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

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

1. Let HCF $=h$, LCM $=5h$
2. $5h^2=20480\Rightarrow h^2=4096\Rightarrow h=64$
3. LCM $=320$

Trick: Check: $64\times 320=20480$ ✓.

Answer: (C) 64 and 320

Concept: If the smallest number divides all the others, it is the HCF.

1. $250=2\times 125$, $750=6\times 125$
2. So $125$ divides all three: HCF $=125$

Trick: First test the smallest number itself.

Answer: (C) $125$

Concept: LCM = product of highest powers of all prime factors.

1. $8=2^3$, $12=2^2\times 3$, $18=2\times 3^2$
2. LCM $=2^3\times 3^2=72$

Trick: $72$ is the first multiple of $18$ that $8$ also divides.

Answer: (C) $72$

Concept: Read the right column for each organisation and add.

1. P + R in 2019 $=69+63=132$
2. Q + R in 2020 $=74+90=164$
3. $132:164=33:41$ (divide by 4)

Trick: Read the year columns carefully – mixing 2019 and 2020 is the usual slip.

Answer: (B) $33:41$

Concept: Expenditure corresponds to $3$ parts; find one part.

1. $3$ parts $=5400\Rightarrow 1$ part $=1800$
2. Income $=7\times 1800=12600$

Trick: Income $=\frac{7}{3}\times$ expenditure.

Answer: (A) ₹12,600

Concept: Turn the equality into a ratio of incomes.

1. $0.3R=0.4S\Rightarrow R=\frac{4}{3}S$
2. $0.5R=\frac{1}{2}\times \frac{4}{3}S=\frac{2}{3}S$
3. $\frac{2}{3}=66\frac{2}{3}\%$

Trick: Scale: $30\%\to 40\%$, so $50\%\to \frac{40}{30}\times 50=66\frac{2}{3}\%$.

Answer: (D) $66\frac{2}{3}\%$

Concept: Time is proportional to the work done (same person).

1. $\frac{2}{3}$ work in $12$ days $\Rightarrow$ full work in $12\times \frac{3}{2}=18$ days
2. $\frac{1}{6}$ work $=\frac{18}{6}=3$ days

Trick: Direct: $12\times \frac{1/6}{2/3}=12\times \frac{1}{4}=3$.

Answer: (A) $3$

Concept: A chord equal to the radius makes an equilateral triangle with the centre, so it subtends ${60}^{\circ }$ at the centre.

1. Angle at a point on the major arc $=\frac{{60}^{\circ }}{2}={30}^{\circ }$
2. Angle at a point on the minor arc $={180}^{\circ }-{30}^{\circ }={150}^{\circ }$ (cyclic quadrilateral)
3. Difference $={150}^{\circ }-{30}^{\circ }={120}^{\circ }$

Trick: Angle in major segment $=\frac{\theta }{2}$, minor segment $={180}^{\circ }-\frac{\theta }{2}$.

Answer: (B) ${120}^{\circ }$

Concept: The line of centres is the perpendicular bisector of the common chord, so half-chord $=12$ forms right triangles with each radius.

1. Radii: $15$ and $13$
2. $O_{1}M=\sqrt{{15}^2-{12}^2}=9$
3. $O_{2}M=\sqrt{{13}^2-{12}^2}=5$
4. $O_1{O}_2=9+5=14\text{cm}$

Trick: Triplets $(9,12,15)$ and $(5,12,13)$ appear instantly.

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

Concept: Compare the 'Total' column (it equals the sum of the five courses).

1. Totals: A $10340$, B $11000$, C $10340$, D $10160$, E $8300$
2. Largest $=11000$ (city B)

Trick: Verify one row: B $=1350+2650+5100+1050+850=11000$ ✓.

Answer: (D) B

Concept: Average $=\frac{\text{sum of the three increments}}{3}$.

1. Kamal: May ₹65, June ₹55, July ₹75
2. Sum $=195$
3. Average $=\frac{195}{3}=65$

Trick: Deviations from 65: $0,-10,+10$ cancel, so average $=65$.

Answer: (D) ₹65

Concept: Volume of material $=\pi (R^2-r^2)h$, inner radius = outer radius − thickness.

1. $R=10$, $r=10-3=7$, $h=49$
2. $V=\frac{22}{7}(100-49)\times 49=\frac{22}{7}\times 51\times 49$
3. $=22\times 51\times 7=7854{\text{cm}}^3$

Trick: $R^2-r^2=(R+r)(R-r)=17\times 3=51$.

Answer: (D) $7854{\text{cm}}^3$

Concept: Use standard values $\tan {45}^{\circ }=1$, $\cos {30}^{\circ }=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$.

1. $2{\tan }^2{45}^{\circ }=2$
2. ${\cos }^2{30}^{\circ }-{\sin }^2{60}^{\circ }=\frac{3}{4}-\frac{3}{4}=0$
3. Value $=2$

Trick: $\cos {30}^{\circ }=\sin {60}^{\circ }$ (complementary angles), so those terms cancel.

Answer: (A) $2$

Concept: Compare the money received with the cost of the quantity actually given.

1. Gives $800\text{g}$ for ₹40
2. Cost of $800\text{g}=0.8\times 36=28.8$
3. Profit $\%=\frac{40-28.8}{28.8}\times 100=\frac{11.2}{28.8}\times 100\approx 38.89\%$

Trick: $(\frac{40}{36}\times \frac{1000}{800}-1)\times 100=(\frac{25}{18}-1)\times 100=38.89\%$.

Answer: (D) $38.89\%$

Concept: Apply Pythagoras: $P{R}^2=P{Q}^2+Q{R}^2$ to relate $x$ and $y$.

1. $(x+y{)}^2=(x-y{)}^2+x^2\Rightarrow 4xy=x^2\Rightarrow x=4y$
2. $PQ=3y$, $QR=4y$
3. $QS=PQ=3y$, $SR=4y-3y=y$
4. $QS:SR=3:1$

Trick: Sides become $3y,4y,5y$ – the 3-4-5 triangle.

Answer: (A) $3:1$

Concept: Add the required cells and simplify the ratio.

1. A + B in January $=212+168=380$
2. D + E in March $=255+245=500$
3. $380:500=19:25$

Trick: Divide both by 20.

Answer: (D) $19:25$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$; take total distance $=100\text{km}$.

1. $\frac{25}{25}=1\text{h}$, $\frac{50}{50}=1\text{h}$, $\frac{25}{12.5}=2\text{h}$
2. Total time $=4\text{h}$
3. Average $=\frac{100}{4}=25\text{km/h}$

Trick: Choose 100 km for percentage distances.

Answer: (C) $25\text{km/h}$

Concept: $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$ and $ab=\text{HCF}\times \text{LCM}$.

1. $ab=5\times 60=300$
2. $\frac{1}{a}+\frac{1}{b}=\frac{60}{300}=\frac{1}{5}$

Trick: Sum of reciprocals $=\frac{\text{sum}}{\text{HCF}\times \text{LCM}}$.

Answer: (D) $\frac{1}{5}$

Concept: Values are in hundreds – convert the difference to actual numbers.

1. 7AH in 2021 $=40$, in 2022 $=45$ (hundreds)
2. Difference $=5$ hundreds $=500$

Trick: Always apply the unit ('in hundreds') at the end.

Answer: (D) $500$

Concept: Numbers in ratio $2:7$ with HCF $9$ are $2\times 9$ and $7\times 9$.

1. Numbers: $18$ and $63$
2. Check LCM: $9\times 2\times 7=126$ ✓
3. Larger $=63$

Trick: HCF × ratio terms gives the numbers directly.

Answer: (B) $63$