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

Concept: A sector’s angle is the same fraction of ${360}^{\circ }$ as its value is of the total.

1. Total hours $=6+5+3+8+2=24$
2. Sleep $=8$ h, so fraction $=\frac{8}{24}=\frac{1}{3}$
3. Percentage $=33.33\%$ (angle $={120}^{\circ }$)

Trick: Percentage of central angle = percentage of the data itself – no need to find degrees.

Answer: (B) 33.33%

Concept: Area of rectangle $=l\times b$ (area in ${\text{m}}^2$).

1. $l=1.2b=\frac{6b}{5}$
2. $\frac{6b}{5}\times b=480\Rightarrow b^2=400\Rightarrow b=20\text{m}$
3. $l=1.2\times 20=24\text{m}$

Trick: Ratio $l:b=6:5$; $6k\times 5k=30k^2=480\Rightarrow k=4$, $l=24$.

Answer: (C) 24 m

Concept: If a tangent from Q touches at P and a secant from Q cuts the circle at B and A, then $Q{P}^2=QB\times QA$.

1. ${12}^2=8\times QA\Rightarrow QA=18\text{cm}$
2. $AB=QA-QB=18-8=10\text{cm}$

Trick: Don’t stop at QA = 18 – the question asks for the chord AB.

Answer: (C) 10 cm

Concept: Downstream $=B+S$, upstream $=B-S$.

1. $B=\frac{15+8}{2}=11.5\text{km/h}$
2. $S=\frac{15-8}{2}=3.5\text{km/h}$

Trick: Boat = half the sum, stream = half the difference; boat speed is always the larger one.

Answer: (C) 11.5 and 3.5

Concept: Decrease $x\%$ then increase $x\%$ gives a net loss of $\frac{x^2}{100}\%$.

1. Net loss $=\frac{{28}^2}{100}=7.84\%$
2. $7.84\%$ of $N=784\Rightarrow N=10000$

Trick: $0.72\times 1.28=0.9216$, so loss $=0.0784N=784$.

Answer: (C) 10000

Concept: Work with percentages first, convert to numbers at the end.

1. Mathematics $=27\%$; all other subjects $=100-27=73\%$
2. Difference $=73-27=46\%$
3. $46\%$ of $350=161$

Trick: Never compute 27% and 73% separately – take the gap of percentages directly.

Answer: (D) 161

Concept: Order: Vinculum (bar) → Brackets → "of" → Division → Multiplication → Addition → Subtraction.

1. Bar: $15-21=-6$
2. Bracket: $27-(-6)=33$
3. "of": $5\times 33=165$, so $19÷165=\frac{19}{165}$
4. $\frac{19}{165}+37=\frac{19+6105}{165}=\frac{6124}{165}$

Trick: "of" is done before ÷: $19÷5\text{ of }33=\frac{19}{165}$, not $\frac{19}{5}\times 33$.

Answer: (A) $\frac{6124}{165}$

Concept: Final value $=\text{Initial}\times (1+r_1)(1-r_2)$.

1. $1000\times 1.3=1300$ after year 1
2. $1300\times 0.8=1040$ after year 2

Trick: Net change $=30-20-\frac{30\times 20}{100}=+4\%$ → ₹1,040.

Answer: (B) ₹1,040

Concept: Product of two numbers = HCF × LCM.

1. Other $=\frac{660\times 5}{55}=\frac{3300}{55}=60$
2. Check: HCF(55, 60) = 5, LCM = 660 ✓

Trick: Both numbers must be multiples of the HCF 5 – 12 is ruled out at once.

Answer: (D) 60

Concept: Number of bottles $=\frac{\text{Volume of hemisphere}}{\text{Volume of one cylinder}}$.

1. Hemisphere: $r=18$, $V=\frac{2}{3}\pi (18{)}^3=3888\pi$
2. Bottle: $r=3$, $h=6$, $V=\pi (3{)}^2(6)=54\pi$
3. Bottles $=\frac{3888\pi }{54\pi }=72$

Trick: $\pi$ cancels – never substitute $\frac{22}{7}$ in a ratio of volumes.

Answer: (B) 72

Concept: Together time $=\frac{ab}{a+b}$.

1. $\frac{7\times 14}{7+14}=\frac{98}{21}=\frac{14}{3}$ days
2. $\frac{14}{3}=4\frac{2}{3}$ days

Trick: Total work = 14 units: Tom 2/day, Jay 1/day → $\frac{14}{3}$ days.

Answer: (B) $4\frac{2}{3}$ days

Concept: Same direction: relative speed $=$ difference; distance $=$ sum of lengths.

1. Relative speed $=68-50=18\text{km/h}=18\times \frac{5}{18}=5\text{m/s}$
2. Distance $=200+250=450\text{m}$
3. Time $=\frac{450}{5}=90\text{s}$

Trick: $18\text{km/h}=5\text{m/s}$ – a pair worth remembering.

Answer: (A) 90

Concept: Speed $=\frac{\text{Distance}}{\text{Time}}$; form an equation in original time $t$.

1. $\frac{1000}{t}-\frac{1000}{t+3}=75$
2. $\frac{3000}{t(t+3)}=75\Rightarrow t(t+3)=40$
3. $t^2+3t-40=0\Rightarrow (t+8)(t-5)=0\Rightarrow t=5$

Trick: Option check: $t=5$ → speeds 200 and $\frac{1000}{8}=125$; difference 75 ✓.

Answer: (B) 5 hours

Concept: Use ${\sin }^{2}A+{\cos }^{2}A=1$ to reduce to one ratio.

1. $7{\sin }^{2}A+3{\cos }^{2}A=3({\sin }^{2}A+{\cos }^{2}A)+4{\sin }^{2}A=3+4{\sin }^{2}A$
2. $3+4{\sin }^{2}A=4\Rightarrow {\sin }^{2}A=\frac{1}{4}\Rightarrow A={30}^{\circ }$
3. $\cot {30}^{\circ }=\sqrt{3}$

Trick: Split the bigger coefficient: $7=3+4$, so only $4{\sin }^{2}A=1$ remains.

Answer: (A) $\sqrt{3}$

Concept: ${\sec }^{2}A-{\tan }^{2}A=1\Rightarrow (\sec A-\tan A)(\sec A+\tan A)=1$.

1. $\sec A-\tan A=p\Rightarrow \sec A+\tan A=\frac{1}{p}$
2. Add: $2\sec A=p+\frac{1}{p}=\frac{p^2+1}{p}$
3. $\sec A=\frac{p^2+1}{2p}$

Trick: Add the two equations for sec, subtract them for tan $(\tan A=\frac{1-p^2}{2p})$.

Answer: (D) $\frac{p^2+1}{2p}$

Concept: $\sec ({90}^{\circ }-\theta )=\csc \theta$, $\csc ({90}^{\circ }-\theta )=\sec \theta$, $\tan ({90}^{\circ }-\theta )=\cot \theta$.

1. $\sin 1^{\circ }\sec {89}^{\circ }=\sin 1^{\circ }\csc 1^{\circ }=1$
2. $\cos {11}^{\circ }\csc {79}^{\circ }=\cos {11}^{\circ }\sec {11}^{\circ }=1$
3. $\tan {21}^{\circ }\tan {69}^{\circ }=\tan {21}^{\circ }\cot {21}^{\circ }=1$
4. Sum $=2+3+5=10$

Trick: Angles adding to ${90}^{\circ }$ with "co" partners give 1 – just add the coefficients.

Answer: (D) 10

Concept: Largest equal piece length = HCF of the lengths → least number of pieces.

1. HCF(20, 30, 40) $=10\text{m}$
2. Pieces $=\frac{20}{10}+\frac{30}{10}+\frac{40}{10}=2+3+4=9$

Yaad rakhein: Least pieces ↔ longest piece ↔ HCF.

Answer: (C) 9

Concept: In a triangle, the side opposite the smallest angle is shortest and opposite the largest angle is longest.

1. $15k={180}^{\circ }\Rightarrow k=12$: $\angle P={36}^{\circ }$, $\angle Q={48}^{\circ }$, $\angle R={96}^{\circ }$
2. Smallest angle P → opposite side QR (shortest)
3. Largest angle R → opposite side PQ (longest)

Trick: Side opposite a vertex = the two other letters: opposite P is QR, opposite R is PQ.

Answer: (D) QR and PQ

Concept: With $\angle B={90}^{\circ }$, AC is the hypotenuse; $\angle C={45}^{\circ }$ makes the triangle isosceles.

1. $BC=AC\cos {45}^{\circ }=4\times \frac{1}{\sqrt{2}}$
2. $=\frac{4}{\sqrt{2}}=2\sqrt{2}\text{cm}$

Yaad rakhein: In a ${45}^{\circ }$-${45}^{\circ }$-${90}^{\circ }$ triangle each leg $=\frac{\text{hypotenuse}}{\sqrt{2}}$.

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

Concept: Number in a school $=$ its percentage of 2000.

1. A $=13\%$ of 2000 $=260$
2. C $=55\%$ of 2000 $=1100$
3. Average $=\frac{260+1100}{2}=680$

Trick: Average of percentages first: $\frac{13+55}{2}=34\%$ of 2000 $=680$.

Answer: (D) 680