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

Concept: Take tank capacity = LCM(10, 20, 25) = 100 units; A, B, C fill 10, 5, 4 units per hour.

1. C works the first 2 h: $2\times 4=8$ units.
2. B works till 4 h: $4\times 5=20$ units.
3. A fills the rest: $100-8-20=72$ units $=72\%$ of the tank.

Trick: Work of A = total − work of the taps that were closed.

Answer: (C) $72\%$

Concept: East and North are perpendicular, so the displacement is the hypotenuse (Pythagoras).

1. Distance $=\sqrt{{24}^2+{10}^2}$.
2. $=\sqrt{576+100}=\sqrt{676}=26\text{m}$.

Trick: $(5,12,13)\times 2=(10,24,26)$ – a Pythagorean triplet.

Answer: (C) $26\text{m}$

Concept: $x+\frac{1}{x}=2$ holds only for $x=1$.

1. $x^2-2x+1=0\Rightarrow (x-1{)}^2=0\Rightarrow x=1$.
2. $x^4+\frac{1}{x^4}=1+1=2$.

Yaad rakhein: If $x+\frac{1}{x}=2$, then $x^n+\frac{1}{x^n}=2$ for every $n$.

Answer: (B) $2$

Concept: Angle at the centre is twice the angle at any point of the major arc (angle in the major segment).

1. Chord $=$ radius $=9$, so triangle OAB is equilateral: $\angle AOB={60}^{\circ }$.
2. Angle in the major segment $=\frac{{60}^{\circ }}{2}={30}^{\circ }$.

Trick: Chord = radius ⇒ ${30}^{\circ }$ in major segment, ${150}^{\circ }$ in minor segment.

Answer: (C) ${30}^{\circ }$

Concept: Half-yearly: rate is halved and time (in half-years) is doubled.

1. Rate $=4\%$ per half-year; $n=3$.
2. $A=7500\times (1.04{)}^3=7500\times 1.124864$.
3. $A=₹8436.48$.

Trick: $(1.04{)}^3=1.124864$ – useful to remember.

Answer: (A) Rs. $8,436.48$

Concept: Rearrange into perfect squares; a sum of squares is zero only if each square is zero.

1. $a^2-2a+1+b^2=0\Rightarrow (a-1{)}^2+b^2=0$.
2. So $a=1,b=0$.
3. $a^4+b^7=1+0=1$.

Trick: Square + square $=0$ ⇒ both are 0.

Answer: (A) $1$

Concept: Earning $=5\frac{1}{2}\%$ of total sales + $\frac{1}{2}\%$ of sales above ₹10,000.

1. $0.055S+0.005(S-10000)=1990$.
2. $0.06S-50=1990\Rightarrow 0.06S=2040$.
3. $S=₹34,000$.

Trick: Check options: $34000\times 0.055=1870$; $24000\times 0.005=120$; total 1990 ✓.

Answer: (C) Rs. $34,000$

Concept: All subjects have the same maximum (100), so the highest column total gives the highest overall percentage.

1. CS $=520$; Physics $=490$; Chemistry $=500$.
2. Maths $=470$; History $=515$; Geography $=80+85+95+95+95+90=540$.
3. Highest: Geography.

Trick: Equal maximum marks ⇒ compare plain totals.

Answer: (B) Geography

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 21=1386$.
2. $132r=1386$.
3. $r=10.5\text{cm}$.

Trick: $2\pi h=132$ here, so $r=\frac{1386}{132}$.

Answer: (A) $10.5\text{cm}$

Concept: Profit % $=\frac{\text{Revenue}-\text{Expenditure}}{\text{Expenditure}}\times 100$.

1. Highest profit % is 30% (2016).
2. Revenue $=50\times \frac{130}{100}=65$ lakhs.

Trick: Revenue = Expenditure × (1 + profit%).

Answer: (C) ₹65 lakhs

Concept: Evaluate each bracket separately, then multiply and divide.

1. $7^3-7^2+6=343-49+6=300$.
2. $6^3-2^3-3^2+1=216-8-9+1=200$.
3. ${21}^2+{20}^2-41=441+400-41=800$.
4. $\frac{300\times 200}{800}=75$.

Trick: Cancel early: $\frac{300\times 200}{800}=300\times \frac{1}{4}=75$.

Answer: (A) $75$

Concept: Subtract each remainder from its number; the required divisor is the HCF of the results.

1. $554-43=511$, $714-57=657$, $213-67=146$.
2. $511=7\times 73$, $657=9\times 73$, $146=2\times 73$.
3. HCF $=73$ (and $73>67$, the largest remainder ✓).

Trick: The divisor must exceed every remainder.

Answer: (D) $73$

Concept: LCM = product of highest powers of the primes.

1. $25=5^2$, $30=2\times 3\times 5$, $50=2\times 5^2$, $75=3\times 5^2$.
2. LCM $=2\times 3\times 5^2=150$.

Trick: 150 is divisible by 25, 30, 50 and 75 – check the smallest such option.

Answer: (B) $150$

Concept: Express $\sin \theta$ in terms of $\cos \theta$ and rationalise.

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

Yaad rakhein: If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$, then $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$ (swap sin and cos).

Answer: (A) $\sqrt{2}\sin \theta$

Concept: Total work in man-days $=60\times 40=2400$; subtract work done in each 10-day block.

1. Days 1–10: $60\times 10=600$; days 11–20: $55\times 10=550$.
2. Days 21–30: $500$; days 31–40: $450$. Done $=2100$.
3. Left $=300$ with 40 men: $\frac{300}{40}=7.5$ days.
4. Total $=40+7.5=47.5$ days.

Trick: Keep a running total of man-days until it reaches 2400.

Answer: (A) $47.5$

Concept: A number is divisible by 9 if its digit sum is a multiple of 9.

1. Digits without x: $8+7+6+4+5=30$.
2. Digit sum of x must make $30+s$ a multiple of 9: $s=6$ or $15$.
3. Least two-digit number with digit sum 6 is 15 (06 is not a two-digit number).

Trick: Need sum $36$ ⇒ digits of x add to 6 ⇒ smallest: 15.

Answer: (A) $15$

Concept: Use the difference / Euclid method: HCF divides the difference of the numbers.

1. $724-364=360$.
2. $\text{HCF}(364,360)$ divides $364-360=4$.
3. Both are divisible by 4, so HCF $=4$.

Trick: HCF always divides the difference of the two numbers.

Answer: (A) $4$

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}$.
2. $=\frac{3abc}{abc}=3$.

Trick: Put $a=1,b=1,c=-2$: $-\frac{1}{2}-\frac{1}{2}+4=3$.

Answer: (A) $3$

Concept: Compare cost price and selling price per orange.

1. CP of 1 orange $=\frac{90}{18}=₹5$.
2. SP of 1 orange $=\frac{105}{15}=₹7$.
3. Gain $\%=\frac{7-5}{5}\times 100=40\%$.

Trick: Per-unit prices make 'buy x for ₹a, sell y for ₹b' questions easy.

Answer: (D) $40\%$

Concept: Savings = Income − Expenditure. Take incomes $x,2x$ and expenditures $y,3y$.

1. $x-y=4000$ and $2x-3y=4000$.
2. Double the first: $2x-2y=8000$; subtract: $y=4000$.
3. $x=8000$; Shiv's income $=2x=₹16,000$.

Trick: Equal savings: $x-y=2x-3y\Rightarrow x=2y$.

Answer: (A) ₹16,000