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

Concept: In a right triangle $\tan x=\frac{\text{perpendicular}}{\text{base}}$ and $\sin x=\frac{\text{perpendicular}}{\text{hypotenuse}}$.

1. $\tan x=5=\frac{5}{1}$, so perpendicular $=5$, base $=1$.
2. Hypotenuse $=\sqrt{5^2+1^2}=\sqrt{26}$.
3. $\sin x=\frac{5}{\sqrt{26}}$.

Trick: From $\tan x=t$: $\sin x=\frac{t}{\sqrt{1+t^2}}=\frac{5}{\sqrt{26}}$.

Answer: (B) $\frac{5}{\sqrt{26}}$

Concept: Downstream speed $=b+s$, upstream speed $=b-s$, where $b$ = boat speed in still water, $s$ = stream speed.

1. Downstream speed $=\frac{104}{8}=13\text{km/h}$.
2. Boat speed $b=13-2=11\text{km/h}$.
3. Upstream speed $=11-2=9\text{km/h}$.
4. Time for 13 km upstream $=\frac{13}{9}=1\frac{4}{9}$ hours.

Trick: Upstream speed $=$ downstream $-2s=13-4=9$ directly.

Answer: (B) $1\frac{4}{9}$ hours

Concept: Follow BODMAS: innermost brackets first, then division/multiplication from left to right.

1. $\{(57-49)÷2\}=8÷2=4$; $\{(85-60)÷5\}=25÷5=5$; $\{(28-19)÷3\}=9÷3=3$.
2. Big bracket $=4+5-3=6$.
3. $72÷8\times 6=9\times 6=54$.

Trick: Division and multiplication have equal priority – work left to right: $72÷8$ first, then $\times 6$.

Answer: (C) 54

Concept: In right $△ABC$ ($\angle B={90}^{\circ }$) with $BD⟂AC$: $△ADB~△ABC$, so $A{B}^2=AD\times AC$.

1. $△ADB$ and $△ABC$ share $\angle A$ and both have a right angle, so they are similar.
2. $\frac{AD}{AB}=\frac{AB}{AC}\Rightarrow AD=\frac{A{B}^2}{AC}$.
3. $AD=\frac{8^2}{17}=\frac{64}{17}\approx 3.76\text{cm}$.

Trick: Leg${}^2$ = (its adjacent segment of hypotenuse) $\times$ hypotenuse. Here $BC=15$ is not needed.

Answer: (C) $3.76\text{cm}$

Concept: When two objects move towards each other, relative speed = sum of speeds; distance = relative speed $\times$ time.

1. Time from 7 a.m. to 3:30 p.m. $=8.5\text{h}$.
2. Relative speed $=85+67=152\text{km/h}$.
3. Distance $=152\times 8.5=1292\text{km}$.

Trick: Unit digit check: $152\times 8.5=1216+76=1292$ – only option (B) fits.

Answer: (B) 1292 km

Concept: $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$.

1. ${10}^2=48+2(ab+bc+ca)$.
2. $2(ab+bc+ca)=100-48=52$.
3. $ab+bc+ca=26$.

Trick: $ab+bc+ca=\frac{(\text{sum}{)}^2-(\text{sum of squares})}{2}=\frac{100-48}{2}$.

Answer: (B) 26

Concept: Percentage share $=\frac{\text{part}}{\text{total}}\times 100$.

1. Total $=74.30+75.00+77.50+69.05+70.00+67.52=433.37$.
2. Rakesh's contribution $=75.00$.
3. Share $=\frac{75}{433.37}\times 100\approx 17.30\%$.

Trick: Six values are all near 72, so each share is near $\frac{100}{6}\approx 16.7\%$; Rakesh (slightly above average) gets a little more – 17.30%.

Answer: (A) 17.30%

Concept: $(a+b{)}^2=a^2+2ab+b^2$.

1. $1801=1800+1$.
2. ${1801}^2={1800}^2+2\times 1800\times 1+1=3240000+3600+1$.
3. $=3243601$.

Trick: Last two digits of ${1801}^2$ must be 01 and the number starts with ${18}^2=324$ – only 3243601 fits.

Answer: (B) 3243601

Concept: Percentage share $=\frac{\text{state value}}{\text{column total}}\times 100$.

1. Total wheat $=113+96+42+47+32+25+18+38=411$ lakh tons.
2. State P wheat $=113$.
3. Share $=\frac{113}{411}\times 100\approx 27.49\%\approx 27.50\%$.

Trick: $\frac{113}{411}$ is just over $\frac{1}{4}$ (as $411÷4\approx 103$) – only 27.50% fits.

Answer: (B) 27.50 percent

Concept: In a chase (same direction), relative speed = difference of speeds; time to catch = gap $÷$ relative speed.

1. Relative speed $=7-5=2\text{m/s}$.
2. Time to catch $=\frac{150}{2}=75\text{s}$.
3. Distance run by thief $=5\times 75=375\text{m}$.

Trick: Thief's distance $=\frac{\text{gap}\times v_t}{v_p-v_t}=\frac{150\times 5}{2}=375$.

Answer: (C) 375 m

Concept: Rearranging digits keeps the digit sum (so divisibility by 3 or 9 stays), but changes the last digits (so divisibility by 2, 4, 8, 11 may be lost).

1. Digits in increasing order: 245679.
2. Last digit 9 is odd, so it is not divisible by 2 – hence not by 22, 44 or 66.
3. Digit sum $=2+4+5+6+7+9=33$, divisible by 3, so 245679 is divisible by 3.

Yaad rakhein: Divisibility by 3 depends only on the digit sum, never on the order of digits.

Answer: (D) 03 (i.e. 3)

Concept: Loss of $x\%$ means $\text{SP}=\text{CP}\times \frac{100-x}{100}$; work backwards from the final selling price.

1. Anuj's CP $=\frac{554400}{0.88}=630000$.
2. Anuj's CP = Aman's SP, which is 90% of Aman's CP.
3. Aman's CP $=\frac{630000}{0.9}=700000$.

Trick: $\text{CP}=\frac{554400}{0.9\times 0.88}=\frac{554400}{0.792}=700000$ in one step.

Answer: (A) Rs.7,00,000

Concept: Take total work = LCM of the days; one-day work = work $÷$ days.

1. $2\frac{1}{2}=\frac{5}{2}$ days. Total work = LCM of 10 and $\frac{5}{2}$ = 10 units.
2. A's rate $=1$ unit/day; (A+B)'s rate $=10÷\frac{5}{2}=4$ units/day.
3. B's rate $=4-1=3$ units/day, so B alone takes $\frac{10}{3}$ days.

Trick: $\frac{1}{B}=\frac{2}{5}-\frac{1}{10}=\frac{3}{10}$, so $B=\frac{10}{3}$ days.

Answer: (B) $\frac{10}{3}$ days

Concept: Read the 'Highest' bars of each subject and compare the two sums.

1. Highest: Accounts $=96$, Business Studies $=96$ → sum $=192$.
2. Highest: Economics $=100$, English $=100$ → sum $=200$.
3. $200-192=8$, so Economics and English is better by 8.

Trick: Each of Eco/Eng beats each of Acc/BSt by 4, so the pair wins by $2\times 4=8$.

Answer: (D) Economics and English by 8

Concept: Find the group size from the percentage, then split it in the given ratio.

1. Below poverty line in B $=60\%$ of $3000=1800$.
2. Boys : girls (below) $=2:4$, so girls $=\frac{4}{6}\times 1800$.
3. Girls $=1200$.

Trick: $3000\times \frac{60}{100}\times \frac{4}{6}=3000\times 0.4=1200$.

Answer: (A) 1200

Concept: $2\sin A\cos B=\sin (A+B)+\sin (A-B)$, $2\cos A\cos B=\cos (A+B)+\cos (A-B)$, $2\sin A\sin B=\cos (A-B)-\cos (A+B)$.

1. Numerator $\times 2$: $(\sin 9\theta +\sin 7\theta )-(\sin 9\theta +\sin 3\theta )=\sin 7\theta -\sin 3\theta =2\cos 5\theta \sin 2\theta$.
2. Denominator $\times 2$: $(\cos 3\theta +\cos \theta )-(\cos \theta -\cos 7\theta )=\cos 7\theta +\cos 3\theta =2\cos 5\theta \cos 2\theta$.
3. Ratio $=\frac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta$.

Trick: Quick check: for very small $\theta$ the expression $\approx \frac{8\theta -6\theta }{1}=2\theta$, which matches $\tan 2\theta$.

Answer: (D) $\tan 2\theta$

Concept: Difference of two sectors = (difference of their percentages) of the total.

1. D $=17\%$, E $=18\%$.
2. Difference $=18\%-17\%=1\%$.
3. $1\%$ of $60000=600$.

Trick: Never find both populations separately – subtract percentages first.

Answer: (D) 600

Concept: Volume of a hemisphere $=\frac{2}{3}\pi r^3$.

1. $r=21\text{cm}$, $r^3=9261$.
2. $V=\frac{2}{3}\times \frac{22}{7}\times 9261=\frac{2\times 22\times 1323}{3}$.
3. $V=44\times 441=19404{\text{cm}}^3$.

Trick: With $r=21$: $\frac{2}{3}\times \frac{22}{7}\times {21}^3=44\times {21}^2=44\times 441$.

Answer: (D) $19404{\text{cm}}^3$

Concept: If a value rises by $20\%$, new $=1.2\times$ old, so old $=\frac{\text{new}}{1.2}$.

1. $1.2\times P=30942$.
2. $P=\frac{30942}{1.2}=25785$.

Trick: $20\%=\frac{1}{5}$: old : new $=5:6$, so old $=\frac{30942}{6}\times 5=5157\times 5=25785$.

Answer: (B) 25785

Concept: Apply each spending step to the correct base: first on income, then on the remainder.

1. After education and food: $100\%-35\%=65\%$ of income remains.
2. Sports $40\%$ + transport $30\%$ of this remainder, so $30\%$ of it is left.
3. Left $=0.65\times 0.30\times I=0.195I=10257$.
4. $I=\frac{10257}{0.195}=52600$.

Trick: Check: $52600\times 0.65=34190$; $30\%$ of $34190=10257$ ✓

Answer: (C) Rs.52,600