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

Concept: Number of votes cast = registered voters × percentage who voted.

1. Centre 1: $26000\times 70\%=18200$; Centre 2: $28000\times 60\%=16800$
2. Centre 3: $25000\times 80\%=20000$; Centre 4: $30000\times 76\%=22800$
3. Maximum = Centre 4 (22800)

Trick: Centre 4 has both the most voters and a high $\%$ — check it against Centre 3 only.

Answer: (D) Centre 4

Concept: Convert each ratio to a decimal (or fraction) and compare.

1. $2:11=0.18$, $3:5=0.6$
2. $3:4=0.75$, $7:2=3.5$
3. Ascending: $2:11<3:5<3:4<7:2$

Trick: Any ratio with first term bigger than second ($7:2$) exceeds 1 — it is the largest here.

Answer: (B) 2 : 11 < 3 : 5 < 3 : 4 < 7 : 2

Concept: Take total work = LCM of days and use per-day efficiencies.

1. Work $=\text{LCM}(30,40)=120$ units; P $=4$, Q $=3$ units/day
2. In 4 days together: $4\times 7=28$ units
3. Remaining $=120-28=92$ units
4. P alone: $\frac{92}{4}=23$ days

Trick: Fraction method: $4(\frac{1}{30}+\frac{1}{40})=\frac{7}{30}$ done, $\frac{23}{30}$ left → P needs 23 days.

Answer: (B) 23 days

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Times: $\frac{72}{45}=1.6$ h, $\frac{28}{4}=7$ h, $\frac{160}{20}=8$ h
2. Total time $=16.6$ h; total distance $=72+28+160=260\text{km}$
3. Average $=\frac{260}{16.6}=15.66\text{km/h}$

Trick: The slow 7-hour bicycle leg drags the average far below the bike/car speeds.

Answer: (A) 15.66

Concept: Apply BODMAS on both sides, then solve the linear equation.

1. LHS: $14+3p-2+4=16+3p$
2. RHS: $7+3=10$
3. $16+3p=10\Rightarrow 3p=-6\Rightarrow p=-2$

Trick: Substitute options: $p=-2$ gives $14-6-2+4=10$ ✓.

Answer: (D) −2

Concept: Volume of cone $=\frac{1}{3}\pi r^{2}h$ and slant height $l=\sqrt{r^2+h^2}$.

1. $150\pi =\frac{1}{3}\pi r^2\times 18=6\pi r^2$
2. $r^2=25\Rightarrow r=5$ cm
3. $l=\sqrt{25+324}=\sqrt{349}\approx 18.68$ cm

Trick: $\sqrt{349}$ is just above $\sqrt{324}=18$ — only 18.68 fits.

Answer: (D) 18.68

Concept: Curved surface area of a cone $=\pi rl$.

1. $\frac{22}{7}\times r\times 28=308$
2. $88r=308$
3. $r=3.5$ cm

Yaad rakhein: $\frac{22}{7}\times 28=88$; then $308÷88=3.5$.

Answer: (A) 3.5 cm

Concept: The four result categories are mutually exclusive and cover every student, so total = sum of all entries.

1. Section A: $25+20+12+15=72$
2. Section B: $22+25+15+20=82$
3. Section C: $18+24+8+22=72$; Section D: $20+12+10+18=60$
4. Total $=72+82+72+60=286$

Yaad rakhein: Fail-fail, fail-pass, pass-fail, pass-pass = all possibilities; no overlap, so just add.

Answer: (A) 286

Concept: Convert speed to m/s: $1\text{km/h}=\frac{5}{18}\text{m/s}$.

1. Speed $=\frac{72}{40/60}=108\text{km/h}$
2. $108\times \frac{5}{18}=30\text{m/s}$
3. Distance in 6 s $=30\times 6=180$ m

Trick: 72 km in 2400 s → $\frac{72000}{2400}=30$ m per second directly.

Answer: (D) 180

Concept: Growth at $r\%$ per year: $A=P{(1+\frac{r}{100})}^n$.

1. $A=64000\times (1.1{)}^3$
2. $(1.1{)}^3=1.331$
3. $A=64000\times 1.331=85184$

Trick: Year by year: $64000\to 70400\to 77440\to 85184$.

Answer: (C) 85184

Concept: Do ÷ and × from left to right before + and −, separately in numerator and denominator.

1. Numerator: $3÷3\times 2=2$, $6÷3=2$ → $2-2+7=7$
2. Denominator: $5÷5\times 6÷6=1$ → $1+3+3=7$
3. Value $=\frac{7}{7}=1$

Trick: Left to right: $5÷5\times 6÷6=1\times 6÷6=1$, not $5÷30÷6$.

Answer: (A) 1

Concept: Use $a^3+b^3=(a+b)(a^2-ab+b^2)$ and ${\sin }^{2}A+{\cos }^{2}A=1$.

1. ${\sin }^{3}A+{\cos }^{3}A=(\sin A+\cos A)({\sin }^{2}A+{\cos }^{2}A-\sin A\cos A)$
2. $=(\sin A+\cos A)(1-\sin A\cos A)$
3. Divide by $(\sin A+\cos A)$: $1-\sin A\cos A$

Trick: Check $A=0^{\circ }$: expression $=\frac{1}{1}=1$, and $1-\sin 0\cos 0=1$ ✓.

Answer: (C) $1-\sin A\cos A$

Concept: Selling price $=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. SP $=5980\times \frac{82}{100}$
2. $=4903.6$
3. To the nearest rupee: ₹4,904

Trick: $18\%$ of $5980=1076.4$; $5980-1076.4=4903.6$.

Answer: (D) 4,904

Concept: Split the mixture into milk and water, add the new quantities, and match the target ratio.

1. Milk $=\frac{5}{6}\times 18=15$ L, water $=3$ L
2. After adding 3 L milk: milk $=18$ L
3. For $9:2$: water $=\frac{18\times 2}{9}=4$ L
4. Water to add $=4-3=1$ L

Trick: Milk 18 = 9 parts → 1 part = 2 L → water 2 parts = 4 L.

Answer: (C) 1

Concept: Crossing a man (a point) in a train moving the same way: length = relative speed × time, relative speed = difference.

1. Relative speed $=60-42=18\text{km/h}$
2. $18\times \frac{5}{18}=5\text{m/s}$
3. Length $=5\times 36=180$ m

Yaad rakhein: Same direction → subtract speeds; opposite → add.

Answer: (C) 180 m

Concept: Find the grand total over five years and compare $20\%$ of it with each yearly total.

1. Grand total $=115+108+149+102+101=575$ thousand
2. $20\%$ of $575=115$ thousand
3. Year with total 115 thousand = 1985

Trick: $20\%=\frac{1}{5}$, so look for the year equal to the average of the five totals ($575÷5=115$).

Answer: (A) 1985

Concept: Effective CP includes cartage. Target SP = CP × 1.42; the last lot must cover what is still missing.

1. Total CP $=100\times 19+100=₹2000$, i.e. ₹20 per kg; target SP $=2000\times 1.42=₹2840$
2. 60 kg at $50\%$: $60\times 30=₹1800$; 20 kg at $40\%$: $20\times 28=₹560$
3. 10 kg at $25\%$: $10\times 25=₹250$; total so far ₹2610
4. Last 10 kg must fetch $2840-2610=₹230$ → ₹23 per kg
5. Profit $=\frac{23-20}{20}=0.15$, i.e. $15\%$

Trick: Weighted-profit check: $\frac{60(50)+20(40)+10(25)+10x}{100}=42\Rightarrow x=15$.

Answer: (B) 0.15 (i.e. 15%)

Concept: Apply BODMAS from the innermost bracket outwards; simplify the denominator separately.

1. $\frac{1}{4}÷\frac{15}{20}=\frac{1}{4}\times \frac{4}{3}=\frac{1}{3}$; $\frac{1}{3}+2=\frac{7}{3}$; $3\times \frac{7}{3}=7$
2. $\{3+7-18\times \frac{1}{3}\}=10-6=4$; numerator $=2\times 4=8$
3. Denominator $=2(4+5)=18$
4. Value $=\frac{8}{18}=\frac{4}{9}$

Trick: $\frac{15}{20}=\frac{3}{4}$ first — makes the division instant.

Answer: (C) $\frac{4}{9}$

Concept: With a false weight, the trader gives less goods for the same price: SP of 1 unit of goods $=\frac{1+\text{profit}}{1-\text{shortfall}}$ × CP.

1. Take CP of 1 kg = ₹100; he sells at $32\%$ profit, i.e. charges ₹132
2. But he gives only $0.8$ kg (CP ₹80)
3. Gain $=\frac{132-80}{80}\times 100=65\%$

Trick: $\frac{132}{80}=1.65$ → gain $65\%$.

Answer: (D) 65%

Concept: Actual activity = percentage × total activities of that city.

1. R in B $=30\%\times 500=150$
2. R in C $=40\%\times 400=160$
3. Difference $=160-150=10$

Yaad rakhein: Percentages of different totals can't be compared directly — convert to actual numbers first.

Answer: (B) 10