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

Concept: Take total work = LCM of the given days and work in units per day.

1. Total work $=\text{LCM}(42,27)=378$ units
2. Jeevan $=\frac{378}{42}=9$/day; Rishi + Jeevan $=\frac{378}{27}=14$/day
3. Rishi $=14-9=5$/day ⇒ $\frac{378}{5}=75\frac{3}{5}$ days

Trick: Direct: $\frac{42\times 27}{42-27}=\frac{1134}{15}=75.6$ days.

Answer: (D) $75\frac{3}{5}$

Concept: Same time ⇒ distance $\propto$ speed.

1. Time $=\frac{28}{8}=3.5\text{h}$
2. Distance at $6\text{km/h}=6\times 3.5=21\text{km}$

Trick: Ratio method: $28\times \frac{6}{8}=21$.

Answer: (D) 21

Concept: Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$, with $\sqrt{3}\approx 1.732$.

1. $\frac{1.732}{4}{a}^2=173.2$
2. $a^2=\frac{173.2\times 4}{1.732}=400$
3. $a=20\text{cm}$

Trick: $173.2=100\sqrt{3}$ ⇒ $\frac{a^2}{4}=100$ ⇒ $a=20$.

Answer: (A) 20 cm

Concept: Length of direct common tangent $=\sqrt{d^2-(r_1-r_2{)}^2}$.

1. $d=17$, $r_1=15$, $r_2=7$
2. $PQ=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15\text{cm}$

Trick: Direct tangent → minus $(r_1-r_2)$; transverse tangent → plus $(r_1+r_2)$. 8-15-17 is a triplet.

Answer: (A) 15 cm

Concept: Convert everyone to one unit (boys) using the given equivalence.

1. 4 men = 6 boys ⇒ 1 man $=1.5$ boys
2. Total work $=6\times 20=120$ boy-days
3. 6 men + 11 boys $=9+11=20$ boys ⇒ $\frac{120}{20}=6$ days

Trick: $M_1{D}_1=M_2{D}_2$ after converting to the same unit.

Answer: (B) 6

Concept: Net profit = profit on X − loss on Y; use CP of X = CP of Y − 90.

1. Let CP of Y $=y$, CP of X $=y-90$
2. $0.23(y-90)-0.13y=180\Rightarrow 0.10y-20.7=180$
3. $0.1y=200.7\Rightarrow y=2007$

Trick: Check: $0.23\times 1917=440.91$, $0.13\times 2007=260.91$; difference $=180$.

Answer: (B) 2,007

Concept: $\cos \theta =\sin ({90}^{\circ }-\theta )$.

1. $\cos {15}^{\circ }=\sin {75}^{\circ }$
2. $\sin 2A=\sin {75}^{\circ }\Rightarrow 2A={75}^{\circ }\Rightarrow A={37.5}^{\circ }$

Trick: If $\sin X=\cos Y$ then $X+Y={90}^{\circ }$: $2A+{15}^{\circ }={90}^{\circ }$.

Answer: (B) ${37.5}^{\circ }$

Concept: Find valid votes first; then use sum and difference of the two candidates' votes.

1. Votes cast $=80\%$ of 9000 $=7200$; valid $=80\%$ of 7200 $=5760$
2. Ajay + Mohan $=5760$, Ajay − Mohan $=1152$
3. Ajay $=\frac{5760+1152}{2}=3456$

Trick: Winner = (sum + difference) ÷ 2; loser = (sum − difference) ÷ 2 (= 2304, a trap option).

Answer: (A) 3456

Concept: Third proportional to $a$ and $b$ is $c$ where $a:b=b:c$, i.e. $c=\frac{b^2}{a}$.

1. $c=\frac{{20}^2}{16}=\frac{400}{16}=25$

Trick: Third proportional $=\frac{(\text{second}{)}^2}{\text{first}}$; mean proportional $=\sqrt{ab}$.

Answer: (A) 25

Concept: Area of a sector $=\frac{1}{2}\times l\times r$ (l = arc length).

1. $l=4.6\text{cm}$, $r=8\text{cm}$
2. Area $=\frac{1}{2}\times 4.6\times 8=18.4{\text{cm}}^2$

Trick: No angle needed when arc length is given — just $\frac{lr}{2}$ (like a triangle with base $l$, height $r$).

Answer: (C) $18.4{\text{cm}}^2$

Concept: Percentage change $=\frac{\text{2021}-\text{2020}}{\text{2020}}\times 100$.

1. Pen $\frac{3}{15}=20\%$, Pencil $\frac{5}{16}=31.25\%$, Notebook $\frac{10}{25}=40\%$
2. Pages $\frac{6}{20}=30\%$, Erasers $\frac{6}{18}=33.3\%$, Charts $\frac{10}{27}=37\%$, Geometry $\frac{12}{32}=37.5\%$
3. Highest = Notebook (40%).

Trick: Geometry has the largest absolute rise (12) but not the largest % rise — divide by the base.

Answer: (B) Notebook

Concept: Volume is conserved: number of pieces = volume of sphere ÷ volume of one hemisphere.

1. Sphere $=\frac{4}{3}\pi (12{)}^3$, hemisphere $=\frac{2}{3}\pi (4{)}^3$
2. $n=\frac{\frac{4}{3}\pi \times 1728}{\frac{2}{3}\pi \times 64}=2\times 27=54$

Trick: $n=2\times {\left(\frac{R}{r}\right)}^3=2\times 3^3=54$; option 27 forgets the factor 2.

Answer: (A) 54

Concept: Hemisphere: $\text{CSA}=2\pi r^2$, $\text{TSA}=3\pi r^2$.

1. $\text{CSA}=2\times \frac{22}{7}\times 49=308{\text{cm}}^2$
2. $\text{TSA}=3\times \frac{22}{7}\times 49=462{\text{cm}}^2$

Trick: With $r=7$: $\pi r^2=154$; CSA $=2\times 154$, TSA $=3\times 154$.

Answer: (D) CSA – $308{\text{cm}}^2$; TSA – $462{\text{cm}}^2$

Concept: Value of a sector $=\frac{\text{angle}}{{360}^{\circ }}\times \text{total}$.

1. UK − Australia $={125}^{\circ }-{75}^{\circ }={50}^{\circ }$
2. $\frac{50}{360}\times 14400=50\times 40=2000$ billion

Trick: $1^{\circ }=\frac{14400}{360}=40$ billion; multiply the angle difference by 40.

Answer: (B) 2,000

Concept: Surface area $=4\pi r^2\propto r^2$.

1. Radius doubled ⇒ area becomes $2^2=4$ times.
2. Increase $=4-1=3$ times $=300\%$

Trick: Increase % $=(k^2-1)\times 100$; 400% is the new area, not the increase.

Answer: (C) $300\%$

Concept: Add the required cells, form the ratio and reduce it.

1. D (Friday + Sunday) = 206 + 326 = 532
2. A (Friday + Saturday) = 350 + 420 = 770
3. $532:770$ — divide by 14 → $38:55$

Trick: Keep the asked order: D first, A second (55 : 38 is the reversed trap).

Answer: (B) 38 : 55

Concept: Third proportional to $a$ and $b$ $=\frac{b^2}{a}$.

1. $a=23$, $b=31$
2. $c=\frac{{31}^2}{23}=\frac{961}{23}$

Trick: Square the second number, divide by the first.

Answer: (B) $\frac{961}{23}$

Concept: Total time $=\frac{d}{B+S}+\frac{d}{B-S}$.

1. $\frac{35}{20+S}+\frac{35}{20-S}=8$
2. Try $S=15$: $\frac{35}{35}+\frac{35}{5}=1+7=8$ ✓
3. (Algebra: $\frac{1400}{400-S^2}=8\Rightarrow S^2=225\Rightarrow S=15$)

Trick: Option testing is fastest; $S$ must be less than 20, which rules out (A).

Answer: (C) $15\text{km/h}$

Concept: HCF = product of the common prime factors with the lowest powers.

1. $12=2^2\times 3$, $18=2\times 3^2$, $42=2\times 3\times 7$
2. $\text{HCF}=2\times 3=6$

Trick: HCF can never exceed the smallest number (12) — and 12 does not divide 18.

Answer: (C) 6

Concept: When only milk is added, the quantity of water stays fixed.

1. Water $=60\%$ of 60 $=36\text{L}$
2. In new mixture water $=40\%$ ⇒ total $=\frac{36}{0.4}=90\text{L}$
3. Milk to add $=90-60=30\text{L}$

Trick: Shortcut: $\text{added}=\frac{60\times (60-40)}{100-60}=30$.

Answer: (A) 30 litres