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

Concept: Percentage more $=\frac{\text{difference}}{\text{base}}\times 100$, base = the class compared with.

1. Cars at 70–80 $\text{km/h}$ $=90$; at 50–60 $\text{km/h}$ $=70$
2. $\frac{90-70}{70}\times 100=\frac{200}{7}\approx 28.6\%$

Trick: $\frac{2}{7}=28.57\%$ – remember the sevenths.

Answer: (C) $28.6\%$

Concept: Order: Brackets, then 'of', then $÷$, $\times$, then $+$.

1. $2\frac{1}{2}\text{ of }1\frac{3}{5}=\frac{5}{2}\times \frac{8}{5}=4$
2. $3\frac{1}{3}÷4=\frac{10}{3}\times \frac{1}{4}=\frac{5}{6}$
3. Bracket: $\frac{3}{8}+\frac{1}{7}\times \frac{7}{4}=\frac{3}{8}+\frac{2}{8}=\frac{5}{8}$
4. Total $=\frac{5}{6}+\frac{5}{8}=\frac{20+15}{24}=\frac{35}{24}$

Yaad rakhein: 'of' is done before $÷$.

Answer: (B) $\frac{35}{24}$

Concept: If each of $n$ students gives ₹$n$, total $=n^2$.

1. $n^2=2209$
2. $n=\sqrt{2209}=47$ (since ${47}^2=2209$)

Trick: Unit digit 9 → root ends in 3 or 7; ${47}^2=2209$ fits.

Answer: (A) $47$

Concept: Express all in terms of C and use sum $=3\times \text{average}$.

1. $A=\frac{C}{6}$, $B=2A=\frac{C}{3}$
2. $\frac{C}{6}+\frac{C}{3}+C=90\Rightarrow \frac{3C}{2}=90\Rightarrow C=60$
3. $A=10$; $C-A=50$

Trick: Ratio $A:B:C=1:2:6$, 9 parts $=90$ → 1 part $=10$; difference $=5$ parts $=50$.

Answer: (C) $50$

Concept: Distance in one revolution $=$ circumference $=\pi d$.

1. $\pi d=\frac{22}{7}\times 42=132\text{cm}$
2. 12 revolutions $=132\times 12=1584\text{cm}=15.84\text{m}$

Yaad rakhein: Distance $=$ revolutions $\times \pi d$.

Answer: (C) $15.84$

Concept: Write all shares in terms of one share and equate to the total.

1. Let B $=x$; A $=x+500$; C $=A+300=x+800$
2. $3x+1300=8200\Rightarrow x=2300$
3. A $=2300+500=2800$

Trick: Check: $2800+2300+3100=8200$.

Answer: (C) ₹$2800$

Concept: Employees in a department $=\text{its }\%\times \text{total}$.

1. Department B $=14\%$
2. $\frac{14}{100}\times 450=63$

Trick: $1\%$ of $450=4.5$, so $14\times 4.5=63$.

Answer: (D) $63$

Concept: Bisectors of the two exterior angles at B and C meet at the excentre O, and $\angle BOC={90}^{\circ }-\frac{\angle A}{2}$.

1. $\angle BOC={90}^{\circ }-\frac{{66}^{\circ }}{2}$
2. $={90}^{\circ }-{33}^{\circ }={57}^{\circ }$

Yaad rakhein: Internal bisectors: ${90}^{\circ }+\frac{A}{2}$; external bisectors: ${90}^{\circ }-\frac{A}{2}$.

Answer: (C) ${57}^{\circ }$

Concept: $\tan \theta =\frac{\text{height}}{\text{shadow}}$; from $\cos \theta =\frac{12}{13}$, the triangle is 5–12–13.

1. $\sin \theta =\frac{5}{13}$, so $\tan \theta =\frac{5}{12}$
2. $h=18\times \frac{5}{12}=7.5\text{m}$

Trick: Triplet $(5,12,13)$: base $12k=18\Rightarrow k=1.5$, height $5k=7.5$.

Answer: (A) $7.5$

Concept: Add girls and boys over all five schools and simplify the ratio.

1. Girls $=500+550+650+600+650=2950$
2. Boys $=600+450+750+700+800=3300$
3. $2950:3300=59:66$ (divide by 50)

Trick: Order matters: girls first, so $59:66$, not $66:59$.

Answer: (C) $59:66$

Concept: Net price factor $=$ product of $(1-d)$ for each discount.

1. $0.92\times 0.85\times 0.88$
2. $0.92\times 0.85=0.782$; $0.782\times 0.88=0.68816$
3. Single discount $=1-0.68816=0.31184=31.184\%$

Trick: $68.816\%$ is the price paid, not the discount – a common trap.

Answer: (D) $31.184\%$

Concept: From $x^2-5x+1=0$, dividing by $x$: $x+\frac{1}{x}=5$.

1. $\frac{x^4+\frac{1}{x^2}}{x^2+1}$: divide numerator and denominator by $x$ → $\frac{x^3+\frac{1}{x^3}}{x+\frac{1}{x}}$
2. $x^3+\frac{1}{x^3}=5^3-3\times 5=110$
3. Value $=\frac{110}{5}=22$

Trick: $\frac{x^3+\frac{1}{x^3}}{x+\frac{1}{x}}=x^2-1+\frac{1}{x^2}=(25-2)-1=22$.

Answer: (B) $22$

Concept: Expenditure constant: % reduction in consumption $=\frac{\text{increase}}{\text{new price}}\times 100$.

1. Increase $=29-25=4$
2. Reduction $=\frac{4}{29}\times 100\approx 13.79\%\approx 14\%$

Yaad rakhein: Divide by the NEW price ($29$), not the old one ($25$ would give $16\%$).

Answer: (A) $14\%$

Concept: Ratio of areas of similar triangles $=$ (ratio of corresponding sides)${}^2$. In $△ABC~△RPQ$: $A\leftrightarrow R$, $B\leftrightarrow P$, $C\leftrightarrow Q$.

1. Side ratio $=\sqrt{\frac{4}{9}}=\frac{2}{3}$
2. PQ corresponds to BC $=4$
3. $\frac{BC}{PQ}=\frac{2}{3}\Rightarrow PQ=6$

Trick: Match letters by position, not by name – PQ pairs with BC, not AB.

Answer: (D) $6$

Concept: $x^3+y^3+z^3-3xyz=(x+y+z)[(x+y+z{)}^2-3(xy+yz+zx)]$.

1. $=19\times (361-342)=19\times 19=361$
2. $x^3+y^3+z^3=361+3\times 216=1009$
3. Required $=1009+216=1225$

Trick: Required $=361+4xyz=361+864=1225$.

Answer: (A) $1225$

Concept: Savings $=$ Salary $-(E+T+C)$, computed before and after the raise.

1. Before: $E=30000$, $T=6000$, $C=900$; savings $=50000-36900=13100$
2. After: salary $=70000$, $E=42000$, $T=12600$, $C=2520$
3. Savings $=70000-57120=12880$
4. Difference $=13100-12880=220$

Trick: Savings $=S(1-0.6-0.6t-0.6tc)$: $50000\times 0.262$ vs $70000\times 0.184$.

Answer: (B) ₹$220$

Concept: Take CP $=100x$ and compare the two selling prices.

1. Actual SP $=110x$
2. New CP $=80x$, new SP $=80x\times 1.4=112x$
3. $112x-110x=2x=1000\Rightarrow x=500$
4. CP $=100x=50000$

Trick: Difference in SP $=2\%$ of CP $=1000$ → CP $=50000$.

Answer: (A) ₹$50000$

Concept: Central angle $=\frac{\text{percentage}}{100}\times {360}^{\circ }$.

1. Department A $=20\%$
2. Angle $=\frac{20}{100}\times {360}^{\circ }={72}^{\circ }$

Trick: $1\%={3.6}^{\circ }$, so $20\%={72}^{\circ }$.

Answer: (A) ${72}^{\circ }$

Concept: The tops and the horizontal line from the shorter top form a right triangle.

1. Vertical gap $=16-9=7\text{cm}$; horizontal gap $=24\text{cm}$
2. Distance $=\sqrt{{24}^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\text{cm}$

Trick: $(7,24,25)$ is a Pythagorean triplet.

Answer: (A) $25$

Concept: Net rate $=$ sum of filling rates $-$ emptying rate.

1. Net per hour $=\frac{1}{15}+\frac{1}{4}-\frac{1}{12}$
2. $=\frac{4+15-5}{60}=\frac{14}{60}=\frac{7}{30}$
3. Time $=\frac{30}{7}$ hours

Trick: Tank $=60$ units: $4+15-5=14$ units/h → $\frac{60}{14}=\frac{30}{7}$.

Answer: (C) $\frac{30}{7}$