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

Concept: Find CP and SP of one orange (or of LCM $60$ oranges).

1. CP of 1 $=\frac{50}{10}=5$
2. SP of 1 $=\frac{74}{12}=\frac{37}{6}$
3. Gain $\%=\frac{\frac{37}{6}-5}{5}\times 100=\frac{7}{30}\times 100=23\frac{1}{3}\%$

Trick: For $60$ oranges: CP $=300$, SP $=370$, gain $=\frac{70}{300}=23\frac{1}{3}\%$.

Answer: (A) $23\frac{1}{3}\%$ gain

Concept: Sides are proportional to the sines of opposite angles (sine rule).

1. $x+2x+3x={180}^{\circ }\Rightarrow {30}^{\circ },{60}^{\circ },{90}^{\circ }$
2. $\sin {30}^{\circ }:\sin {60}^{\circ }:\sin {90}^{\circ }=\frac{1}{2}:\frac{\sqrt{3}}{2}:1$
3. $=1:\sqrt{3}:2$

Yaad rakhein: ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle: sides $1:\sqrt{3}:2$.

Answer: (A) $1:\sqrt{3}:2$

Concept: Inside brackets: $÷$ and $\times$ left to right, then $+$/$-$.

1. $48÷8\times 3=6\times 3=18$; $82+18=100$
2. $100+5\times 6=130$
3. $130-7.5\times 2=130-15=115$

Trick: Do $÷$ before $\times$ when $÷$ comes first (left to right).

Answer: (A) $115$

Concept: Multiply the factors: $0.9\times 1.3\times 0.8$.

1. $0.9\times 1.3=1.17$
2. $1.17\times 0.8=0.936$
3. Net decrease $=1-0.936=0.064=6.4\%=6\frac{2}{5}\%$

Trick: Start with $100$: $100\to 90\to 117\to 93.6$.

Answer: (C) $6\frac{2}{5}\%$

Concept: Write ${\cot }^2\theta =\frac{1}{{\tan }^2\theta }$ and simplify.

1. Let $t={\tan }^2\theta$: numerator $=1-\frac{1}{t}=\frac{t-1}{t}$
2. $\frac{(t-1)/t}{t-1}=\frac{1}{t}$
3. $={\cot }^2\theta$

Trick: Put $\theta ={60}^{\circ }$: $\frac{1-\frac{1}{3}}{3-1}=\frac{1}{3}={\cot }^2{60}^{\circ }$.

Answer: (C) ${\cot }^2\theta$

Concept: Add the three pair rates to get $2(R+S+M)$.

1. $R+S=\frac{1}{2}$, $S+M=\frac{1}{4}$, $R+M=\frac{1}{2}$
2. $2(R+S+M)=\frac{5}{4}\Rightarrow R+S+M=\frac{5}{8}$
3. $R=\frac{5}{8}-\frac{1}{4}=\frac{3}{8}$
4. Ravi alone $=\frac{8}{3}$ days

Trick: Total work $=4$ units: pairs do $2,1,2$; all three $2.5$; Ravi $=1.5$ units/day.

Answer: (B) $\frac{8}{3}$ days

Concept: SP $=$ cost $+$ margin; the discount slab depends on the bill total.

1. Shirt $=900\times 1.09=981$; jacket $=1900\times 1.06=2014$
2. Total $=2995<3000$, so discount $=12\%$
3. Pay $=2995\times 0.88=2635.60$

Trick: Check the slab first: $2995$ is just below $3000$.

Answer: (D) ₹$2,635.60$

Concept: Capacity of hemispherical bowl $=\frac{2}{3}\pi r^3$.

1. $r^3=9^3=729$
2. $V=\frac{2}{3}\times \frac{22}{7}\times 729=\frac{32076}{21}$
3. $\approx 1527.43{\text{cm}}^3$

Trick: $\frac{2}{3}\times 729=486$; $486\times \frac{22}{7}=1527.43$.

Answer: (D) $1527.43{\text{cm}}^3$

Concept: $\frac{\sec A+\tan A}{\sec A-\tan A}=\frac{1+\sin A}{1-\sin A}$; then componendo–dividendo.

1. $2\frac{51}{79}=\frac{209}{79}$
2. $\frac{1+\sin A}{1-\sin A}=\frac{209}{79}$
3. $\sin A=\frac{209-79}{209+79}=\frac{130}{288}=\frac{65}{144}$

Trick: If $\frac{1+x}{1-x}=\frac{p}{q}$, then $x=\frac{p-q}{p+q}$.

Answer: (B) $\frac{65}{144}$

Concept: “of” is done before $÷$ and $\times$.

1. $5\text{ of }4=20$: $20÷20\times 8=8$
2. $24\text{ of }3=72$: $6÷72\times 48=4$
3. $6÷18\times 9=3$
4. $8+4-3=9$

Trick: Resolve every “of” first, then go left to right.

Answer: (C) $9$

Concept: Volume of a cube $=a^3$.

1. $a=8\text{cm}$
2. $V=8^3=512{\text{cm}}^3$

Trick: $8^3=2^9=512$.

Answer: (C) $512{\text{cm}}^3$

Concept: Extra quantity $\times$ new price $=$ money saved by the reduction.

1. Saving $=7.5\%$ of $7400=555$
2. New price $=\frac{555}{15}=37$ per kg
3. Original price $=\frac{37}{0.925}=40$ per kg
4. SP $=40\times 1.325=53$ per kg

Trick: Reduced price $=\frac{\text{saving}}{\text{extra kg}}$ — no equation needed.

Answer: (D) $53$

Concept: Make the common term (Praveen) equal in both ratios.

1. J : P $=5:8=15:24$; P : K $=3:2=24:16$
2. J : P : K $=15:24:16$, sum $=55$
3. J $=\frac{15}{55}\times 17600=4800$

Trick: One part $=\frac{17600}{55}=320$; J $=15\times 320$.

Answer: (C) ₹$4,800$

Concept: Percentage higher $=\frac{\text{difference}}{\text{base}}\times 100$, base $=2018$.

1. 2018: $25$ crore; 2020: $30$ crore
2. $\frac{30-25}{25}\times 100=20\%$

Trick: Base is the value after “than”.

Answer: (A) $20\%$

Concept: Budget share $=$ percentage $\times$ total.

1. Grocery $+$ Electricity $=20\%+19\%=39\%$
2. Budget $=39\%$ of $58000=22620$
3. Difference $=22620-18875=3745$

Trick: $1\%$ of $58000=580$; $39\times 580=22620$.

Answer: (C) ₹$3,745$

Concept: Passing in Q $=$ candidates of Q $\times \frac{3}{3+7}$.

1. Q passing $=2.26\times \frac{3}{10}=0.678$ lakh
2. Total $=1.10+2.26+2.12+1.46=6.94$ lakh
3. $\frac{0.678}{6.94}\times 100\approx 9.77\%$

Trick: Pass share of Q is $30\%$; $2.26$ is about one-third of $6.94$, so answer $\approx 10\%$.

Answer: (D) $9.77\%$

Concept: Inner brackets first, then $\times$, then $+$/$-$.

1. $48÷12=4$; $8÷4+4=6$
2. $4+2\times 6=16$
3. $29-16=13$

Trick: Never add before multiplying: $4+2\times 6\ne 36$.

Answer: (D) $13$

Concept: Treat it as a quadratic in $\cos A$ and factorise.

1. $2{\cos }^{2}A+4\cos A-\cos A-2=0$
2. $(2\cos A-1)(\cos A+2)=0$
3. $\cos A=-2$ is impossible, so $\cos A=\frac{1}{2}$
4. $A={60}^{\circ }$

Trick: Put options: $\cos {60}^{\circ }=\frac{1}{2}$ gives $2·\frac{1}{4}+\frac{3}{2}-2=0$.

Answer: (B) ${60}^{\circ }$

Concept: Area of sector $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$.

1. $\frac{60}{360}=\frac{1}{6}$
2. $\pi r^2=\frac{22}{7}\times 441=1386$
3. Area $=\frac{1386}{6}=231{\text{cm}}^2$

Trick: A ${60}^{\circ }$ sector is one-sixth of the circle.

Answer: (C) $231$

Concept: Divide by $\cos {13}^{\circ }$: $\frac{1+\tan {13}^{\circ }}{1-\tan {13}^{\circ }}=\tan ({45}^{\circ }+{13}^{\circ })$.

1. $\frac{\cos {13}^{\circ }+\sin {13}^{\circ }}{\cos {13}^{\circ }-\sin {13}^{\circ }}=\frac{1+\tan {13}^{\circ }}{1-\tan {13}^{\circ }}$
2. $=\frac{\tan {45}^{\circ }+\tan {13}^{\circ }}{1-\tan {45}^{\circ }\tan {13}^{\circ }}=\tan {58}^{\circ }$

Trick: $\frac{\cos x+\sin x}{\cos x-\sin x}=\tan ({45}^{\circ }+x)$ always.

Answer: (D) $\tan {58}^{\circ }$