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

Concept: If $a,b,c,d$ are in proportion, $a:b=c:d$, so the fourth proportional $d=\frac{b\times c}{a}$.

1. $22:66=11:d$
2. $d=\frac{66\times 11}{22}=3\times 11=33$

Trick: $66$ is $3$ times $22$, so $d$ is $3$ times $11$.

Answer: (B) $33$

Concept: With successive discounts, the selling price $=\text{MP}\times (1-d_1)(1-d_2)$.

1. After $25\%$ and $28\%$: $\text{SP}=\text{MP}\times \frac{75}{100}\times \frac{72}{100}=0.54\text{MP}$
2. $0.54\text{MP}=10800$
3. $\text{MP}=\frac{10800}{0.54}=20000$

Trick: Net discount $=25+28-\frac{25\times 28}{100}=46\%$, so ₹10,800 is $54\%$ of MP.

Answer: (C) ₹20,000

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)(1-d_3)$.

1. $\text{SP}=15000\times \frac{95}{100}\times \frac{90}{100}\times \frac{80}{100}$
2. $15000\times 0.95=14250$; $14250\times 0.9=12825$; $12825\times 0.8=10260$

Trick: Apply each discount on the new price, never add them ($35\%$ off would give ₹9,750, a trap).

Answer: (A) ₹10,260

Concept: Depreciated value $=P{(1-\frac{r}{100})}^n$.

1. ${(1-\frac{r}{100})}^3=\frac{21970}{40960}=\frac{2197}{4096}={\left(\frac{13}{16}\right)}^3$
2. $1-\frac{r}{100}=\frac{13}{16}\Rightarrow \frac{r}{100}=\frac{3}{16}$
3. $r=\frac{300}{16}=18.75\%$

Trick: Spot cubes: ${13}^3=2197$, ${16}^3=4096$.

Answer: (C) $18.75\%$

Concept: Each worker does $1$ unit a day; the whole work $=36$ units (one person alone needs $36$ days). A rests every 6th day, B every 8th day.

1. In days 1 to 20, A is off on days $6,12,18$ ($17$ units) and B on days $8,16$ ($18$ units): total $35$ units.
2. On day 21 both work, giving $37\ge 36$ units – work ends on day 21.
3. Day 1 = 1 July, so day 21 = 21 July 2021.

Trick: Without holidays the pair needs $18$ days; add the lost days ($5$ in all) and stop as soon as $36$ units are reached.

Answer: (C) 21 July 2021

Concept: When metal is recast, total volume stays the same: $x=\frac{\text{volume of sphere}}{\text{volume of one hemisphere}}$.

1. Sphere: $\frac{4}{3}\pi (3{)}^3=36\pi$
2. Hemisphere ($r=1.5$): $\frac{2}{3}\pi (1.5{)}^3=2.25\pi$
3. $x=\frac{36\pi }{2.25\pi }=16$

Trick: Halving the radius gives $\frac{1}{8}$ the volume, and a hemisphere is half of that: $x=8\times 2=16$.

Answer: (C) $16$

Concept: Volume of the submerged ice balls = volume of the juice that rises in the cylinder.

1. Rise volume $=\pi (9{)}^2\times 40=3240\pi$
2. One ball ($r=3$): $\frac{4}{3}\pi (3{)}^3=36\pi$
3. Number $=\frac{3240\pi }{36\pi }=90$

Trick: Work with radii (9 and 3), not diameters; $\pi$ cancels.

Answer: (D) $90$

Concept: For a cone $r=\sqrt{l^2-h^2}$, $\text{TSA}=\pi r(l+r)$ and $V=\frac{1}{3}\pi r^{2}h$.

1. $r=\sqrt{{29}^2-{20}^2}=\sqrt{441}=21$
2. $\text{TSA}=\pi \times 21\times 50=1050\pi$
3. $V=\frac{1}{3}\pi \times 441\times 20=2940\pi$
4. $1050:2940=5:14$

Trick: Ratio shortcut: $\frac{\text{TSA}}{V}=\frac{3(l+r)}{rh}=\frac{3\times 50}{21\times 20}=\frac{5}{14}$.

Answer: (A) $5:14$

Concept: The largest angle is opposite the largest side ($12$). Its bisector divides that side in the ratio of the other two sides ($8:9$).

1. $BD:DC=8:9$ with $BD+DC=12$
2. Shorter segment $BD=\frac{8}{17}\times 12=\frac{96}{17}$
3. $\frac{96}{17}=5\frac{11}{17}\text{cm}$

Trick: Shorter segment lies next to the shorter adjacent side: $\frac{\text{smaller side}}{\text{sum}}\times \text{opposite side}$.

Answer: (A) $5\frac{11}{17}\text{cm}$

Concept: Write ${\cot }^3\theta \sec \theta ={\cot }^2\theta ·\frac{\cos \theta }{\sin \theta }·\frac{1}{\cos \theta }={\cot }^2\theta \text{cosec}\theta$, then use $1+{\cot }^2\theta ={\text{cosec}}^2\theta$.

1. $\text{cosec}\theta +{\cot }^3\theta \sec \theta =\text{cosec}\theta (1+{\cot }^2\theta )={\text{cosec}}^3\theta$
2. ${\text{cosec}}^2\theta =1+{\cot }^2\theta =1+1-e^2=2-e^2$
3. ${\text{cosec}}^3\theta =(2-e^2{)}^{\frac{3}{2}}$

Trick: Whenever $\cot \theta \sec \theta$ appears, it is simply $\text{cosec}\theta$.

Answer: (D) $(2-e^2{)}^{\frac{3}{2}}$

Concept: Cross-multiply to get a relation between $\sin A$ and $\cos A$, then divide to get $\cot A$.

1. $2\sin A-\cos A=\sin A+\cos A$
2. $\sin A=2\cos A$
3. $\cot A=\frac{\cos A}{\sin A}=\frac{1}{2}$

Trick: Ratio $=1$ means numerator = denominator – just equate them.

Answer: (B) $\frac{1}{2}$

Concept: Take total work $=\text{LCM}(11,20,55)=220$ units. Efficiencies: A $=20$, B $=11$, C $=4$ units/day.

1. Day 1 (A+B): $31$ units; Day 2 (A+C): $24$ units
2. Every 2 days: $55$ units
3. $\frac{220}{55}=4$ cycles $=8$ days, work finishes exactly.

Trick: Check if the work divides exactly by the 2-day total – then no fractional day is needed.

Answer: (C) 8 days

Concept: Average $=\frac{\sum (\text{children}\times \text{families})}{\sum \text{families}}$.

1. Children: $0\times 4+1\times 22+2\times 15+3\times 6+4\times 2=0+22+30+18+8=78$
2. Families: $4+22+15+6+2=49$
3. Average $=\frac{78}{49}\approx 1.59$

Trick: Always divide by the number of families, not by the number of rows (5).

Answer: (B) $1.59$

Concept: Time difference $=\frac{D}{v}-\frac{D}{v+10}$, with $45$ min $=\frac{3}{4}$ h.

1. $\frac{150}{v}-\frac{150}{v+10}=\frac{3}{4}$
2. $\frac{1500}{v(v+10)}=\frac{3}{4}\Rightarrow v(v+10)=2000$
3. $v=40$ (since $40\times 50=2000$)

Trick: Check options: $v=40$ gives $\frac{150}{40}-\frac{150}{50}=3.75-3=0.75$ h ✓.

Answer: (D) $40$

Concept: Split the total in the given ratio, then make the new totals satisfy the new ratio.

1. Boys $=\frac{3}{8}\times 720=270$, girls $=450$
2. Girls now $=450+18=468$
3. Required boys $=\frac{2}{3}\times 468=312$
4. New boys $=312-270=42$

Trick: Fix the side whose change is known (girls) first.

Answer: (A) $42$

Concept: Percentage increase $=\frac{\text{new}-\text{old}}{\text{old}}\times 100$.

1. Expenditure 2017 $=80$ thousand, 2018 $=100$ thousand
2. Increase $=20$
3. $\frac{20}{80}\times 100=25\%$

Trick: Divide by the earlier year's value; $\frac{20}{100}=20\%$ is the common trap.

Answer: (A) $25\%$

Concept: A number divisible by all of them is a multiple of their LCM.

1. $\text{LCM}(15,25,40,75)=2^3\times 3\times 5^2=600$
2. $9999÷600=16$ remainder $399$
3. Greatest 4-digit multiple $=9999-399=9600$

Trick: Check options: only $9600=600\times 16$.

Answer: (C) $9600$

Concept: First six primes: $2,3,5,7,11,13$ ($1$ is not prime).

1. Sum $=2+3+5+7+11+13=41$
2. Average $=\frac{41}{6}=6\frac{5}{6}$

Trick: Remember $2$ is the only even prime – don't miss it.

Answer: (B) $6\frac{5}{6}$

Concept: Find the percentage left as savings, get the total, then take the required percentage.

1. Spent $=25+26+23+19=93\%$, saved $=7\%$
2. $7\%=26600\Rightarrow 1\%=3800$
3. Food + recreation $=45\%=45\times 3800=171000$

Trick: Find $1\%$ once and multiply.

Answer: (B) 1,71,000

Concept: Amount for a sector $=\frac{\text{angle}}{{360}^{\circ }}\times \text{total}$.

1. Housing $={79.2}^{\circ }$
2. $\frac{79.2}{360}=0.22$
3. $0.22\times 56800=12496$

Trick: ${79.2}^{\circ }$ is exactly $22\%$ of the circle (${3.6}^{\circ }=1\%$).

Answer: (D) ₹12,496