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

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

1. Digit sum $=x+5+8+y=13+(x+y)$.
2. Next multiple of 9 after 13 is 18 → $x+y=5$.
3. ($x+y=-4$ is impossible), so least value $=5$.

Trick: $13\to 18$ needs $5$.

Answer: (B) $5$

Concept: $2{\cos }^2\theta -1=\cos 2\theta =1-2{\sin }^2\theta$.

1. Numerator $=\cos \theta (2{\cos }^2\theta -1)=\cos \theta \cos 2\theta$.
2. Denominator $=\sin \theta (1-2{\sin }^2\theta )=\sin \theta \cos 2\theta$.
3. Ratio $=\frac{\cos \theta }{\sin \theta }=\cot \theta$.

Trick: Put $\theta ={30}^{\circ }$: $\frac{2·\frac{3\sqrt{3}}{8}-\frac{\sqrt{3}}{2}}{\frac{1}{2}-\frac{1}{4}}=\sqrt{3}=\cot {30}^{\circ }$.

Answer: (C) $\cot \theta$

Concept: Surface area of a sphere $=4\pi r^2$.

1. $4\pi (5{)}^2=4\pi \times 25$.
2. $=100\pi {\text{cm}}^2$.

Yaad rakhein: Sphere SA is 4 times the area of its great circle.

Answer: (A) $100\pi {\text{cm}}^2$

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. Other number $=\frac{126\times 9}{18}$.
2. $=\frac{1134}{18}=63$.

Trick: Use the product rule directly: $\frac{\text{HCF}\times \text{LCM}}{\text{given number}}$.

Answer: (A) $63$

Concept: Find each average separately, then average them.

1. $J=\frac{1500+1460+1105}{3}=\frac{4065}{3}=1355$.
2. $K=\frac{1320+1180+1170}{3}=\frac{3670}{3}=1223.33$.
3. $\frac{J+K}{2}=\frac{2578.33}{2}=1289.17$.

Trick: $\frac{J+K}{2}=\frac{4065+3670}{6}=\frac{7735}{6}=1289.17$.

Answer: (A) $1289.17$

Concept: Below 36 years = groups up to 15, 16–25 and 26–35; use proportion.

1. Below 36: $15+20.25+16.75=52\%$ ↔ $30$ million.
2. Age 56–65: $5.5\%$.
3. $\frac{5.5}{52}\times 30=\frac{165}{52}\approx 3.17$ million.

Trick: Unitary: $1\%=\frac{30}{52}$ million.

Answer: (B) $3.17$ million

Concept: $(x-y{)}^2=(x+y{)}^2-4xy$.

1. $xy=24$; $(x-y{)}^2=100-96=4$, so $x-y=2$ (as $x>y$).
2. $x=6$, $y=4$.
3. $2x-y=12-4=8$.

Trick: Two numbers with sum 10 and product 24: 6 and 4.

Answer: (B) $8$

Concept: After both start, the gap closes at the sum of speeds.

1. By 9 a.m. train A covers $20$ km; gap $=120-20=100$ km.
2. Relative speed $=20+30=50\text{km/h}$; time $=\frac{100}{50}=2$ h.
3. They meet at $9+2=11$ a.m.

Trick: Head start first, then (remaining gap ÷ sum of speeds).

Answer: (C) 11:00 a.m.

Concept: Greatest length that measures all exactly = HCF of the lengths.

1. $62=2\times 31$, $186=6\times 31$, $279=9\times 31$.
2. HCF$(2,6,9)=1$, so HCF $=31$ m.

Trick: 62 is not a factor of 279, and 93 not of 62 — so the answer must be smaller: 31.

Answer: (C) $31$ m

Concept: In BODMAS, 'of' is done before division.

1. $2000$ of $(0.2{)}^2=2000\times 0.04=80$.
2. $(0.2{)}^3\times 400=0.008\times 400=3.2$.
3. $3.2÷80=0.04=\frac{1}{25}$.

Trick: $\frac{(0.2{)}^3\times 400}{2000\times (0.2{)}^2}=\frac{0.2\times 400}{2000}=\frac{1}{25}$.

Answer: (A) $\frac{1}{25}$

Concept: Exactly 50% = appeared − (less than 50%) − (more than 50%).

1. Less than 50%: $270+120+300+220+200=1110$.
2. More than 50%: A $330$, B $160$, C $75$, D $35$, E $75$.
3. Exactly 50%: A $0$, B $120$, C $0$, D $95$, E $25$; total $240$. Ratio $1110:240=111:24$.

Trick: Convert the % column to numbers first: $\frac{\%}{100}\times \text{appeared}$.

Answer: (B) $111:24$

Concept: TSA of a cuboid $=2(lb+bh+hl)$.

1. $lb=35$, $bh=45$, $hl=63$.
2. TSA $=2(35+45+63)=2\times 143=286{\text{cm}}^2$.

Trick: TSA must be even — options are all even, so compute carefully.

Answer: (C) $286{\text{cm}}^2$

Concept: CSA $=2\pi rh=(\text{circumference})\times h$.

1. $110\times h=2200$.
2. $h=20$ cm.

Trick: CSA is just base perimeter × height — no need to find $r$.

Answer: (C) $20$ cm

Concept: LCM = highest powers of primes; HCF = lowest common powers.

1. $48=2^4\times 3$, $64=2^6$ → LCM $=2^6\times 3=192$.
2. HCF$(12,18)=6$.
3. $\frac{192}{6}=32$.

Trick: $192÷6=32$.

Answer: (B) $32$

Concept: $\cot ({90}^{\circ }-\theta )=\tan \theta$, so $\cot {42}^{\circ }=\tan {48}^{\circ }$.

1. $\cos {48}^{\circ }=\frac{m}{n}$: base $m$, hypotenuse $n$, perpendicular $\sqrt{n^2-m^2}$.
2. $\sec {48}^{\circ }=\frac{n}{m}$, $\tan {48}^{\circ }=\frac{\sqrt{n^2-m^2}}{m}$.
3. Difference $=\frac{n-\sqrt{n^2-m^2}}{m}$.

Trick: Both terms have denominator $m$ (base) — only option (D) fits.

Answer: (D) $\frac{n-\sqrt{n^2-m^2}}{m}$

Concept: Add the one-hour works.

1. $\frac{1}{36}+\frac{1}{18}+\frac{1}{24}=\frac{2+4+3}{72}$.
2. $=\frac{9}{72}=\frac{1}{8}$ → 8 hours.

Trick: LCM method: total work $72$ units; rates $2+4+3=9$; $\frac{72}{9}=8$.

Answer: (D) $8$ hours

Concept: $1-{\sin }^2\theta ={\cos }^2\theta$ and $\sec \theta =\frac{1}{\cos \theta }$.

1. $\sqrt{1-{\sin }^2\theta }=\cos \theta$ (acute $\theta$).
2. ${\sec }^2\theta \times \cos \theta =\frac{\cos \theta }{{\cos }^2\theta }=\frac{1}{\cos \theta }=\sec \theta$.

Trick: ${\sec }^2\theta ·\cos \theta$ = one $\sec \theta$ left.

Answer: (C) $\sec \theta$

Concept: Here 42% is of the votes polled (including invalid). Opponent gets all valid votes except the candidate's.

1. Let polled $=T$. Candidate $=0.42T$; opponent $=T-1400-0.42T=0.58T-1400$.
2. $0.58T-1400-0.42T=3080\Rightarrow 0.16T=4480\Rightarrow T=28000$.
3. Invalid $\%=\frac{1400}{28000}\times 100=5\%$.

Trick: Majority $+$ invalid $=16\%$ of polled: $\frac{3080+1400}{16}\times 100=28000$.

Answer: (A) $5\%$

Concept: Each year: amount $=1.1\times$ balance, then subtract the payment.

1. Year 1: $50000\times 1.1-15000=40000$.
2. Year 2: $40000\times 1.1-15000=29000$.
3. Year 3: $29000\times 1.1-15000=31900-15000=$ Rs. $16900$.

Trick: Work year by year — interest is charged only on the remaining balance.

Answer: (A) Rs. $16900$

Concept: Percentage increase $=\frac{\text{increase}}{\text{2020 value}}\times 100$.

1. Sprite $\frac{400}{5400}\approx 7.4\%$; Coca-Cola $\frac{300}{5300}\approx 5.7\%$.
2. Pepsi $\frac{800}{6000}\approx 13.3\%$; Fanta $\frac{900}{1800}=50\%$.
3. Maximum: Fanta.

Trick: Fanta's small base with the largest rise gives the biggest %.

Answer: (C) Fanta