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

Concept: Solve brackets first, then 'of'/division/multiplication from left to right, then addition/subtraction.

1. Numerator: $\frac{2}{5}\times 75=30$, $30÷6=5$, so $61+5=66$.
2. Bracket: $3\times 6÷\frac{1}{3}=18\times 3=54$; then $54\times \frac{1}{6}=9$.
3. Numerator $=66-9=57$.
4. Denominator: $5+\frac{2}{3}\times 21=5+14=19$.
5. Value $=\frac{57}{19}=3$.

Trick: 57 is a multiple of 19 ($19\times 3$) – check this quickly to confirm.

Answer: (A) $3$

Concept: When equal weights are mixed, the fraction of a metal in the new alloy is the average of its fractions in the two alloys.

1. Alloy 1: A : B : C = 2 : 3 : 1, so B $=\frac{3}{6}=\frac{1}{2}$.
2. Alloy 2: B : C : D = 5 : 4 : 3, so B $=\frac{5}{12}$.
3. Take 12 kg of each (LCM): B $=6+5=11$ kg out of $24$ kg.
4. Part of B $=\frac{11}{24}$.

Trick: For equal weights, make both totals equal to the LCM (here 12) and simply add the B parts.

Answer: (D) $\frac{11}{24}$

Concept: Use $(a+b)(b-a)=b^2-a^2$ and ${\cos }^2\theta =1-{\sin }^2\theta$.

1. Numerator $=(\sin y+\sin x)(\sin y-\sin x)={\sin }^{2}y-{\sin }^{2}x$.
2. Denominator $={\cos }^{2}y-{\cos }^{2}x=(1-{\sin }^{2}y)-(1-{\sin }^{2}x)={\sin }^{2}x-{\sin }^{2}y$.
3. Ratio $=\frac{{\sin }^{2}y-{\sin }^{2}x}{-({\sin }^{2}y-{\sin }^{2}x)}=-1$.

Trick: Put $x=0^{\circ },y={90}^{\circ }$: $\frac{(0+1)(1-0)}{(1+0)(0-1)}=-1$.

Answer: (A) $-1$

Concept: For an equilateral triangle of side $a$: circumradius $R=\frac{a}{\sqrt{3}}$, inradius $r=\frac{a}{2\sqrt{3}}$. Required area $=\pi (R^2-r^2)$.

1. $R=\frac{18}{\sqrt{3}}=6\sqrt{3}$, so $R^2=108$.
2. $r=\frac{18}{2\sqrt{3}}=3\sqrt{3}$, so $r^2=27$.
3. Area $=\frac{22}{7}\times (108-27)=\frac{22}{7}\times 81=\frac{1782}{7}$.
4. $\frac{1782}{7}=254\frac{4}{7}{\text{cm}}^2$.

Trick: $R^2-r^2=\frac{a^2}{4}$ (half-side squared), so area $=\pi \times 9^2=81\pi$ directly.

Answer: (A) $254\frac{4}{7}$

Concept: Increase in total cost = (share of the component) × (increase in its price).

1. Let cost of sweet $=1$; sugar share $=\frac{1}{3}$.
2. New sugar price $=\frac{7}{6}$ of old, i.e. increase $=\frac{1}{6}$.
3. Increase in sweet cost $=\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}$.
4. Percentage $=\frac{1}{18}\times 100=\frac{100}{18}\%$.

Trick: Share × rise: $\frac{1}{3}\times 16\frac{2}{3}\%=5\frac{5}{9}\%$.

Answer: (C) $\frac{100}{18}\%$

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

1. Total $=150+210+170+190=720$, so 1 movie $=\frac{{360}^{\circ }}{720}={0.5}^{\circ }$.
2. Jan-Mar + Jul-Sep $=150+170=320$ movies $={160}^{\circ }$.
3. Apr-Jun + Oct-Dec $=210+190=400$ movies $={200}^{\circ }$.
4. Difference $={200}^{\circ }-{160}^{\circ }={40}^{\circ }$, so the first is ${40}^{\circ }$ less.

Trick: Find the difference in values first ($400-320=80$), then convert once: $80\times {0.5}^{\circ }={40}^{\circ }$.

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

Concept: Invalid votes = total votes × polled % × invalid % (invalid % = 100 − valid %).

1. P: $40\times 0.80\times 0.10=3.2$ thousand.
2. S: $80\times 0.80\times 0.15=9.6$ thousand.
3. Ratio $=3.2:9.6=1:3$.

Trick: Both have 80% polling, so compare $40\times 10:80\times 15=400:1200=1:3$.

Answer: (A) 1 : 3

Concept: 'of' means multiplication and is done before ordinary division/subtraction.

1. Numerator: $\frac{2}{7}\text{ of }63=18$, so $53-18-9=26$.
2. Denominator: $96-31=65$; $\frac{2}{5}\text{ of }65=26$; $72-26=46$.
3. Value $=\frac{26}{46}=\frac{13}{23}$.

Trick: Denominator 23 appears in all options – just simplify the numerator over 46.

Answer: (A) $\frac{13}{23}$

Concept: Profit $\%=\frac{\text{SP of goods given}-\text{CP of goods given}}{\text{CP of goods given}}\times 100$.

1. Let CP of 1 g $=$ ₹1. He charges for 200 g: price of 200 g at 24% loss $=0.76\times 200=$ ₹152.
2. He actually gives only 100 g, costing ₹100.
3. Profit $=152-100=$ ₹52 on ₹100, i.e. $52\%$.

Trick: $\frac{(100-\text{loss}\%)\times \text{claimed wt}}{\text{actual wt}}-100=\frac{76\times 200}{100}-100=52$.

Answer: (B) $52\%$

Concept: If $x-\frac{1}{x}=k$, then $x^2+\frac{1}{x^2}=k^2+2$.

1. $x^2-6x-1=0$; divide by $x$: $x-\frac{1}{x}=6$.
2. $x^2+\frac{1}{x^2}=6^2+2=38$.
3. Required value $=38-8=30$.

Trick: Sign of the constant term: $-1$ gives $x-\frac{1}{x}$, so add 2 after squaring.

Answer: (D) $30$

Concept: Angles on a straight line add to ${180}^{\circ }$; angles opposite equal sides are equal.

1. P, R, S are collinear, so $\angle QRS={180}^{\circ }-{86}^{\circ }={94}^{\circ }$.
2. QR = RS, so in $△QRS$: $\angle RQS=\angle RSQ$.
3. $2\angle RSQ={180}^{\circ }-{94}^{\circ }={86}^{\circ }\Rightarrow \angle RSQ={43}^{\circ }$.
4. $\angle PSQ=\angle RSQ={43}^{\circ }$.

Trick: Exterior angle $\angle PRQ={86}^{\circ }$ equals the sum of two equal angles, so each $=\frac{{86}^{\circ }}{2}={43}^{\circ }$.

Answer: (A) ${43}^{\circ }$

Concept: Convert mixed fractions to improper fractions; do 'of', ÷, × before + and −.

1. $3\frac{1}{7}÷1\frac{4}{7}=\frac{22}{7}\times \frac{7}{11}=2$.
2. $3\text{ of }2\frac{2}{3}=3\times \frac{8}{3}=8$.
3. $2.5\times 2.4÷3\frac{1}{3}=6\times \frac{3}{10}=1.8$.
4. $4\text{ of }3=12$; total $=2+8+1.8-12=-0.2=-\frac{1}{5}$.

Trick: Keep decimals: $-0.2$ instantly matches $-\frac{1}{5}$.

Answer: (C) $-\frac{1}{5}$

Concept: BODMAS: 'of' and multiplication first, then addition/subtraction.

1. $8.16\times 5.35=43.656$.
2. $19.5\text{ of }\frac{1}{15}=1.3$; $\frac{1}{5}=0.2$.
3. $43.656+17.9-1.3+0.2=60.456$.

Trick: Only the product $8.16\times 5.35$ gives the last decimals (…656), so the answer must end in 56 – only option (A) does.

Answer: (A) $60.456$

Concept: Find each CP from SP: $\text{CP}=\frac{\text{SP}\times 100}{100+\text{P}\%}$; overall profit % $=\frac{\text{total profit}}{\text{total CP}}\times 100$.

1. CPs: $\frac{2800}{1.40}=2000$, $\frac{2800}{1.25}=2240$, $\frac{2800}{1.12}=2500$; total CP $=6740$.
2. Total SP $=3\times 2800=8400$; profit $=8400-6740=1660$.
3. Profit $\%=\frac{1660}{6740}\times 100\approx 24.63\%$, which is $0.2463$ in decimal form.

Trick: Same SP each time ⇒ do not average 40, 25, 12; always work through the CPs.

Answer: (D) $0.2463$ (i.e. $24.63\%$)

Concept: If two chords AB and CD intersect inside a circle at L, then $AL\times LB=CL\times LD$.

1. $8\times 6=CL\times 5$.
2. $CL=\frac{48}{5}=9.6\text{cm}$.

Yaad rakhein: Products of the two parts of each chord are equal.

Answer: (D) $9.6\text{cm}$

Concept: Opposite directions ⇒ relative speed = sum of speeds; distance = sum of lengths.

1. Distance $=110+140=250\text{m}$, time $=5\text{s}$.
2. Relative speed $=\frac{250}{5}=50\text{m/s}=50\times \frac{18}{5}=180\text{km/h}$.
3. $70+x=180\Rightarrow x=110\text{km/h}$.

Trick: Convert m/s to km/h by multiplying by $\frac{18}{5}$.

Answer: (C) $110\text{km/h}$

Concept: Division and multiplication have equal priority – do them left to right, then + and −.

1. $45÷5=9$.
2. $9\times 5=45$.
3. $45+10-5=50$.

Trick: Do not compute $5\times 5$ first – left to right gives 45, not 1.8.

Answer: (C) $50$

Concept: $\frac{1}{{\sec }^2\theta }={\cos }^2\theta$, $1+{\cot }^2\theta ={\csc }^2\theta$ so $\frac{1}{1+{\cot }^2\theta }={\sin }^2\theta$.

1. Expression $=17{\cos }^2\theta +10{\sin }^2\theta +7{\sin }^2\theta$.
2. $=17{\cos }^2\theta +17{\sin }^2\theta =17({\cos }^2\theta +{\sin }^2\theta )$.
3. $=17$.

Trick: Put $\theta =0^{\circ }$: $17+0+0=17$.

Answer: (C) $17$

Concept: Open brackets from the innermost: ( ), then { }, then [ ].

1. $(30+15)=45$.
2. $\{50-45\}=5$.
3. $[100-5]=95$.
4. $250+95=345$.

Trick: Work inside-out; never skip a bracket level.

Answer: (D) $345$

Concept: For equal distances at speeds $a$ and $b$: average speed $=\frac{2ab}{a+b}$.

1. $a=30,b=50$.
2. Average $=\frac{2\times 30\times 50}{30+50}=\frac{3000}{80}=37.5\text{km/h}$.

Trick: Average speed for equal distances is always less than the simple mean (here 40), so 40 is a trap.

Answer: (A) $37.5$