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

Concept: $x^3+y^3=(x+y)(x^2-xy+y^2)$.

1. $\frac{x}{y}+\frac{y}{x}=1\Rightarrow x^2+y^2=xy$
2. So $x^2-xy+y^2=0$
3. $x^3+y^3=2\times 0=0$

Trick: $\frac{x}{y}+\frac{y}{x}=1$ always gives $x^3+y^3=0$, whatever $x+y$ is.

Answer: (C) 0

Concept: Volume of cone $=\frac{1}{3}\pi r^{2}h$.

1. $r=\frac{12}{2}=6\text{cm}$, $h=7\text{cm}$
2. $V=\frac{1}{3}\times \frac{22}{7}\times 36\times 7$
3. $=\frac{22\times 36}{3}=264{\text{cm}}^3$

Trick: $h=7$ cancels the $7$ of $\frac{22}{7}$: $V=\frac{22\times 36}{3}$.

Answer: (C) $264{\text{cm}}^3$

Concept: Use total work = LCM of days and find work done in one repeating cycle.

1. Work $=60$ units; A $=3$, B $=2$, C $=1$ unit/day
2. 3-day cycle: $3+3+(3+2+1)=12$ units
3. Cycles $=\frac{60}{12}=5$
4. Days $=5\times 3=15$

Trick: Work finishes exactly at the end of a cycle, so no leftover check is needed.

Answer: (D) 15

Concept: Tourists from A in a month $=\text{Total}\times \text{share}\times \frac{A}{A+B}$.

1. March: $60000\times 20\%=12000$; A $=12000\times \frac{3}{5}=7200$
2. April: $60000\times 25\%=15000$; A $=15000\times \frac{8}{15}=8000$
3. Ratio $=7200:8000=9:10$

Trick: Skip the total: $20\times \frac{3}{5}:25\times \frac{8}{15}=12:\frac{40}{3}=9:10$.

Answer: (D) $9:10$

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

1. $11\times 11=121$, $11÷11=1$
2. Expression $=121+1-121+11+121-11-121$
3. $=1$

Trick: The $121$s and $11$s cancel in pairs, leaving only $1$.

Answer: (A) 1

Concept: Savings $\%=100\%-\text{Expenditure}\%$.

1. Savings $=100-72=28\%$ of monthly salary $=$ ₹$5740$
2. Monthly salary $=\frac{5740}{0.28}=$ ₹$20500$
3. Annual salary $=20500\times 12=$ ₹$246000$
4. $10\%$ of it $=$ ₹$24600$

Trick: $10\%$ of annual $=1.2\times$ monthly salary $=1.2\times 20500$.

Answer: (B) ₹24600

Concept: Population $n$ years ago $=\frac{\text{Present}}{{(1+\frac{r}{100})}^n}$.

1. ${\left(1.15\right)}^2=1.3225$
2. Population 2 years ago $=\frac{352314}{1.3225}$
3. $=266400$

Trick: $1.15=\frac{23}{20}$, so divide by $529$ and multiply by $400$: $\frac{352314}{529}=666$, $666\times 400=266400$.

Answer: (C) 2,66,400

Concept: For equal distances, average speed $=\frac{2xy}{x+y}$.

1. $x=36,y=60$
2. Average $=\frac{2\times 36\times 60}{36+60}=\frac{4320}{96}$
3. $=45\text{km/h}$

Trick: The distance ($35\text{km}$) does not matter when both legs are equal.

Answer: (C) 45

Concept: If numbers are in ratio $a:b$ (co-prime), they are $a\times \text{HCF}$ and $b\times \text{HCF}$.

1. Numbers $=5\times 5=25$ and $7\times 5=35$
2. Check: LCM$(25,35)=175$ ✓
3. Larger number $=35$

Trick: Larger $=$ larger ratio term $\times$ HCF $=7\times 5$.

Answer: (A) 35

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. $726=11\times \text{LCM}$
2. LCM $=\frac{726}{11}=66$

Trick: LCM must be a multiple of HCF $11$ — only $66$ is.

Answer: (C) 66

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$; for two equal distances it is $\frac{2xy}{x+y}$.

1. $\frac{2\times 30\times 45}{30+45}=\frac{2700}{75}$
2. $=36\text{km/h}$
3. Check: time $=\frac{1}{30}+\frac{1}{45}=\frac{1}{18}\text{h}$; $\frac{2}{1/18}=36$

Trick: Average for equal distances is the harmonic mean — always less than the simple average $37.5$.

Answer: (B) $36\text{km/h}$

Concept: Let the job take $T$ days; each person's work = efficiency × days worked; total = whole work.

1. Work $=$ LCM$(36,48,144)=144$; P $=4$, Q $=3$, R $=1$ unit/day
2. $4(T-12)+3(T-8)+1\times T=144$
3. $8T-72=144\Rightarrow T=27$

Trick: Add back the missed work: $(144+4\times 12+3\times 8)÷(4+3+1)=216÷8=27$.

Answer: (B) 27

Concept: $a^3+b^3=(a+b{)}^3-3ab(a+b)$.

1. $(a+b{)}^3={10}^3=1000$
2. $3ab(a+b)=3\times 6\times 10=180$
3. $a^3+b^3=1000-180=820$

Trick: Cube of sum minus $3\times$product$\times$sum.

Answer: (B) 820

Concept: Express $\sin \theta$ in terms of $\cos \theta$ and use $(\sqrt{2}+1)(\sqrt{2}-1)=1$.

1. $\cos \theta =(\sqrt{2}+1)\sin \theta \Rightarrow \sin \theta =\frac{\cos \theta }{\sqrt{2}+1}=(\sqrt{2}-1)\cos \theta$
2. $\cos \theta +\sin \theta =\cos \theta +(\sqrt{2}-1)\cos \theta$
3. $=\sqrt{2}\cos \theta$

Trick: Standard pair: if $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$ then $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$.

Answer: (B) $\sqrt{2}\cos \theta$

Concept: $5:12:13$ is a Pythagorean triplet, so the triangle is right-angled.

1. $5k+12k+13k=30k=90\Rightarrow k=3$
2. Sides $15,36,39$ (right angle between $15$ and $36$)
3. Area $=\frac{1}{2}\times 15\times 36=270{\text{cm}}^2$

Trick: Area of $5$–$12$–$13$ triangle $=30k^2=30\times 9$.

Answer: (B) 270

Concept: Central angles are proportional to percentages, so compare percentages directly.

1. Maths $=20\%$, Science $=30\%$
2. Ratio $=20:30=2:3$

Trick: No need to convert to degrees (${72}^{\circ }:{108}^{\circ }$ gives the same $2:3$).

Answer: (B) $2:3$

Concept: Failed $=\text{Total}\times \frac{\text{Failing}}{\text{Passing}+\text{Failing}}$.

1. City D: total $=2.35$ lakh, ratio $4:6$
2. Failed $=2.35\times \frac{6}{10}=1.41$ lakh

Trick: $\frac{6}{10}=60\%$; $60\%$ of $2.35=1.41$.

Answer: (D) 1.41

Concept: The line joining the centres is the perpendicular bisector of the common chord.

1. Half chord $AM=6\text{cm}$
2. $O_{1}M=\sqrt{{13}^2-6^2}=\sqrt{133}$
3. $O_{2}M=\sqrt{{15}^2-6^2}=\sqrt{189}$
4. $O_1{O}_2=\sqrt{133}+\sqrt{189}\text{cm}$

Trick: Distance $=\sqrt{r_1^2-{\left(\frac{c}{2}\right)}^2}+\sqrt{r_2^2-{\left(\frac{c}{2}\right)}^2}$.

Answer: (D) $\sqrt{133}+\sqrt{189}$

Concept: Percentage $=\frac{\text{Part}}{\text{Total students of that year}}\times 100$ (total includes the failed students).

1. Total in 2010 $=40+60+70+30=200$
2. 2nd division $=60$
3. $\frac{60}{200}\times 100=30\%$

Trick: Do not leave out the 'Fail' bar while finding the total.

Answer: (C) $30\%$

Concept: Write the numbers as $5k$ and $3k$.

1. $5k-3k=2k=34\Rightarrow k=17$
2. Smaller $=3\times 17=51$

Trick: Difference of ratio terms $=2$ parts $=34$, so $1$ part $=17$.

Answer: (A) 51