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

Concept: $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, $s=\frac{a+b+c}{2}$.

1. $s=\frac{8+12+16}{2}=18$
2. Area $=\sqrt{18\times 10\times 6\times 2}=\sqrt{2160}$
3. $=\sqrt{144\times 15}=12\sqrt{15}{\text{cm}}^2$

Trick: Sides $4(2,3,4)$: area of the $2,3,4$ triangle is $\frac{3\sqrt{15}}{4}$; multiply by $4^2=16$ to get $12\sqrt{15}$.

Answer: (D) $12\sqrt{15}$

Concept: Let $\frac{1}{d}=a$, $\frac{1}{u}=b$ (d, u = downstream, upstream speeds); still-water speed $=\frac{d+u}{2}$.

1. $40a+25b=7.5$ and $48a+36b=10$
2. Multiply: $240a+150b=45$, $240a+180b=50\Rightarrow 30b=5,b=\frac{1}{6}$
3. $40a=7.5-\frac{25}{6}=\frac{10}{3}\Rightarrow a=\frac{1}{12}$; so $d=12$, $u=6$
4. Boat speed $=\frac{12+6}{2}=9\text{km/h}$

Trick: Check: $\frac{40}{12}+\frac{25}{6}=3\frac{1}{3}+4\frac{1}{6}=7\frac{1}{2}$ h ✓.

Answer: (C) $9$

Concept: The largest number dividing all given numbers is their HCF.

1. $24=2^3\times 3$, $56=2^3\times 7$, $96=2^5\times 3$
2. Common: $2^3=8$

Trick: HCF divides differences: $56-24=32$, $96-56=40$; $\text{HCF}(32,40)=8$.

Answer: (C) $8$

Concept: TSA of solid hemisphere $=3\pi r^2$; volume is conserved: $\frac{2}{3}\pi r^3=\frac{1}{3}\pi r^{2}h$.

1. $3\times \frac{22}{7}\times r^2=462\Rightarrow r^2=49,r=7\text{cm}$
2. $\frac{2}{3}\pi r^3=\frac{1}{3}\pi r^{2}h\Rightarrow h=2r$
3. $h=14\text{cm}$

Yaad rakhein: Hemisphere $\to$ cone with the same radius always gives $h=2r$.

Answer: (A) $14\text{cm}$

Concept: Read each option's value from the table and pick the maximum.

1. A, 2011 $=700000$; D, 2012 $=700000$; D, 2014 $=700000$
2. B, 2014 $=710000$ — the highest value in the whole table

Trick: Scan for the single largest entry: 710000 occurs only once (B, 2014).

Answer: (B) B, 2014

Concept: Majority $=$ (winner% $-$ loser%) of valid votes.

1. Let voters $=V$. Cast $=0.9V$; valid $=0.9\times 0.9V=0.81V$
2. Margin $=(54\%-46\%)=8\%$ of valid $=0.08\times 0.81V=0.0648V$
3. $0.0648V=1620\Rightarrow V=25000$

Trick: Valid votes $=\frac{1620}{8\%}=20250$; $V=\frac{20250}{0.81}=25000$.

Answer: (A) $25000$

Concept: Apply the discounts one after another, checking the condition after the first.

1. List price $=12\times 750=9000$
2. After 8%: $9000\times 0.92=8280$ ($>5000$, so extra 20% applies)
3. Final $=8280\times 0.8=6624$

Trick: Net factor $=0.92\times 0.8=0.736$; $9000\times 0.736=6624$.

Answer: (B) ₹6,624

Concept: Selling at a fraction of SP gives a known loss; use it to relate SP and CP.

1. $\frac{3}{8}\text{SP}=0.55\text{CP}$ (45% loss)
2. $\frac{\text{SP}}{\text{CP}}=\frac{0.55\times 8}{3}=\frac{4.4}{3}\approx 1.4667$
3. Profit $\approx 46.67\%\approx 47\%$

Trick: Take CP $=300$: $\frac{3}{8}\text{SP}=165\Rightarrow \text{SP}=440$, profit $=\frac{140}{300}\times 100\approx 47\%$.

Answer: (C) 47 Percent

Concept: Rationalise: $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\frac{1-\sin \theta }{\cos \theta }=\sec \theta -\tan \theta$.

1. Multiply inside by $\frac{1-\sin \theta }{1-\sin \theta }$: $\sqrt{\frac{(1-\sin \theta {)}^2}{{\cos }^2\theta }}$
2. $=\frac{1-\sin \theta }{\cos \theta }=\sec \theta -\tan \theta$
3. With $\theta ={45}^{\circ }$: $\sec {45}^{\circ }-\tan {45}^{\circ }$

Trick: Value check: $\sec {45}^{\circ }-\tan {45}^{\circ }=\sqrt{2}-1\approx 0.414>0$; option (B) is negative, so it cannot be a square root.

Answer: (D) $\sec {45}^{\circ }-\tan {45}^{\circ }$

Concept: Simplify each part, then solve the resulting quadratic.

1. $(1÷x)\times (x^3-2x^2)=x^2-2x$
2. $(4^3÷2^5)\times x=\frac{64}{32}x=2x$
3. Sum $=x^2-2x+2x=x^2$; $\frac{x^2}{25}=1\Rightarrow x^2=25$
4. $x=5$ or $x=-5$ (both non-zero, so valid)

Trick: $x^2=25$ always has two roots $\pm 5$ — don't drop the negative one.

Answer: (D) Both 5 and $-5$

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