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

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

1. Subsidy $=21\%$.
2. Angle $=21\times {3.6}^{\circ }={75.6}^{\circ }$.

Trick: $1\%={3.6}^{\circ }$ in a pie chart.

Answer: (D) ${75.6}^{\circ }$

Concept: Let present ages be $3x$ and $4x$.

1. $\frac{3x-12}{4x-12}=\frac{2}{3}\Rightarrow 9x-36=8x-24$.
2. $x=12$: ages $36$ and $48$.
3. Sum $=84$ years.

Trick: Difference of ratio terms stays the same: $(4-3)$ and $(3-2)$ both 1 ⇒ 1 part = 12 years.

Answer: (C) $84$

Concept: $a^3+b^3=(a+b)(a^2-ab+b^2)$.

1. Let $a=4.35$, $b=3.25$; the denominator is $100(a^2+b^2-ab)$.
2. Value $=\frac{(a+b)(a^2-ab+b^2)}{100(a^2-ab+b^2)}=\frac{a+b}{100}$.
3. $=\frac{7.6}{100}=0.076$.

Trick: Watch the decimal shift: denominator numbers are 10 times bigger, squared → 100 times.

Answer: (C) $0.076$

Concept: Add the required bars of each year and simplify.

1. 2018 (B1+B2+B3) $=80+75+95=250$.
2. 2019 (B4+B5+B6) $=95+95+80=270$.
3. Ratio $=250:270=25:27$.

Trick: Mind the order: first 2018 group, then 2019 group.

Answer: (B) 25 : 27

Concept: Total work $=\text{LCM}(35,15)=105$ units; A $=3$, B $=7$ units/day.

1. 8 days together: $8\times 10=80$ units.
2. Remaining $=25$ units; $60\%$ of it $=15$ units.
3. A alone: $\frac{15}{3}=5$ days.

Trick: Use LCM units to avoid fractions.

Answer: (C) $5$

Concept: Equal SI at the same rate ⇒ principal $\times$ time is equal.

1. $A\times 2=B\times 4\Rightarrow A:B=2:1$ → $A=18000$, $B=9000$.
2. SI on A $=18000\times 0.15\times 2=5400$; SI on B $=9000\times 0.15\times 4=5400$.
3. Total $=$ ₹10,800.

Trick: Both interests are equal, so total = $2\times 5400$.

Answer: (D) ₹10,800

Concept: Crossing a platform: distance = train + platform; overtaking a man: relative speed $(x-6)$.

1. $L+300=x\times \frac{5}{18}\times 30$
2. $L=(x-6)\times \frac{5}{18}\times 20$
3. Subtract: $300=\frac{5}{18}(30x-20x+120)\Rightarrow 1080=10x+120$
4. $x=96$.

Trick: Check: $90$ km/h $=25$ m/s, so $L=25\times 20=500$ m; $96$ km/h $=\frac{80}{3}$ m/s and $\frac{80}{3}\times 30=800=500+300$ ✓.

Answer: (B) $96$

Concept: “of” is done before ÷ and ×; then ÷, × left to right; then + and −.

1. $\frac{3}{10}\text{ of }\frac{4}{9}=\frac{2}{15}$, so $\frac{2}{3}÷\frac{2}{15}=5$.
2. $\frac{4}{5}\times \frac{10}{9}÷\frac{8}{15}=\frac{8}{9}\times \frac{15}{8}=\frac{5}{3}$; $\frac{3}{4}÷\frac{1}{2}=\frac{3}{2}$.
3. $5-\frac{5}{3}-\frac{3}{4}+\frac{3}{2}=\frac{60-20-9+18}{12}=\frac{49}{12}$.

Trick: Remember VBODMAS: “of” has higher priority than division.

Answer: (D) $\frac{49}{12}$

Concept: Complete squares; a sum of squares is zero only when each square is zero.

1. $(a^2-4a+4)+(b^2+8b+16)+(c^2-16c+64)=0$ (since $4+16+64=84$).
2. $(a-2{)}^2+(b+4{)}^2+(c-8{)}^2=0\Rightarrow a=2,b=-4,c=8$.
3. $ab-bc+ca=-8+32+16=40$, $\sqrt{40}=2\sqrt{10}$.

Trick: Values = half the coefficients on the right: $a=\frac{4}{2}$, $b=\frac{-8}{2}$, $c=\frac{16}{2}$.

Answer: (D) $2\sqrt{10}$

Concept: Amount $=\frac{\text{percentage}}{100}\times \text{total}$.

1. Education + Health $=6\%+4\%=10\%$.
2. $10\%$ of $1680=168$ billion.

Trick: 10% = move the decimal one place.

Answer: (A) $168$

Concept: The number must be a multiple of the LCM and a perfect square.

1. $\text{LCM}(10,12,15,18,20)=180=2^2\times 3^2\times 5$.
2. Make every power even: multiply by 5 → $900={30}^2$.

Trick: Check options: 400 and 625 are not divisible by 18; 180 is not a square.

Answer: (C) $900$

Concept: Solve innermost brackets first; “of” before ÷.

1. $(4\text{ of }2÷4)=8÷4=2$; $(4÷4\text{ of }2)=4÷8=\frac{1}{2}$.
2. $\{4+2+\frac{1}{2}\}=\frac{13}{2}$; $5+1÷2=\frac{11}{2}$.
3. Bracket $=\frac{11}{2}-\frac{13}{2}=-1$; $-7÷(-1)=7$.

Trick: Negative ÷ negative is positive – that rules out options (A) and (B).

Answer: (D) $7$

Concept: For a fixed distance, time $\propto \frac{1}{\text{speed}}$.

1. Speed ratio $5:9$ ⇒ time ratio $9:5$.
2. Saving $=4$ parts $=40$ min ⇒ 1 part $=10$ min.
3. Usual time $=9\times 10=90$ min.

Trick: Usual time $=\frac{\text{saving}}{1-\frac{5}{9}}=\frac{40}{4/9}=90$.

Answer: (C) $90$

Concept: Use $(a+b{)}^2=a^2+b^2+2ab$ with $a^2+b^2={26}^2$.

1. $a+b=60-26=34$.
2. $2ab={34}^2-{26}^2=1156-676=480\Rightarrow ab=240$.
3. Area $=\frac{1}{2}ab=120{\text{cm}}^2$.

Trick: Triplet 10-24-26 (= 2 × 5-12-13): area $=\frac{1}{2}\times 10\times 24=120$.

Answer: (D) $120$

Concept: % more $=\frac{\text{difference}}{\text{base}}\times 100$, base = students with 400 or more.

1. Less than 350: $30+45+60+35=170$.
2. 400 or more: $35$.
3. More by $\frac{170-35}{35}\times 100=\frac{135}{35}\times 100\approx 385.7\%$.

Trick: $\frac{170}{35}=4.857$ times → $385.7\%$ more (subtract 100%).

Answer: (C) 385.7%

Concept: $6^2+8^2={10}^2$, so $\angle A={90}^{\circ }$; the median to the hypotenuse equals half the hypotenuse.

1. $36+64=100$ ⇒ right angle at A, BC is the hypotenuse.
2. $AD=\frac{BC}{2}=5$ cm.

Trick: Apollonius check: $A{D}^2=\frac{2(36+64)-100}{4}=25$.

Answer: (B) $5$

Concept: Numbers $=\text{HCF}\times$ each co-prime factor.

1. Numbers $=29\times 13=377$ and $29\times 15=435$.
2. Larger $=435$.

Trick: Larger number = HCF × larger factor.

Answer: (A) $435$

Concept: Area of a path around a circle $=\pi (R^2-r^2)=\pi (R+r)(R-r)$, where $R=r+\text{width}$.

1. $r=\frac{10}{2}=5\text{m}$, width $=64\text{cm}=0.64\text{m}$, so $R=5.64\text{m}$.
2. Area $=\frac{22}{7}\times (5.64+5)\times (5.64-5)=\frac{22}{7}\times 10.64\times 0.64$.
3. $=\frac{22}{7}\times 6.8096\approx 21.4{\text{m}}^2$.
4. None of the options 10, 11, 15, 9 is close to $21.4$. (A value near 10 would need a path only $32\text{cm}$ wide: $\frac{22}{7}\times 10.32\times 0.32\approx 10.4$.)

Trick: Use $(R+r)(R-r)$ instead of squaring decimals: $10.64\times 0.64\approx 6.81$, then $\times \frac{22}{7}\approx 21.4$.

Answer: The area of the path as printed is about $21.4{\text{m}}^2$, so no option fits; this question was treated as a bonus question.

Concept: Build the number backwards from the last quotient $k$.

1. Last step: $7k+5$; previous: $4(7k+5)+3=28k+23$.
2. Number $=3(28k+23)+2=84k+71$.
3. Divided by $84$ ($=3\times 4\times 7$) → remainder $71$.

Trick: Remainder $=r_1+d_1{r}_2+d_1{d}_2{r}_3=2+3\times 3+12\times 5=71$.

Answer: (A) $71$

Concept: From the midpoint M both feet are at the same distance $d$, so height $=d\tan \theta$.

1. $h_A=d\tan {30}^{\circ }=\frac{d}{\sqrt{3}}$; $h_B=d\tan {60}^{\circ }=\sqrt{3}d$.
2. $\frac{h_B}{h_A}=\frac{\sqrt{3}}{1/\sqrt{3}}=3$ → $3:1$.

Trick: $\frac{\tan {60}^{\circ }}{\tan {30}^{\circ }}=3$ always.

Answer: (B) 3 : 1