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

Concept: Average $=\frac{\text{sum of values}}{\text{number of values}}$. Read the 2018 (blue) bar of each branch.

1. 2018 sales: $80,75,95,85,75,70$.
2. Total $=80+75+95+85+75+70=480$.
3. Average $=\frac{480}{6}=80$.

Trick: Take 80 as base: deviations $0,-5,+15,+5,-5,-10$ sum to $0$, so the average is exactly $80$.

Answer: (D) $80$

Concept: After successive discounts, $\text{SP}=\text{MP}\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})(1-\frac{d_3}{100})$.

1. $\text{SP}=1250\times 0.95\times 0.85\times 0.80$
2. $1250\times 0.80=1000$; $1000\times 0.85=850$
3. $850\times 0.95=807.50$, so SP = ₹807.50.

Trick: Multiply the easiest factor first ($1250\times 0.8=1000$) – the order of successive discounts does not matter.

Answer: (B) ₹807.50

Concept: Square: $\text{diagonal}=a\sqrt{2}$; triangle area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Diagonal $6\sqrt{2}\Rightarrow$ side $=6$ cm, perimeter $=4\times 6=24$ cm = base.
2. Area $144\Rightarrow$ side $=\sqrt{144}=12$ cm = height.
3. Area $=\frac{1}{2}\times 24\times 12=144{\text{cm}}^2$.

Trick: Diagonal $a\sqrt{2}$ → just drop the $\sqrt{2}$ to get the side.

Answer: (C) $144$

Concept: Let the side of the square be $x$; the rectangle is $(x+10)$ by $(x-5)$ (breadth is 5 cm less than the side). Equate the areas.

1. $x^2=(x+10)(x-5)=x^2+5x-50$
2. $5x=50\Rightarrow x=10$ cm
3. Rectangle: $20$ cm $\times 5$ cm, perimeter $=2(20+5)=50$ cm.

Trick: Check: square $10\times 10=100$ and rectangle $20\times 5=100$ – equal areas confirm the answer.

Answer: (A) $50$

Concept: One-day work of Q $=\frac{1}{12}-\frac{1}{18}$.

1. Q's one-day work $=\frac{3-2}{36}=\frac{1}{36}$ → Q alone takes $36$ days.
2. Two-thirds of the work: $\frac{2}{3}\times 36=24$ days.

Trick: LCM method: total work $=36$ units, P+Q $=3$/day, P $=2$/day ⇒ Q $=1$/day; $\frac{2}{3}$ of 36 $=24$ units → 24 days.

Answer: (C) $24$

Concept: “Less than 250” means all class intervals that end at or before 250.

1. Intervals below 250: $150\text{–}200$ and $200\text{–}250$.
2. Students $=30+45=75$.

Trick: Do not include the $250\text{–}300$ bar – it starts at 250, so those students scored 250 or more.

Answer: (B) $75$

Concept: If $A+B={90}^{\circ }$, then $\sin B=\cos A$, so ${\sin }^{2}A+{\sin }^{2}B=1$ and ${\cos }^{2}A+{\cos }^{2}B=1$.

1. ${52}^{\circ }+{38}^{\circ }={90}^{\circ }$: numerator $=({\sin }^2{52}^{\circ }+{\sin }^2{38}^{\circ })+2=1+2=3$.
2. ${43}^{\circ }+{47}^{\circ }={90}^{\circ }$: denominator $=4({\cos }^2{43}^{\circ }+{\cos }^2{47}^{\circ })-5=4-5=-1$.
3. Value $=\frac{3}{-1}=-3$.

Trick: Spot angle pairs adding to ${90}^{\circ }$ – each ${\sin }^2$ or ${\cos }^2$ pair becomes 1.

Answer: (D) $-3$

Concept: Write both selling prices as fractions of the marked price $M$.

1. 35% discount: ${\text{SP}}_1=0.65M$.
2. 20% and 15%: ${\text{SP}}_2=0.80\times 0.85M=0.68M$.
3. $0.65M+0.68M=1.33M=1995\Rightarrow M=\frac{1995}{1.33}=1500$.

Trick: Option check: $1500\times 1.33=1995$ ✓.

Answer: (C) ₹1,500

Concept: Convert “$Y\%$ of 50” to a multiple of $Y$, then substitute $Y=0.49X$.

1. $Y\%\text{ of }50=\frac{Y}{100}\times 50=\frac{Y}{2}$
2. $\frac{Y}{2}=\frac{0.49X}{2}=0.245X$
3. $0.245X=24.5\%$ of $X$.

Trick: $a\%$ of $b=b\%$ of $a$: $Y\%$ of 50 $=50\%$ of $Y$ = half of $Y$.

Answer: (D) 24.5% of X

Concept: Tangent-secant theorem: $P{C}^2=PA\times PB$.

1. ${12}^2=PA\times 10\Rightarrow PA=\frac{144}{10}=14.4$ cm.
2. B lies between A and P, so $AB=PA-PB=14.4-10=4.4$ cm.

Trick: Square of tangent = (whole secant) × (outside part).

Answer: (D) $4.4$

Concept: The 60° is the angle the ladder makes at its foot (with the ground); the distance from the wall is the adjacent side, so $\cos {60}^{\circ }=\frac{\text{base}}{\text{ladder}}$.

1. $\cos {60}^{\circ }=\frac{3.6}{L}$
2. $\frac{1}{2}=\frac{3.6}{L}\Rightarrow L=7.2$ m.

Trick: In a 30-60-90 triangle the side next to the ${60}^{\circ }$ angle is half the hypotenuse.

Answer: (D) $7.2$

Concept: Leak rate = pump rate − net rate.

1. Pump: $\frac{1}{3}$ per hour; net with leak: $\frac{1}{10/3}=\frac{3}{10}$ per hour.
2. Leak $=\frac{1}{3}-\frac{3}{10}=\frac{10-9}{30}=\frac{1}{30}$.
3. Leak empties the tank in $30$ hours.

Trick: Leak time $=\frac{a\times b}{b-a}=\frac{3\times \frac{10}{3}}{\frac{10}{3}-3}=\frac{10}{1/3}=30$.

Answer: (C) $30$

Concept: For the incentre I: $\angle BIC={90}^{\circ }+\frac{\angle A}{2}$.

1. $\angle BIC={90}^{\circ }+\frac{{68}^{\circ }}{2}$
2. $={90}^{\circ }+{34}^{\circ }={124}^{\circ }$.

Yaad rakhein: Incentre: ${90}^{\circ }+\frac{A}{2}$; excentre: ${90}^{\circ }-\frac{A}{2}$; circumcentre: $2A$.

Answer: (A) ${124}^{\circ }$

Concept: If SP of $x$ articles = CP of $y$ articles, then $\text{result}\%=\frac{y-x}{x}\times 100$ (negative = loss).

1. Let CP of 1 article = ₹1: CP of 50 = ₹50, SP of 50 = ₹42.
2. Loss = ₹8 on ₹50.
3. Loss% $=\frac{8}{50}\times 100=16\%$.

Trick: Fewer articles' CP received (42 < 50) means loss.

Answer: (B) 16% loss

Concept: Compare the percentages directly – the total (₹1,680 billion) cancels.

1. Defence $=12\%$, Education $=6\%$.
2. More by $\frac{12-6}{6}\times 100=100\%$.

Trick: Defence is exactly double Education → 100% more.

Answer: (D) 100%

Concept: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.

1. $xy+yz+zx=\frac{{13}^2-133}{2}=\frac{36}{2}=18$.
2. $847-3xyz=13(133-18)=13\times 115=1495$.
3. $-3xyz=648\Rightarrow xyz=-216$, so $\sqrt[3]{xyz}=-6$.

Trick: A negative product is possible, so a negative cube root is not a mistake: $(-6{)}^3=-216$.

Answer: (D) $-6$

Concept: Profit $\propto$ capital $\times$ time, so capital $\propto \frac{\text{profit}}{\text{time}}$.

1. Capitals $=\frac{8}{7}:\frac{7}{8}:\frac{5}{14}$.
2. Multiply by LCM 56: $64:49:20$.

Trick: Largest profit with the shortest time ⇒ largest capital – only option (D) starts with the largest number 64.

Answer: (D) 64 : 49 : 20

Concept: Multiply the multiplying factors of each change.

1. Factor $=1.40\times 0.75\times 1.15\times 0.80$.
2. $1.40\times 0.75=1.05$; $1.15\times 0.80=0.92$.
3. $1.05\times 0.92=0.966$ → change $=-3.4\%$.

Trick: Start with 100: $100\to 140\to 105\to 120.75\to 96.6$ → 3.4% decrease.

Answer: (D) 3.4% decrease

Concept: Convert averages to sums; the common days cancel.

1. $M+W+F=3\times 41=123$; $W+F+Th=3\times 42=126$.
2. Subtract: $Th-M=3\Rightarrow M=43-3={40}^{\circ }\text{C}$.

Trick: Difference of averages $\times$ 3 = difference of the odd days: $(42-41)\times 3=3$.

Answer: (D) ${40}^{\circ }\text{C}$

Concept: “Less than 2019” → divide the difference by the 2019 total.

1. 2018 total $=80+75+95+85+75+70=480$.
2. 2019 total $=105+65+110+95+95+80=550$.
3. Less by $\frac{70}{550}\times 100\approx 12.7\%$.

Trick: Base = the quantity after “than” (here 2019).

Answer: (A) 12.7%