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

Concept: When two sectors of the same pie are compared, compare their percentages directly – the total cancels out.

1. Rent $=20\%$, Others $=10\%$ of the total.
2. Rent is more by $20-10=10$ percentage points.
3. $\text{Required}\%=\frac{20-10}{10}\times 100=100\%$

Trick: Rent is exactly double Others, and "double" always means 100% more.

Answer: (D) 100%

Concept: If two articles are sold at the same selling price, one at $x\%$ gain and the other at $x\%$ loss, there is always a loss of $\frac{x^2}{100}\%$.

1. Here $x=20$.
2. $\text{Loss}\%=\frac{{20}^2}{100}=4\%$
3. Check: CPs are ₹50,000 and ₹75,000 (total ₹1,25,000); SP ₹1,20,000; loss ₹5,000 $=4\%$.

Trick: Same SP, same $\pm x\%$ ⇒ always a LOSS of $\frac{x^2}{100}\%$ – never "no profit no loss".

Answer: (B) Loss 4%

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

1. $96=2^5\times 3$
2. $132=2^2\times 3\times 11$
3. $438=2\times 3\times 73$
4. $\text{LCM}=2^5\times 3\times 11\times 73=96\times 803=77088$

Trick: The LCM must be divisible by 96, so it must be even and divisible by 3 – 67055 and 85046 fail at once (odd / digit sum not a multiple of 3).

Answer: (D) 77088

Concept: Winner's share $=\frac{\text{winner's votes}}{\text{total valid votes}}\times 100$.

1. Total $=1136+7636+11628=20400$
2. Winner got 11628 votes.
3. $\frac{11628}{20400}\times 100=57\%$

Trick: 57% of 20400 $=204\times 57=11628$ – verify the option directly instead of dividing.

Answer: (D) 57

Concept: Convert each ratio hens : cows to a decimal (hens ÷ cows) and compare.

1. Ram $=\frac{12}{5}=2.4$
2. Shyam $=\frac{18}{9}=2$
3. Preetham $=\frac{4}{8}=0.5$; Vinay $=\frac{10}{7}\approx 1.43$
4. Greatest is 2.4 ⇒ Ram.

Trick: Shyam has the most hens, but the ratio depends on hens per cow – Ram has the most hens per cow.

Answer: (C) Ram

Concept: Sphere: $\text{TSA}=4\pi r^2$, $V=\frac{4}{3}\pi r^3$. Solid hemisphere: $\text{TSA}=3\pi R^2$, $V=\frac{2}{3}\pi R^3$.

1. $4\pi r^2=3\pi R^2\Rightarrow \frac{r}{R}=\frac{\sqrt{3}}{2}$
2. $\frac{V_1}{V_2}=\frac{\frac{4}{3}\pi r^3}{\frac{2}{3}\pi R^3}=2{\left(\frac{r}{R}\right)}^3$
3. $=2\times \frac{3\sqrt{3}}{8}=\frac{3\sqrt{3}}{4}$

Trick: A solid hemisphere includes its flat circular base, so its surface is $3\pi R^2$, not $2\pi R^2$.

Answer: (C) $3\sqrt{3}:4$

Concept: Curved surface area $=2\pi rh$; volume $=\pi r^{2}h$.

1. $2\times \frac{22}{7}\times r\times 15=660$
2. $r=\frac{660\times 7}{2\times 22\times 15}=7\text{cm}$
3. $V=\frac{22}{7}\times 7^2\times 15=22\times 7\times 15=2310{\text{cm}}^3$

Trick: $V=\text{CSA}\times \frac{r}{2}=660\times \frac{7}{2}=2310$ – one step once $r$ is known.

Answer: (C) 2310

Concept: Slant height $l=\sqrt{r^2+h^2}$; total surface area $=\pi r(l+r)$.

1. $l=\sqrt{3^2+4^2}=5\text{cm}$
2. $\text{TSA}=\pi \times 3\times (5+3)=24\pi$
3. $=24\times \frac{22}{7}=\frac{528}{7}{\text{cm}}^2$

Trick: 3-4-5 is a Pythagorean triplet – the slant height is 5 at sight.

Answer: (D) $\frac{528}{7}{\text{cm}}^2$

Concept: Moving in opposite directions, the gap grows at the sum of the speeds.

1. Let Soma $=x\text{km/h}$, Julie $=x+4$.
2. $(x+x+4)\times 4=40\Rightarrow 2x+4=10$
3. $x=3$ ⇒ Soma 3 km/h, Julie 7 km/h.

Trick: Sum of speeds $=\frac{40}{4}=10$; only option B has speeds adding to 10 with a difference of 4.

Answer: (B) Julie – 7 km/h; Soma – 3 km/h

Concept: Alligation: quantities are in the ratio (higher − mean) : (mean − lower).

1. 9% part : 19% part $=(19-15):(15-9)=4:6=2:3$
2. 9% part $=\frac{2}{5}\times 650=260\text{kg}$

Trick: The mean 15% is closer to 19%, so the 19% part is larger – the 9% part must be less than half of 650 (rules out 455, 390).

Answer: (A) 260

Concept: Make the decimal places equal, find the LCM of the whole numbers and put back the same number of decimal places.

1. 0.15, 0.18, 0.45 → 15, 18, 45 (2 decimal places).
2. $15=3\times 5$, $18=2\times 3^2$, $45=3^2\times 5$
3. $\text{LCM}=2\times 3^2\times 5=90$ ⇒ 0.90

Trick: The LCM can never be smaller than the largest number 0.45, so 0.09 is out at once.

Answer: (A) 0.9

Concept: To cross a tunnel a train covers (its own length + tunnel length).

1. Distance $=180+500=680\text{m}$
2. Speed $=\frac{680}{40}=17\text{m/s}$
3. $17\times \frac{18}{5}=61.2\text{km/h}$

Trick: m/s → km/h: multiply by $\frac{18}{5}$ (i.e. 3.6).

Answer: (B) 61.2

Concept: $\sin ({90}^{\circ }-\theta )=\cos \theta$, $\cos ({90}^{\circ }-\theta )=\sin \theta$ and $\tan \theta \tan ({90}^{\circ }-\theta )=1$.

1. Numerator $={\sin }^2{39}^{\circ }+{\cos }^2{39}^{\circ }=1$
2. Denominator $={\cos }^2{35}^{\circ }+{\sin }^2{35}^{\circ }=1$, so the fraction $=1$.
3. The tangent pair is meant to be complementary: $\tan {25}^{\circ }\tan {65}^{\circ }=1$, giving $3\times 1=3$.
4. Value $=1+3=4$

Yaad rakhein: Two angles adding to ${90}^{\circ }$: $\tan A\tan B=1$. (Note: $\tan {25}^{\circ }\tan {75}^{\circ }\approx 1.74$, which matches no option.)

Answer: (B) 4

Concept: Write each ratio as a fraction/decimal and compare.

1. $2:5=0.4$
2. $7:10=0.7$
3. $5:6\approx 0.833$
4. $1:3\approx 0.333$ ⇒ greatest is $5:6$.

Trick: Only $5:6$ is within one part of being equal ($\frac{5}{6}$ is closest to 1), so it is the largest.

Answer: (C) 5 : 6

Concept: LCM = product of the highest powers of the prime factors.

1. $20=2^2\times 5$, $30=2\times 3\times 5$
2. $45=3^2\times 5$, $65=5\times 13$
3. $\text{LCM}=2^2\times 3^2\times 5\times 13=2340$

Trick: The LCM must be a multiple of 13 and of 9 – only 2340 qualifies ($2340=13\times 180$).

Answer: (B) 2340

Concept: The greatest length that measures all lengths exactly = HCF of the lengths (in the same unit).

1. Lengths: 315 cm, 500 cm, 685 cm.
2. $315=3^2\times 5\times 7$, $500=2^2\times 5^3$, $685=5\times 137$
3. Common factor only 5 ⇒ HCF $=5\text{cm}$.

Trick: 500 has only factors 2 and 5, and 315, 685 are odd – so the answer is 5 straight away.

Answer: (D) 5 cm

Concept: Express $x$ in terms of $y$ and substitute.

1. $\frac{x}{4y}=\frac{3}{4}\Rightarrow x=3y$
2. $2x+3y=6y+3y=9y$
3. $x-2y=3y-2y=y$
4. Value $=\frac{9y}{y}=9$

Trick: Put $y=1$, $x=3$: $\frac{6+3}{3-2}=9$.

Answer: (B) 9

Concept: Two successive discounts $a\%$ and $b\%$ = single discount $(a+b-\frac{ab}{100})\%$.

1. Single 30% on ₹5,000 = ₹1,500.
2. Successive: $15+25-\frac{15\times 25}{100}=36.25\%$ ⇒ ₹1,812.50.
3. Difference $=1812.50-1500=$ ₹312.50

Trick: Difference $=(36.25-30)\%$ of 5000 $=6.25\%\times 5000=$ ₹312.50.

Answer: (C) ₹312.50

Concept: Circles touching internally: distance between centres = difference of radii. A chord at distance $d$ from the centre has length $2\sqrt{r^2-d^2}$.

1. $PQ=6-4=2\text{cm}$
2. The perpendicular bisector of PQ passes through M with $PM=1\text{cm}$; AB is a chord of the bigger circle at distance 1 from P.
3. $AM=\sqrt{6^2-1^2}=\sqrt{35}$
4. $AB=2\sqrt{35}\text{cm}$

Trick: Internal touch ⇒ subtract radii; external touch ⇒ add radii.

Answer: (B) $2\sqrt{35}\text{cm}$

Concept: Savings = Income − Expenditure; % increase $=\frac{\text{increase}}{\text{original}}\times 100$.

1. Old saving $=26500-22000=$ ₹4,500.
2. New income $=26500\times 1.2=31800$; new expenditure $=22000\times 1.15=25300$.
3. New saving $=31800-25300=$ ₹6,500; increase ₹2,000.
4. $\frac{2000}{4500}\times 100=\frac{400}{9}=44\frac{4}{9}\%$

Trick: Increase in saving $=5300-3300=2000$ (20% of income − 15% of expenditure).

Answer: (D) $44\frac{4}{9}\%$