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

Concept: Direct common tangent $=\sqrt{d^2-(r_1-r_2{)}^2}$; for circles touching externally $d=r_1+r_2$, so it becomes $2\sqrt{r_1{r}_2}$.

1. $r_1=18$, $r_2=8$, $d=AB=26\text{cm}$.
2. $CD=\sqrt{{26}^2-{10}^2}=\sqrt{676-100}=\sqrt{576}$.
3. $CD=24\text{cm}$ (same as $2\sqrt{18\times 8}=2\times 12$).

Trick: Touching circles: tangent $=2\sqrt{r_1{r}_2}=2\sqrt{144}=24$.

Answer: (C) $24\text{cm}$

Concept: $(r+1{)}^3=r^3+3r^2+3r+1$, so $r(r^2+3r+3)=(r+1{)}^3-1$.

1. $r=55\Rightarrow r+1=56$.
2. ${56}^3=175616$.
3. Value $=175616-1=175615$.

Trick: Spot the cube: $r^3+3r^2+3r$ is $(r+1{)}^3$ minus $1$. Option $175616$ is the trap.

Answer: (D) $175615$

Concept: Total work in man-days stays the same: $M_1{D}_1=M_2{D}_2$.

1. Let the original number be $M$: $M\times 20=(M-5)\times 40$.
2. $20M=40M-200\Rightarrow 20M=200$.
3. $M=10$.

Trick: Days doubled $\Rightarrow$ men halved; so the $5$ absent men are half the group: $M=10$.

Answer: (B) $10$

Concept: Add all 2020 (blue) bars, then add the 2021 Geography bar.

1. 2020 total $=75+100+60+125+75=435$.
2. Geography in 2021 $=75$.
3. Sum $=435+75=510$.

Trick: Option $435$ is only the 2020 total – remember to add Geography 2021.

Answer: (B) $510$

Concept: Divisible by both $2$ and $7$ means divisible by $\text{LCM}(2,7)=14$.

1. Count multiples of $14$ from $1$ to $1000$: $\lfloor \frac{1000}{14}\rfloor$.
2. $14\times 71=994\le 1000$, $14\times 72=1008>1000$.
3. Count $=71$.

Trick: Option $142$ counts multiples of $7$ only – "both" means LCM.

Answer: (B) $71$

Concept: Convert mixed fractions, then follow BODMAS (innermost bracket first).

1. $\frac{1}{2}-\frac{1}{7}=\frac{5}{14}$; $3\frac{1}{2}÷\frac{5}{14}=\frac{7}{2}\times \frac{14}{5}=\frac{49}{5}$.
2. $1\frac{2}{7}\times \frac{49}{5}=\frac{9}{7}\times \frac{49}{5}=\frac{63}{5}$.
3. $4\frac{1}{5}\times 1\frac{1}{2}=\frac{21}{5}\times \frac{3}{2}=\frac{63}{10}$.
4. $\frac{63}{5}÷\frac{63}{10}=2$.

Trick: Both parts reduce to $63$ over $5$ and $10$ – the ratio is $2$.

Answer: (C) $2$

Concept: Third proportional to $a$ and $b$ $=\frac{b^2}{a}$.

1. $a=0.024$, $b=0.12$.
2. $\frac{(0.12{)}^2}{0.024}=\frac{0.0144}{0.024}$.
3. $=0.6$, i.e. $0.600$.

Trick: $\frac{b}{a}=\frac{0.12}{0.024}=5$, so the third term $=b\times 5=0.6$ (a geometric sequence).

Answer: (C) $0.600$

Concept: For $x=a\cos \theta +b\sin \theta$ and $y=a\sin \theta -b\cos \theta$, squaring and adding removes $\theta$: $x^2+y^2=a^2+b^2$.

1. Directly, $x+y=(4-3)\cos \theta +(3+4)\sin \theta =\cos \theta +7\sin \theta$, which changes with $\theta$, so a fixed value comes only from $x^2+y^2$.
2. $x^2+y^2=16({\cos }^2\theta +{\sin }^2\theta )+9({\sin }^2\theta +{\cos }^2\theta )$ (the $24\sin \theta \cos \theta$ terms cancel).
3. $x^2+y^2=16+9=25$.

Trick: Same pair of coefficients $4,3$ in both expressions $\Rightarrow$ sum of squares $=4^2+3^2=25$.

Answer: (C) $25$

Concept: Solve numerator and denominator separately with BODMAS ("of" before ÷ and ×).

1. Numerator: $\frac{5}{8}-\{\frac{3}{8}-\frac{2}{8}\}=\frac{4}{8}=\frac{1}{2}$; $\frac{1}{2}$ of $8.8=4.4$; $4.4-1.2=3.2$.
2. Denominator: $\frac{1}{6}$ of $60=10$; $\frac{25}{6}÷2.5=\frac{5}{3}$; $\times 2=\frac{10}{3}$; $÷10=\frac{1}{3}$.
3. $\frac{3}{4}-\frac{3}{8}=\frac{3}{8}$; denominator $=\frac{1}{3}+\frac{3}{8}=\frac{17}{24}$.
4. Value $=3.2\times \frac{24}{17}=\frac{76.8}{17}=\frac{384}{85}=4\frac{44}{85}$.

Trick: $\frac{76.8}{17}\approx 4.52$; only $4\frac{44}{85}\approx 4.52$ matches.

Answer: (C) $4\frac{44}{85}$

Concept: Odd divisors use only the odd prime factors. Sum of divisors of $p^n$ $=1+p+\dots +p^n$.

1. $216=2^3\times 3^3$.
2. Odd divisors come from $3^3$ only: $1,3,9,27$.
3. Sum $=1+3+9+27=40$.

Trick: Ignore the power of $2$; sum of odd divisors $=\frac{3^4-1}{3-1}=\frac{80}{2}=40$. ($600$ is the sum of all divisors.)

Answer: (C) $40$

Concept: A number is divisible by 3 if its digit sum is divisible by 3, and by 11 if (sum of alternate digits) difference is $0$ or a multiple of $11$.

1. Let the number be $476ab0$ with $a$ = hundreds digit, $b$ = tens digit.
2. Test option (D) $a=8,b=5$: number $476850$; digit sum $=4+7+6+8+5+0=30$, divisible by $3$ ✓.
3. For 11 (from right): $(0+8+7)-(5+6+4)=15-15=0$ ✓.
4. Check (A) $476230$: sum $=22$, not divisible by $3$ ✗; (C) $476580$: $11$-test gives $(0+5+7)-(8+6+4)=-6$ ✗.

Trick: Option testing is fastest: first apply the 3-test (digit sum), then the 11-test on the survivors.

Answer: (D) 8 and 5

Concept: For a cylinder, $\text{CSA}=2\pi rh$ and $\text{TSA}=2\pi r(r+h)$.

1. $2\pi r\times 7=70\pi \Rightarrow r=5\text{cm}$.
2. $\text{TSA}=2\pi \times 5\times (5+7)=120\pi {\text{cm}}^2$.

Trick: $\text{TSA}=\text{CSA}+2\pi r^2=70\pi +50\pi =120\pi$.

Answer: (D) $120\pi {\text{cm}}^2$

Concept: Volume stays the same on recasting; cone $\text{CSA}=\pi rl$, $l=\sqrt{r^2+h^2}$.

1. Cylinder volume $=\pi \times 6^2\times 8=288\pi$; one cone (same $r,h$) $=\frac{1}{3}\times 288\pi =96\pi$, so exactly $3$ cones.
2. Slant height $l=\sqrt{6^2+8^2}=10\text{cm}$.
3. CSA of one cone $=\pi \times 6\times 10=60\pi$; total $=3\times 60\pi =180\pi {\text{cm}}^2$.

Trick: A cone with the same base and height is $\frac{1}{3}$ of the cylinder — hence 3 cones.

Answer: (A) $180\pi {\text{cm}}^2$

Concept: $V=\frac{1}{3}\pi r^{2}h$; perimeter (circumference) of base $=2\pi r$, so radii are in the same ratio as base perimeters.

1. $r_1:r_2=5:3$, $h_1:h_2=1:5$.
2. $V_1:V_2=r_1^2{h}_1:r_2^2{h}_2=25\times 1:9\times 5=25:45$.
3. $=5:9$.

Trick: Ratio of volumes $=(\text{radius ratio}{)}^2\times \text{height ratio}$.

Answer: (C) 5 : 9

Concept: Trains in opposite directions: relative speed $=$ sum of speeds; distance to clear $=$ sum of lengths.

1. Relative speed $=40+32=72\text{km/h}=72\times \frac{5}{18}=20\text{m/s}$.
2. Distance $=121+99=220\text{m}$.
3. Time $=\frac{220}{20}=11\text{s}$.

Trick: $\text{km/h}\to \text{m/s}$: multiply by $\frac{5}{18}$.

Answer: (A) 11 sec

Concept: In "buy 4 get 1 free" the shopkeeper hands over 5 articles but is paid for only 4.

1. Cost of 5 articles $=5\times 1200=₹6000$.
2. Money received $=4\times 1800=₹7200$.
3. Profit $\%=\frac{7200-6000}{6000}\times 100=20\%$.

Trick: Compare SP of 4 with CP of 5: $\frac{7200}{6000}=1.2\Rightarrow 20\%$.

Answer: (B) $20\%$

Concept: For 3 years: $\text{CI}-\text{SI}=P{\left(\frac{r}{100}\right)}^2(3+\frac{r}{100})$.

1. $P=15625$, $\frac{r}{100}=\frac{1}{25}$.
2. $\text{Diff}=15625\times \frac{1}{625}\times (3+\frac{1}{25})=25\times \frac{76}{25}$.
3. $=76$, i.e. Rs. 76.

Trick: $15625={25}^3$, so with $r=4\%$ (i.e. $\frac{1}{25}$) the numbers cancel neatly.

Answer: (A) Rs.76

Concept: A continuous increase means every year’s value is higher than the previous year’s.

1. Barley: $2500\to 4400\to 5200\to 5800\to 6100$ — rises every year ✓.
2. Rice falls in 2018 ($8000\to 6800$); Wheat falls in 2018 ($4200\to 3500$); Others falls in 2017 ($4500\to 3000$).

Trick: Look for the first drop in each column — eliminate it at once.

Answer: (A) Barley

Concept: $\overline{x}=\frac{\sum fx}{\sum f}$, where $x$ is the class mid-value.

1. Mid-values: $5,15,25,35,45,55,65$; $\sum f=2+4+12+21+6+3+2=50$.
2. $\sum fx=10+60+300+735+270+165+130=1670$.
3. $\overline{x}=\frac{1670}{50}=33.4$.

Trick: Assumed mean $35$: deviations $\times f$ give $-60-80-120+0+60+60+60=-80$; $35-\frac{80}{50}=33.4$.

Answer: (A) 33.4

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

1. Time becomes $\frac{4}{5}$ of original (20% less).
2. Speed becomes $\frac{5}{4}$ of original.
3. Increase $=\frac{1}{4}\times 100=25\%$.

Trick: Decrease of $\frac{1}{5}$ in time $\Rightarrow$ increase of $\frac{1}{4}$ in speed.

Answer: (A) $25\%$