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

Concept: B is common with the same value 4, so A : B : C $=3:4:5$.

1. C − A $=5-3=2$ parts $=30$
2. 1 part $=15$
3. B $=4\times 15=60$

Trick: Equalise the common term first, then use the given difference.

Answer: (B) $60$

Concept: BODMAS: brackets, then 'of', then division, then addition/subtraction.

1. $2\frac{1}{5}-2=\frac{1}{5}$; $\frac{1}{5}\times 5=1$
2. $\frac{2}{5}÷1=\frac{2}{5}$
3. $\frac{2}{3}\text{ of }\frac{3}{5}=\frac{2}{5}$
4. $\frac{1}{4}+\frac{2}{5}-\frac{2}{5}=\frac{1}{4}$

Trick: The two $\frac{2}{5}$ terms cancel.

Answer: (B) $\frac{1}{4}$

Concept: Non-trained $\%=100-$ trained $\%$.

1. College A male: $56\%$ of $225=126$
2. College B female: $75\%$ of $128=96$
3. Total $=126+96=222$

Trick: $\frac{3}{4}\times 128=96$ is quick.

Answer: (A) $222$

Concept: $\cot \theta \sin \theta =\cos \theta$; max of $a\sin \theta +b\cos \theta$ is $\sqrt{a^2+b^2}$.

1. Expression $=19\sin \theta +6\cos \theta$
2. Max $=\sqrt{{19}^2+6^2}=\sqrt{361+36}=\sqrt{397}$

Trick: Simplify $\cot \theta \sin \theta$ first, then apply $\sqrt{a^2+b^2}$.

Answer: (B) $\sqrt{397}$

Concept: Volume of a hollow cylinder $=\pi (R^2-r^2)h$.

1. $R=7$ cm, $r=4$ cm, $h=42$ cm
2. $V=\frac{22}{7}\times (49-16)\times 42$
3. $=22\times 33\times 6=4356{\text{cm}}^3$

Trick: $R^2-r^2=(R+r)(R-r)=11\times 3=33$.

Answer: (C) $4356$

Concept: $\times$ and $÷$ are done left to right, then $+$.

1. $22\times 10=220$
2. $220÷5=44$
3. $44\times 2=88$
4. $88+12=100$

Trick: Do not compute $5\times 2$ first — go strictly left to right.

Answer: (C) $100$

Concept: Compare P + Q with R + S for each year.

1. 1989: 24 vs 25; 1990: 30 vs 23; 1991: 30 vs 26
2. 1992: 29 vs 31
3. 1993: $21+12=33$ and $13+20=33$ ✔

Trick: Check only the option years.

Answer: (C) 1993

Concept: Value of a sector $\propto$ its angle.

1. Savings ${60}^{\circ }=₹8000$
2. Expenditure $={360}^{\circ }-{60}^{\circ }={300}^{\circ }$
3. ${300}^{\circ }=\frac{300}{60}\times 8000=₹40000$

Trick: Expenditure is 5 times savings (${300}^{\circ }:{60}^{\circ }$).

Answer: (B) 40000

Concept: Area $\propto r^2$, volume $\propto r^3$.

1. $r_1:r_2=\sqrt{9}:\sqrt{49}=3:7$
2. Volumes $=3^3:7^3=27:343$

Trick: Square root, then cube.

Answer: (A) $27:343$

Concept: Do all $\times$ and $÷$ first, then $+$ and $-$ from left to right.

1. $81+1-81+9+81-9-81$
2. $=1$

Trick: The $81$s and $9$s cancel in pairs, leaving $9÷9=1$.

Answer: (C) $1$

Concept: A radius is perpendicular to the tangent at the point of contact; with vertically opposite angles at E, the two triangles are similar.

1. $DA⟂AC$ and $BC⟂AC$, so $\angle DAE=\angle BCE={90}^{\circ }$.
2. $\angle AED=\angle CEB$ (vertically opposite), hence $△DAE~△BCE$.
3. $\frac{DA}{AE}=\frac{BC}{EC}\Rightarrow \frac{8}{12}=\frac{x}{18}$
4. $x=\frac{8\times 18}{12}=12$

Trick: Radii are in the same ratio as the tangent pieces: $8:x=12:18$.

Answer: (A) $12$

Concept: If $a:b=c:d$, then $d$ is the fourth proportional: $d=\frac{b\times c}{a}$.

1. $10:12=15:x$
2. $x=\frac{12\times 15}{10}=\frac{180}{10}=18$

Trick: Fourth proportional = (2nd × 3rd) ÷ 1st.

Answer: (D) $18$

Concept: Total surface area of a solid hemisphere = curved surface $2\pi r^2$ + flat circular base $\pi r^2$ = $3\pi r^2$.

1. $r=11\text{cm}$
2. $\text{TSA}=3\pi r^2=3\pi \times 121=363\pi {\text{cm}}^2$

Yaad rakhein: Hemisphere: CSA = $2\pi r^2$, TSA = $3\pi r^2$.

Answer: (C) $363\pi {\text{cm}}^2$

Concept: Solve the innermost bracket first, then curly, then square brackets; ÷ and × left to right.

1. $(14\times 14)=196$
2. $\{13\times 14÷196\}=\frac{182}{196}=\frac{13}{14}$
3. $[14÷13\times \frac{13}{14}]=\frac{14}{13}\times \frac{13}{14}=1$
4. $5\times 1=5$

Trick: Everything inside the square bracket cancels to $1$, so the answer is just the $5$ outside.

Answer: (C) $5$

Concept: Valid votes = votes cast − blank votes; winner's + rival's votes = valid votes.

1. Let voters $=N$. Votes cast $=0.9N$; valid $=0.9N-60$
2. Winner $=0.46N$; rival $=0.46N-410$
3. $0.92N-410=0.9N-60\Rightarrow 0.02N=350\Rightarrow N=17500$

Trick: Check: $0.46\times 17500=8050$, rival $7640$, sum $15690=15750-60$ ✓.

Answer: (A) $17500$

Concept: Look down each month's column and check whether all four types (X, Y, Z, T) have the same value.

1. Jan: X = Y = Z = T = $25$ → all equal.
2. May: $20,30,30,30$ → X differs; March and June are clearly unequal.
3. Only January has equal production of all types ($4\times 25=100$).

Trick: If the total is divisible by 4, check that month first: Jan total $100=4\times 25$.

Answer: (A) January

Concept: Efficiency is inversely proportional to time; 4 times as fast means one-fourth the time.

1. Suresh: $20$ days, Arvind: $\frac{20}{4}=5$ days.
2. One day work together $=\frac{1}{5}+\frac{1}{20}=\frac{4+1}{20}=\frac{1}{4}$
3. Time together $=4$ days.

Trick: Efficiencies $4:1$, total work $=20\times 1=20$ units, together $\frac{20}{4+1}=4$ days.

Answer: (D) $4$

Concept: Arc length $l=\frac{\theta }{{360}^{\circ }}\times 2\pi r$.

1. $\frac{{72}^{\circ }}{{360}^{\circ }}=\frac{1}{5}$
2. $l=\frac{1}{5}\times 2\times \frac{22}{7}\times 35=\frac{1}{5}\times 220=44\text{cm}$

Trick: ${72}^{\circ }$ is one-fifth of the circle, so arc = circumference ÷ 5 $=220÷5$.

Answer: (C) $44\text{cm}$

Concept: Use $\beta =-(\alpha +\gamma )$ and expand.

1. ${\beta }^2={\alpha }^2+{\gamma }^2+2\alpha \gamma$
2. Numerator $=3({\alpha }^2+{\gamma }^2+2\alpha \gamma )+{\alpha }^2+{\gamma }^2=4{\alpha }^2+4{\gamma }^2+6\alpha \gamma$
3. Denominator $=2({\alpha }^2+{\gamma }^2+2\alpha \gamma )-\alpha \gamma =2{\alpha }^2+2{\gamma }^2+3\alpha \gamma$
4. Ratio $=\frac{2(2{\alpha }^2+2{\gamma }^2+3\alpha \gamma )}{2{\alpha }^2+2{\gamma }^2+3\alpha \gamma }=2$

Trick: Put $\alpha =1,\gamma =1,\beta =-2$: $\frac{12+1+1}{8-1}=\frac{14}{7}=2$.

Answer: (D) $2$

Concept: In a degree pie chart, amounts are proportional to angles, so the percentage can be found directly from degrees.

1. Tennis + Basketball $={35}^{\circ }+{40}^{\circ }={75}^{\circ }$
2. Hockey + Cricket $={94}^{\circ }+{135}^{\circ }={229}^{\circ }$
3. $\frac{75}{229}\times 100=32.75\%$ (the ₹21,000 for Football is not needed).

Trick: Compare angles, not rupees – the given amount is a distractor.

Answer: (D) $32.75\%$