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

Concept: BODMAS: 'of', then $÷$/$\times$ left to right, then $+$/$-$.

1. $7.2\times 4.1=29.52$; $29.52÷12.3=2.4$
2. $22.5\text{ of }\frac{1}{25}=0.9$
3. $2.4+0.9-0.1=3.2$

Trick: $\frac{4.1}{12.3}=\frac{1}{3}$, so first term $=\frac{7.2}{3}=2.4$.

Answer: (D) $3.2$

Concept: $a^3+b^3+c^3-3abc=(a+b+c)[(a+b+c{)}^2-3(ab+bc+ca)]$.

1. $(a+b+c{)}^2=225$
2. $225-3\times 35=120$
3. $15\times 120=1800$

Yaad rakhein: $a^2+b^2+c^2-ab-bc-ca=(a+b+c{)}^2-3(ab+bc+ca)$.

Answer: (C) $1800$

Concept: Simplify each bracket, then $\times$ and $÷$ left to right.

1. $\frac{3}{4}+\frac{5}{4}=2$
2. $\frac{9}{5}-\frac{4}{5}=1$
3. $\frac{3}{2}+\frac{1}{2}=2$
4. $2\times 1÷2=1$

Trick: Each bracket gives a whole number – no LCM needed.

Answer: (A) $1$

Concept: SP $=\text{MP}\times (1-d_1)(1-d_2)$.

1. SP $=\text{MP}\times 0.76\times 0.70=0.532\text{MP}$
2. $0.532\text{MP}=7700.70$
3. MP $=\frac{7700.70}{0.532}=14475$

Trick: Option check: $14475\times 0.532=7700.70$ ✔

Answer: (C) ₹14475

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

1. $d=13$, $r_1+r_2=5$
2. $\sqrt{169-25}=\sqrt{144}=12\text{cm}$

Yaad rakhein: Transverse uses $(r_1+r_2)$; direct common tangent uses $(r_1-r_2)$.

Answer: (D) $12\text{cm}$

Concept: $\%\text{ increase}=\frac{\text{increase}}{\text{old value}}\times 100$.

1. 2019: $32$ lakh, 2020: $34$ lakh
2. Increase $=2$
3. $\frac{2}{32}\times 100=6.25\%=6\frac{1}{4}\%$

Trick: $\frac{1}{16}=6.25\%$.

Answer: (A) $6\frac{1}{4}\%$

Concept: Remainder on dividing $p(x)$ by $(x-5)$ is $p(5)$; add its negative.

1. $p(5)=125-75+20-75=-5$
2. Add $5$ to make remainder $0$

Trick: Remainder $-5$ → add $+5$.

Answer: (D) $5$

Concept: Start with $D=100$ and move up the chain.

1. $C=120$
2. $B=120\times 0.7=84$
3. $A=84\times 1.25=105$
4. A is $5\%$ more than D

Trick: $1.2\times 0.7\times 1.25=1.05$.

Answer: (A) $5\%$

Concept: Volume of cone $=\frac{1}{3}\pi r^{2}h$.

1. $\frac{1}{3}\times \frac{22}{7}\times 7^2\times 9$
2. $=22\times 7\times 3=462$ cubic ft

Trick: Cone is one third of the cylinder: $\frac{1386}{3}=462$ (1386 is the cylinder trap).

Answer: (B) $462$ cubic ft.

Concept: Lateral (curved) surface area of a cone $=\pi rl$.

1. $\frac{22}{7}\times r\times 4.2=13.2$
2. $13.2r=13.2\Rightarrow r=1$

Trick: $\frac{22}{7}\times 4.2=13.2$ exactly, so $r=1$.

Answer: (D) 1 unit

Concept: In any triangle, the straight side between two vertices is the shortest path, so going around through the third vertex is always longer.

1. For sides $a,b,c$: $a+b>c$, $b+c>a$, $c+a>b$.
2. Equality $a+b=c$ would flatten the triangle into a straight line, so it is not allowed.
3. Hence the sum of any two sides is always greater than the third side.

Trick: Sum of two sides $>$ third side $>$ difference of the two sides.

Answer: (B) greater than the third side

Concept: Marks in a subject $=\frac{\text{percentage}}{100}\times \text{maximum marks}$; add over the five subjects.

1. Hindi: $82\%\times 150=123$; English: $64\%\times 100=64$
2. Geography: $56\%\times 175=98$; Mathematics: $70\%\times 120=84$
3. Physics: $80\%\times 75=60$
4. Aggregate $=123+64+98+84+60=429$

Trick: Use friendly forms: $56\%$ of $175=0.56\times 175=98$, $80\%$ of $75=60$ — no long multiplication needed.

Answer: (B) 429

Concept: When a solid is melted and recast, volume is conserved. Flat surfaces: cylinder has two circles, each cone has one circular base.

1. Cylinder volume $=\pi \times 7^2\times 12=588\pi$; its flat area $=2\pi \times 7^2=98\pi$.
2. Cone 1: $\frac{1}{3}\pi r_1^2\times 12=\frac{1}{3}\times 588\pi \Rightarrow r_1^2=49$.
3. Cone 2: $\frac{1}{3}\pi r_2^2\times 12=\frac{2}{3}\times 588\pi \Rightarrow r_2^2=98$.
4. Flat area of cones $=\pi (49+98)=147\pi$.
5. Change $=\frac{147\pi -98\pi }{98\pi }\times 100=50\%$ (increase).

Trick: With the same height, a cone needs 3 times the base area of a cylinder of equal volume: $3\times 49\pi =147\pi$ vs $98\pi$, i.e. $\frac{3}{2}$ times.

Answer: (C) 50%

Concept: Perimeter of a triangle = sum of its three sides.

1. Sides: 13 cm, 12 cm, 5 cm.
2. Perimeter $=13+12+5=30\text{cm}$.
3. (Since $5^2+{12}^2={13}^2$, it is a right triangle — a useful check.)

Yaad rakhein: 5–12–13 is a Pythagorean triplet; its perimeter is always 30 units.

Answer: (C) 30 cm

Concept: BODMAS: brackets first; inside, 'of' acts as multiplication; then ÷ and × from left to right, then + and −.

1. $(4.2\times 2.5÷3.5)=10.5÷3.5=3$
2. $(3.6÷2.4+2.5\times 3÷7.5)=1.5+1=2.5$
3. $(6-3\times 2+4\times 3÷6)=6-6+2=2$
4. Total $=3-2.5+2=2.5$

Trick: Look for neat pairs: $10.5÷3.5=3$, $7.5÷7.5=1$ — decimals cancel quickly.

Answer: (D) 2.5

Concept: Area of trapezium $=\frac{1}{2}\times (\text{sum of parallel sides})\times \text{height}$.

1. $1188=\frac{1}{2}\times (a+b)\times 18$
2. $1188=9(a+b)$
3. $a+b=\frac{1188}{9}=132\text{cm}$

Trick: Divide area by half the height directly: $1188÷9=132$.

Answer: (C) 132 cm

Concept: Use known values: $\sin {90}^{\circ }=1$ and $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$.

1. $\sin (x+y)=1\Rightarrow x+y={90}^{\circ }$
2. $\cos (x-y)=\frac{\sqrt{3}}{2}\Rightarrow x-y={30}^{\circ }$
3. Subtracting: $2y={60}^{\circ }\Rightarrow y={30}^{\circ }$ (and $x={60}^{\circ }$).

Trick: Sum and difference given → $y=\frac{\text{sum}-\text{difference}}{2}=\frac{{90}^{\circ }-{30}^{\circ }}{2}$.

Answer: (B) ${30}^{\circ }$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (not the average of the speeds).

1. Distance 1 $=8\times 4=32\text{km}$; distance 2 $=4\times 8=32\text{km}$
2. Total distance $=64\text{km}$, total time $=8+4=12$ h
3. Average speed $=\frac{64}{12}=\frac{16}{3}=5\frac{1}{3}\text{km/h}$

Trick: When times are given, average speed $=\frac{t_1{s}_1+t_2{s}_2}{t_1+t_2}=\frac{32+32}{12}$.

Answer: (B) $5\frac{1}{3}$

Concept: Required $\%=\frac{\text{Pune (2018)}}{\text{total of all towns (2018)}}\times 100$.

1. Total 2018 $=874+850+179+857+820=3580$
2. Pune 2018 $=179$
3. $\frac{179}{3580}\times 100=5\%$

Trick: Check with options: $5\%$ of $3580=179$ exactly.

Answer: (D) 5%

Concept: Remainder theorem: remainder of $f(x)$ divided by $(x-2)$ is $f(2)$.

1. $f(2)=2(8)+4a+6-5=17+4a$
2. $g(2)=8+4-4+a=8+a$
3. Equal remainders: $17+4a=8+a\Rightarrow a=-3$
4. Remainder $=8+a=8-3=5$

Trick: Put $x=2$ in both, equate, solve for $a$, then substitute back into the simpler one.

Answer: (D) 5