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

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) $25\%$

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. $8+6+y+5=19+y$.
2. Smallest $y$ making $19+y$ a multiple of 3: $y=2$ (sum $21$).
3. $y=0$ gives $19$, $y=1$ gives $20$ — not divisible.

Trick: $19$ leaves remainder $1$ on division by $3$; add $2$.

Answer: (D) 2

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

1. Expression $={\csc }^2\theta \times {\sin }^2\theta$.
2. $=\frac{1}{{\sin }^2\theta }\times {\sin }^2\theta =1$.

Trick: Put $\theta ={90}^{\circ }$: $(1+0)(1-0)(1+0)=1$.

Answer: (A) 1

Concept: Length of direct common tangent $=\sqrt{d^2-(r_1-r_2{)}^2}$.

1. Draw $O_{2}C\parallel AB$ meeting $O_{1}A$ at $C$; then $O_{1}C=8-3=5$.
2. In right $△O_{1}C{O}_2$: $AB=O_{2}C=\sqrt{{13}^2-5^2}=\sqrt{144}$.
3. $AB=12\text{cm}$.

Yaad rakhein: Direct tangent uses $(r_1-r_2)$; transverse tangent uses $(r_1+r_2)$. $5,12,13$ is a Pythagorean triple.

Answer: (B) 12 cm

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

1. New radius $=1.1r$; new area factor $={1.1}^2=1.21$.
2. Increase $=21\%$.

Trick: Successive change: $10+10+\frac{10\times 10}{100}=21\%$.

Answer: (B) $21\%$

Concept: Read the Activated Carbon row and find equal values.

1. Activated Carbon: April $250$, May $260$, June $250$, July $330$, August $300$, September $350$.
2. April and June both $=250$ million.

Trick: Scan only the asked row; no calculation needed.

Answer: (A) April, June

Concept: In a cyclic quadrilateral, an exterior angle equals the interior opposite angle; opposite angles sum to ${180}^{\circ }$.

1. Let $\angle SMN=x$ and $\angle QNM=y$. In cyclic $MSQN$: $\angle MSQ={180}^{\circ }-y$.
2. $△RMN$: ${68}^{\circ }+x+y={180}^{\circ }$ … (1)
3. $△MSP$: ${54}^{\circ }+x+({180}^{\circ }-y)={180}^{\circ }\Rightarrow y=x+{54}^{\circ }$ … (2)
4. From (1): $68+x+x+54=180\Rightarrow 2x=58\Rightarrow x={29}^{\circ }$.

Trick: For this figure, $\angle M=\frac{{180}^{\circ }-(\angle R+\angle P)}{2}=\frac{180-122}{2}={29}^{\circ }$.

Answer: (A) ${29}^{\circ }$

Concept: On a circular track the faster runner meets the slower one each time he gains one full lap.

1. Speeds: A $=\frac{1200}{300}=4\text{m/s}$, B $=\frac{1200}{400}=3\text{m/s}$; gain $=1\text{m/s}$.
2. One meeting every $\frac{1200}{1}=1200\text{s}$.
3. A finishes $12\text{km}$ (10 laps) in $10\times 300=3000\text{s}$; meetings within this: $1200,2400\text{s}$.
4. Last meeting at $2400\text{s}$.

Trick: Last meeting $=\lfloor \frac{3000}{1200}\rfloor \times 1200=2400\text{s}$.

Answer: (A) 2400 seconds

Concept: For similar triangles, ratio of perimeters $=$ ratio of sides, and ratio of areas $=$ (ratio of sides)${}^2$.

1. Sides ratio $=36:24=3:2$.
2. Areas ratio $=3^2:2^2=9:4$.

Yaad rakhein: Perimeter, median, altitude all follow side ratio; area follows its square.

Answer: (C) 9 : 4

Concept: Compare the cost and selling price of the same quantity sold.

1. CP of one notebook $=\frac{700}{35}=₹20$; CP of 28 $=₹560$.
2. SP of 28 $=₹600$; gain $=₹40$.
3. Gain $\%=\frac{40}{560}\times 100=\frac{50}{7}=7\frac{1}{7}\%$.

Trick: $\frac{40}{560}=\frac{1}{14}$ and $\frac{1}{14}=7\frac{1}{7}\%$.

Answer: (A) $7\frac{1}{7}\%$

Concept: Required number $=$ (largest multiple of LCM within limit) $+$ remainder.

1. $\text{LCM}(16,24,72,84)=2^4\times 3^2\times 7=1008$.
2. $999999÷1008$ leaves remainder $63$; largest 6-digit multiple $=999999-63=999936$.
3. Required number $=999936+15=999951$.

Trick: Check: $999951-15=999936=1008\times 992$.

Answer: (B) 999951