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

Concept: Savings = Income − Expenditure; take income $=100$.

1. Old: income $100$, expenditure $65$, savings $35$.
2. New: income $120$, expenditure $65\times 1.1=71.5$, savings $48.5$.
3. Increase $=\frac{48.5-35}{35}\times 100=\frac{13.5}{35}\times 100\approx 38.57\%$, i.e. about $38.5\%$.

Trick: Increase in savings $=20-6.5=13.5$ (rise in income − rise in expenditure) on a base of $35$.

Answer: (A) $38.5\%$

Concept: Arc length $l=\frac{\theta }{{360}^{\circ }}\times 2\pi r\Rightarrow \theta =\frac{l}{r}\times \frac{{180}^{\circ }}{\pi }$.

1. $l=19\text{cm}$, $r=30\text{cm}$, $\pi =\frac{22}{7}$.
2. $\theta =\frac{19}{30}\times 180\times \frac{7}{22}=\frac{19\times 6\times 7}{22}=\frac{798}{22}$.
3. $\theta ={36.27}^{\circ }$ (approx.).

Trick: $1$ radian $\approx {57.27}^{\circ }$ (with $\pi =\frac{22}{7}$); $\frac{19}{30}\times 57.27\approx {36.27}^{\circ }$.

Answer: (C) $36.27$

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a-b{)}^2={10}^2-4\times 9=100-36=64$.
2. $a-b=\sqrt{64}=8$ (taking the positive value).
3. Check: $a=9,b=1$ gives $a+b=10$, $ab=9$, $a-b=8$.

Trick: Guess two numbers with sum $10$ and product $9$: $9$ and $1$, so $a-b=8$.

Answer: (C) $8$

Concept: Production $\propto$ working hours (manpower fixed at the new level). To restore after an $r\%$ fall, increase by $\frac{r}{100-r}\times 100\%$.

1. Production falls from $100$ to $65$.
2. Required increase $=\frac{35}{65}\times 100=\frac{700}{13}\%$.
3. $=53.85\%$ (approx.).

Trick: Divide the fall by what remains: $\frac{35}{65}=\frac{7}{13}\approx 0.5385$.

Answer: (C) $53.85\%$

Concept: Write each number as a sum of two perfect squares and find the common term $b^2$.

1. $82=81+1$, so $\{a^2,b^2\}=\{81,1\}$.
2. $65=64+1=49+16$. The common square must be $b^2=1\Rightarrow b=1$.
3. Then $a^2=81\Rightarrow a=9$ and $c^2=64\Rightarrow c=8$.
4. $2a+7b-3c=18+7-24=1$.

Trick: If $b=9$, then $c^2=65-81<0$ – impossible, so $b=1$.

Answer: (D) $1$

Concept: Trains moving towards each other close the gap at the sum of their speeds; the gap just before meeting = relative speed × time left.

1. Relative speed $=80+120=200\text{km/h}$.
2. $10$ minutes $=\frac{1}{6}$ hour.
3. Distance $10$ min before crossing $=200\times \frac{1}{6}=33.33\text{km}$.

Trick: The initial $500\text{km}$ is not needed – the gap just before meeting depends only on relative speed and the time left.

Answer: (B) $33.33\text{km}$

Concept: $\sin ({90}^{\circ }-\theta )=\cos \theta$ and $\sec ({90}^{\circ }-\theta )=\csc \theta$.

1. ${54}^{\circ }+{36}^{\circ }={90}^{\circ }\Rightarrow \sin {54}^{\circ }=\cos {36}^{\circ }$, so $\frac{\sin {54}^{\circ }}{\cos {36}^{\circ }}=1$.
2. ${46}^{\circ }+{44}^{\circ }={90}^{\circ }\Rightarrow \sec {46}^{\circ }=\csc {44}^{\circ }$, so $\frac{\sec {46}^{\circ }}{\csc {44}^{\circ }}=1$.
3. Sum $=1+1=2$.

Trick: When the two angles add up to ${90}^{\circ }$ and the functions are co-functions, each ratio is $1$.

Answer: (C) $2$

Concept: "Beats by 200 m or 20 s" means the loser covers the last $200\text{m}$ in $20\text{s}$.

1. Shyam's speed $=\frac{200}{20}=10\text{m/s}$.
2. When Ram finishes $1200\text{m}$, Shyam has run $1000\text{m}$, taking $\frac{1000}{10}=100\text{s}$.
3. Ram's speed $=\frac{1200}{100}=12\text{m/s}$.

Trick: Speeds are in the ratio of distances covered in the same time: Ram : Shyam $=1200:1000=6:5$, so Ram $=\frac{6}{5}\times 10=12$.

Answer: (B) $12\text{m/sec}$

Concept: In a rhombus the diagonals bisect at right angles: ${\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2={\text{side}}^2$; area $=\frac{1}{2}{d}_1{d}_2$.

1. Perimeter $4P\Rightarrow$ side $=P$, so $d_1^2+d_2^2=4P^2$.
2. $(d_1+d_2{)}^2=L^2\Rightarrow 2d_1{d}_2=L^2-4P^2$.
3. Area $=\frac{1}{2}{d}_1{d}_2=\frac{1}{4}(L^2-4P^2)$.

Trick: Test a square of side $1$: $P=1$, $L=2\sqrt{2}$, area $=1$; $\frac{1}{4}(8-4)=1$ ✓.

Answer: (B) $\frac{1}{4}(L^2-4P^2)$

Concept: A number is divisible by $8$ if the number formed by its last three digits is divisible by $8$.

1. Last three digits of $246p48$ are $p48$.
2. For $p=0$: $048=48=8\times 6$ – divisible.
3. $0$ is the least possible digit, so $p=0$ (also $p=2,4,6,8$ work).

Trick: $48$ is already a multiple of $8$ and $200$ is too, so any even $p$ works – the least is $0$.

Answer: (D) $0$

Concept: HCF = product of the prime factors common to all numbers (lowest powers).

1. $69=3\times 23$, $96=2^5\times 3$, $99=3^2\times 11$.
2. Only $3$ is common to all three.
3. HCF $=3$.

Trick: HCF divides the differences: $96-69=27$, $99-96=3$; so HCF divides $3$, and $3$ divides all three numbers.

Answer: (A) $3$

Concept: Take convenient values and compare the combined totals.

1. Let Manju $=100$, then Raju $=125$; combined $=225$.
2. New: Raju $=125\times 1.25=156.25$, Manju $=100\times 1.40=140$; combined $=296.25$.
3. Increase $=\frac{71.25}{225}\times 100=31.67\%$.

Trick: Weighted average: $\frac{125\times 25+100\times 40}{225}=\frac{7125}{225}=31.67\%$.

Answer: (B) $31.67\%$

Concept: Add each column for the whole country, then take the ratio.

1. Rice $=55+58+69+51+47+78+67+48=473$ lakh tons.
2. Wheat $=113+96+42+47+32+25+18+38=411$ lakh tons.
3. Rice : Wheat $=473:411$.

Trick: Only one option begins with $473$; checking the rice total alone is enough.

Answer: (B) $473:411$

Concept: Population after $n$ years at $r\%$ decrease $=P{(1-\frac{r}{100})}^n$.

1. $P=45000$, $r=2$, $n=2$.
2. Population $=45000\times (0.98{)}^2=45000\times 0.9604$.
3. $=43218$.

Trick: Two successive $2\%$ falls $=-2-2+\frac{4}{100}=-3.96\%$; $3.96\%$ of $45000=1782$; $45000-1782=43218$.

Answer: (D) $43218$

Concept: If $x+\frac{1}{x}=k$, then $x^3+\frac{1}{x^3}=k^3-3k$.

1. $x(x-5)=-1\Rightarrow x^2-5x+1=0\Rightarrow x+\frac{1}{x}=5$.
2. $x^3+\frac{1}{x^3}=125-15=110$.
3. Multiply by $x^3$: $x^6+1=110x^3\Rightarrow x^6-110x^3=-1$.
4. So $x^3(x^3-110)=-1$.

Trick: The pattern $x^3(x^3-k)$ with $k=x^3+\frac{1}{x^3}$ always equals $-1$.

Answer: (B) $-1$

Concept: Angles on a straight line add to ${180}^{\circ }$; angles opposite equal sides are equal.

1. BCD is a straight line: $\angle ACB={180}^{\circ }-{125}^{\circ }={55}^{\circ }$.
2. $AB=AC\Rightarrow \angle ABC=\angle ACB={55}^{\circ }$.
3. $\angle BAC={180}^{\circ }-{55}^{\circ }-{55}^{\circ }={70}^{\circ }$.

Trick: Exterior angle $=\angle A+\angle B$: ${125}^{\circ }=\angle A+{55}^{\circ }\Rightarrow \angle A={70}^{\circ }$.

Answer: (D) ${70}^{\circ }$

Concept: Volume of a cylinder $=\pi r^{2}h$; with $h=r$ it becomes $\pi r^3$.

1. Capacity $=64\pi$ litres $=64\pi \times 1000=64000\pi {\text{cm}}^3$.
2. $\pi r^3=64000\pi \Rightarrow r^3=64000$.
3. $r=\sqrt[3]{64000}=40\text{cm}$.

Trick: Convert litres to ${\text{cm}}^3$ first – skipping it gives the wrong $4\text{cm}$.

Answer: (B) $40\text{cm}$

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$. The larger effective discount is better for the customer.

1. I: $25+15-\frac{375}{100}=36.25\%$.
2. II: $35+5-\frac{175}{100}=38.25\%$.
3. III: $38\%$.
4. Largest is II ($38.25\%$).

Trick: For the same total ($40$), the more unequal pair loses less in the $\frac{ab}{100}$ term, so II beats I.

Answer: (C) Only II

Concept: Percentage increase $=\frac{\text{GK}-\text{Science}}{\text{Science}}\times 100$ (Science is the base).

1. GK total $=182+200+120+130+183=815$.
2. Science total $=100+100+150+200+80=630$.
3. Increase $=\frac{815-630}{630}\times 100=\frac{185}{630}\times 100=29.37\%$.

Trick: Dividing by the GK total ($815$) gives $22.7\%$, which is not offered – the base must be Science.

Answer: (A) $29.37\%$

Concept: LCM = product of all prime factors with highest powers.

1. $28=2^2\times 7$, $92=2^2\times 23$.
2. LCM $=2^2\times 7\times 23=644$.
3. Check: $\text{HCF}\times \text{LCM}=4\times 644=2576=28\times 92$ ✓.

Trick: $\text{LCM}=\frac{28\times 92}{\text{HCF}}=\frac{2576}{4}=644$.

Answer: (B) $644$