Answer: (B) $\frac{2}{9}$

Concept: Bells ring together again after the LCM of their intervals.

1. $\text{LCM}(3,5,8,9,10)=2^3\times 3^2\times 5=360$ seconds.
2. $360\text{s}=\frac{360}{60}=6$ minutes.

Trick: Take highest powers: 8 gives $2^3$, 9 gives $3^2$, and 5.

Answer: (A) 6 minutes

Concept: Total = average × number.

1. Old total $=40\times 16=640$; new total $=50\times 15=750$.
2. 10 new students' total $=750-640=110$.
3. 5 of them $=5\times 11=55$; remaining 5 $=110-55=55$, average $=\frac{55}{5}=11$.

Trick: Sum of new 10 = 110, so their average is also 11.

Answer: (D) 11

Concept: Central angle $=\frac{\text{percentage}}{100}\times {360}^{\circ }$.

1. Interest payment $=15\%$.
2. Angle $=\frac{15}{100}\times {360}^{\circ }={54}^{\circ }$.

Trick: $1\%={3.6}^{\circ }$, so $15\%=15\times 3.6={54}^{\circ }$.

Answer: (C) ${54}^{\circ }$

Concept: If A is $r\%$ more than B, B is $\frac{r}{100+r}\times 100\%$ less than A.

1. Let B = 100, then A = 135.
2. $\%\text{ less}=\frac{35}{135}\times 100\approx 25.93\%\approx 26\%$.

Trick: $35\%=\frac{7}{20}$; B less than A by $\frac{7}{27}\approx 26\%$.

Answer: (D) $26\%$

Concept: $SI=\frac{P\times R\times T}{100}$; equal interest means $P_A{T}_A=P_B{T}_B$.

1. $A\times 2=B\times 4\Rightarrow A:B=2:1$.
2. $A=\frac{2}{3}\times 36000=24000$.
3. Interest from A $=\frac{24000\times 15\times 2}{100}=7200$.

Trick: Same rate: principal ratio = inverse of time ratio $=4:2=2:1$.

Answer: (C) ₹7,200

Concept: Use the LCM of days as total work.

1. Total work $=\text{LCM}(12,36)=36$ units; P+Q = 3/day, P = 1/day.
2. Q $=3-1=2$ units/day.
3. $\frac{2}{3}$ of work $=24$ units $\Rightarrow \frac{24}{2}=12$ days.

Trick: Q alone does whole work in 18 days; $\frac{2}{3}\times 18=12$.

Answer: (A) 12

Concept: Overall pass % $=\frac{\text{total passed}}{\text{total appeared}}\times 100$.

1. Passed: $5+9+14+18+30+75=151$.
2. Appeared: $25+30+40+45+60+100=300$.
3. Pass % $=\frac{151}{300}\times 100\approx 50.3\%\approx 50\%$.

Trick: Never average the percentages directly — weight them by students.

Answer: (D) $50\%$