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

Concept: Surface area of sphere $=4\pi r^2$.

1. $4\times \frac{22}{7}\times r^2=528$.
2. $r^2=\frac{528\times 7}{88}=42\Rightarrow r=\sqrt{42}\text{cm}$.

Trick: $\frac{528}{88}=6$, then $6\times 7=42$.

Answer: (B) $\sqrt{42}\text{cm}$

Concept: A diameter perpendicular to a chord bisects it; also $PC·CQ=AC·CB$.

1. $AB=14$, $AC:BC=4:3\Rightarrow AC=8$, $BC=6$.
2. $P{C}^2=AC\times CB=48\Rightarrow PC=4\sqrt{3}$.
3. $PQ=2\times 4\sqrt{3}=8\sqrt{3}\text{cm}$.

Trick: Intersecting chords: $PC·CQ=AC·CB$ with $PC=CQ$.

Answer: (D) $8\sqrt{3}$

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$.

1. Times: $\frac{1200}{80}=15$, $\frac{1600}{40}=40$, $\frac{1000}{400}=2.5$, $\frac{200}{50}=4$; total $=61.5$ h.
2. Distance $=4000$ km.
3. Average $=\frac{4000}{61.5}=65.04\text{km/h}$.

Trick: The slow ship leg (40 h) dominates, pulling the average down.

Answer: (A) $65.04\text{km/h}$

Concept: Brackets first, then divide.

1. $9\times 81=729$, $142÷2=71$ → $800$.
2. $800+350-155+5=1000$.
3. $(25\times 25)-125=500$; $1000÷500=2$.

Trick: Numerator $=1000$, denominator $=500$ – neat round numbers.

Answer: (D) $2$

Concept: Required $\%=\frac{\text{Total}-\text{Pop}}{\text{Total}}\times 100$ for age 31-50.

1. Total $=62$, Pop $=16$ → others $=46$.
2. $\frac{46}{62}\times 100=74.19\%$.

Trick: $\frac{23}{31}\approx 0.742$.

Answer: (C) 74.19%

Concept: Total work = LCM$(21,28,15)=420$ units.

1. Efficiencies: Manish $20$, Nakul $15$, Pintoo $28$ units/day.
2. First 5 days (M + P): $5\times 48=240$; left $=180$.
3. All three: $63$/day → $\frac{180}{63}=\frac{20}{7}=2\frac{6}{7}$ days.

Trick: Nakul works only for the remaining part – divide remainder by combined rate.

Answer: (B) $2\frac{6}{7}$

Concept: $\tan \theta ·\cot \theta =1$, so $(\tan \theta +\cot \theta {)}^2={\tan }^2\theta +{\cot }^2\theta +2$.

1. $6^2=36={\tan }^2\theta +{\cot }^2\theta +2$.
2. ${\tan }^2\theta +{\cot }^2\theta =34$.

Trick: If $x+\frac{1}{x}=k$ then $x^2+\frac{1}{x^2}=k^2-2$.

Answer: (C) $34$

Concept: $|a-b|<x<a+b$.

1. $12-7<x<12+7\Rightarrow 5<x<19$.
2. Integers: $6,7,\dots ,18$ → $18-6+1=13$ values.

Yaad rakhein: Count $=(a+b)-(a-b)-1=2b-1=2\times 7-1=13$ (b = smaller side).

Answer: (A) $13$

Concept: Net change $=a+b+\frac{ab}{100}$ (decrease negative).

1. $a=-3.125$, $b=+7.4$.
2. $-3.125+7.4-\frac{3.125\times 7.4}{100}=4.275-0.23125=4.04375$.
3. ≈ $4.04\%$ more.

Trick: Multiplier: $0.96875\times 1.074=1.0404$.

Answer: (B) 4.04% more

Concept: Final SP $=$ tag price after discount $-$ cashback $=$ CP $\times 1.1$.

1. SP $=1200\times 1.1=₹1320$.
2. $0.9T-250=1320\Rightarrow 0.9T=1570$.
3. $T=\frac{1570}{0.9}=₹1744.44$.

Trick: Add the cashback back before dividing by $0.9$.

Answer: (C) 1744.44

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{75\times 12}=\sqrt{900}=30$
2. $30-13=17$

Trick: $75\times 12=900$ is a perfect square, so the root is $30$ instantly.

Answer: (C) $17$

Concept: BODMAS: brackets first, then “of”, then $÷$ and $\times$ left to right, then $+$ and $-$.

1. $8\text{ of }3=24$, so $(24-16)=8$
2. $8÷4\times 20=2\times 20=40$
3. $[44-40]=4$
4. $64÷4=16$

Trick: Treat “of” as multiplication done before $÷$.

Answer: (B) $16$

Concept: Solve the innermost bracket first; convert mixed fractions to improper fractions with LCM $120$.

1. $\frac{3}{5}\times \frac{1}{10}÷\frac{4}{15}=\frac{3}{50}\times \frac{15}{4}=\frac{9}{40}$
2. $\frac{5}{6}-\frac{9}{40}=\frac{100-27}{120}=\frac{73}{120}$
3. $\frac{7}{2}-\frac{73}{120}=\frac{420-73}{120}=\frac{347}{120}$
4. $\frac{21}{4}-\frac{347}{120}=\frac{630-347}{120}=\frac{283}{120}=2\frac{43}{120}$

Trick: Use one common denominator ($120$) for all steps to avoid repeated LCMs.

Answer: (A) $2\frac{43}{120}$

Concept: The table is cumulative (“and above”). Students below a mark = total − students at or above it.

1. Aggregate “$40$ and above” $=78$ students
2. Total students $=100$
3. Below $40\%$ $=100-78=22$

Trick: “Less than $x$” $=$ total $-$ “$x$ and above” — read only the Aggregate row.

Answer: (D) $22$

Concept: Average speed $=\frac{\text{distance}}{\text{time}}$, with distance in km and time in hours.

1. Distance $=37\text{km}500\text{m}=37.5\text{km}$
2. Time $=90\text{min}=1.5\text{h}$
3. Speed $=\frac{37.5}{1.5}=25\text{km/h}$

Trick: $90$ min is $\frac{3}{2}$ h, so speed $=37.5\times \frac{2}{3}=25$.

Answer: (A) $25$

Concept: Angle sum of a triangle is ${180}^{\circ }$; at most one angle can be $\ge {90}^{\circ }$.

1. A: all angles $<{60}^{\circ }$ gives sum $<{180}^{\circ }$ — false.
2. B: one obtuse angle (e.g. ${110}^{\circ }$) with two small acute angles — true.
3. C: two right angles already make ${180}^{\circ }$, third angle $0^{\circ }$ — false.
4. D: every triangle has at least two acute angles — true.

Yaad rakhein: Equilateral triangle (${60}^{\circ }$ each) shows angles can never all be below ${60}^{\circ }$.

Answer: (B) B and D

Concept: Open brackets from inside out; product of two negatives is positive.

1. $-4(-a)=4a$, so $x[-5\{4a\}]=-20ax$
2. $-3(-a)=3a$, so $6[-3\{3a\}]=6(-9a)=-54a$
3. $-20ax-54a=6a\Rightarrow -20ax=60a$
4. $x=-3$

Trick: Count minus signs: even count gives $+$, odd gives $-$.

Answer: (A) $-3$

Concept: If $x+y+z=0$, then $x^3+y^3+z^3=3xyz$.

1. Let $x=a$, $y=2b$, $z=3c$; then $x+y+z=0$
2. $a^3+(2b{)}^3+(3c{)}^3=3·a·2b·3c$
3. $=18abc$

Trick: Multiply $3$ by all the coefficients: $3\times 1\times 2\times 3=18$.

Answer: (A) $18abc$

Concept: Required $\%=\frac{\text{F total}}{\text{C total}}\times 100$.

1. C: $95+110=205$ thousand
2. F: $70+80=150$ thousand
3. $\frac{150}{205}\times 100=73.17\%$

Trick: $\frac{150}{205}=\frac{30}{41}$; $\frac{30}{41}\approx 0.7317$.

Answer: (C) $73.17\%$

Concept: Add the pair rates: $2(A+B+C)$ = sum of pair rates. Then $D=(A+B+C+D)-(A+B+C)$.

1. $2(A+B+C)=\frac{1}{25}+\frac{1}{36}+\frac{1}{30}=\frac{36+25+30}{900}=\frac{91}{900}$
2. $A+B+C=\frac{91}{1800}$
3. $D=\frac{1}{18}-\frac{91}{1800}=\frac{100-91}{1800}=\frac{9}{1800}=\frac{1}{200}$
4. D alone takes $200$ days

Trick: Total work $=1800$ units: A+B+C do $91$ units/day, all four do $100$ units/day, so D does $9$ units/day $\Rightarrow \frac{1800}{9}=200$ days.

Answer: (C) $200$