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

Concept: Apply BODMAS: division and multiplication left to right first, then addition and subtraction.

1. $72÷8\times 4=9\times 4=36$
2. $35÷5=7$
3. $36+40-7+15=84$

Trick: Do all $÷$ / $\times$ blocks first, then add/subtract in one sweep.

Answer: (B) $84$

Concept: Write everything in $\sin \theta ,\cos \theta$ and use $(a+b)(a-b)=a^2-b^2$ with ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $\tan \theta +\sec \theta +1=\frac{\sin \theta +1+\cos \theta }{\cos \theta }$, $\cot \theta -\csc \theta +1=\frac{\cos \theta -1+\sin \theta }{\sin \theta }$
2. Product of numerators $=(\sin \theta +\cos \theta {)}^2-1=2\sin \theta \cos \theta$
3. Divide by $\sin \theta \cos \theta$: value $=2$

Trick: Put $\theta ={45}^{\circ }$: $(1+\sqrt{2}+1)(1-\sqrt{2}+1)=4-2=2$.

Answer: (B) $2$

Concept: Convert each bracket to $\sin ,\cos$: $\csc \theta -\sin \theta =\frac{{\cos }^2\theta }{\sin \theta }$, $\sec \theta -\cos \theta =\frac{{\sin }^2\theta }{\cos \theta }$, $\tan \theta +\cot \theta =\frac{1}{\sin \theta \cos \theta }$.

1. Product of first two $=\frac{{\cos }^2\theta {\sin }^2\theta }{\sin \theta \cos \theta }=\sin \theta \cos \theta$
2. Multiply by $\frac{1}{\sin \theta \cos \theta }$
3. Value $=1$

Trick: Put $\theta ={45}^{\circ }$: $\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\times 2=1$.

Answer: (A) $1$

Concept: Volume of a cuboid $=l\times b\times h$. Express all sides through the height.

1. Let height $=h$; breadth $b=2h$; breadth is half the length, so $l=4h$
2. $4h\times 2h\times h=8h^3=1728\Rightarrow h^3=216\Rightarrow h=6$
3. Length $=4\times 6=24\text{m}$

Trick: $1728={12}^3$, so $8h^3={12}^3\Rightarrow 2h=12$.

Answer: (D) $24\text{m}$

Concept: Angles opposite equal sides are equal; angle sum of a triangle is ${180}^{\circ }$.

1. $AB=AC\Rightarrow \angle B=\angle C$
2. $\angle B+\angle C={180}^{\circ }-{70}^{\circ }={110}^{\circ }$
3. $\angle C={55}^{\circ }$

Trick: Base angle $={90}^{\circ }-\frac{\text{vertex}}{2}={90}^{\circ }-{35}^{\circ }$.

Answer: (D) ${55}^{\circ }$

Concept: Net change of two successive changes $a\%$ and $b\%$ $=a+b+\frac{ab}{100}$.

1. $25+25+\frac{25\times 25}{100}=50+6.25$
2. Net increase $=56.25\%$

Trick: ${1.25}^2=1.5625$, i.e. $56.25\%$ increase.

Answer: (D) $56.25\%$

Concept: Curved surface area of a cone $=\pi rl$, where $l=\sqrt{r^2+h^2}$.

1. $l=\sqrt{3^2+4^2}=5\text{cm}$
2. CSA $=\pi \times 3\times 5=15\pi {\text{cm}}^2$

Trick: $(3,4,5)$ is a Pythagorean triplet – slant height is $5$ at once.

Answer: (C) $15\pi$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$. The second part earns for only $5-2=3$ years.

1. First part $x$: SI $=\frac{x\times 15\times 5}{100}=0.75x$
2. Second part $y$: SI $=\frac{y\times 10\times 3}{100}=0.3y$
3. $\frac{0.75x}{0.3y}=\frac{5}{3}\Rightarrow 2.25x=1.5y\Rightarrow y=1.5x$
4. $x+1.5x=48500\Rightarrow x=19400$, $y=29100$

Trick: $x:y=2:3$, so second part $=\frac{3}{5}\times 48500=29100$.

Answer: (A) ₹29,100

Concept: Convert mixed fractions to improper fractions, take LCM, then multiply/divide left to right.

1. $\frac{5}{2}+\frac{7}{4}-\frac{1}{6}=\frac{30+21-2}{12}=\frac{49}{12}$
2. $\frac{49}{12}\times 16=\frac{196}{3}$
3. $\frac{196}{3}÷49=\frac{4}{3}=1\frac{1}{3}$

Trick: $49$ cancels directly: $\frac{49}{12}\times \frac{16}{49}=\frac{16}{12}$.

Answer: (C) $1\frac{1}{3}$

Concept: Loan $=$ sum of present values of the instalments: $P=\frac{x}{1.1}+\frac{x}{{1.1}^2}$.

1. $35000=x(\frac{1}{1.1}+\frac{1}{1.21})=x\times \frac{2.1}{1.21}$
2. $x=\frac{35000\times 1.21}{2.1}=20166.67$
3. Each instalment $\approx 20167$

Trick: Rate $10\%=\frac{1}{10}$: principal parts ${11}^2:11\times 10\Rightarrow$ $x=\frac{35000\times 121}{210}$.

Answer: (C) ₹20,167

Concept: Average $=\frac{\text{sum}}{\text{count}}$; then reduce the ratio.

1. Test 2 average $=\frac{94+79+85}{3}=\frac{258}{3}=86$
2. Maths average $=\frac{80+94+88+92+96}{5}=\frac{450}{5}=90$
3. Ratio $=86:90=43:45$

Trick: Only $43:45$ has terms near $86:90$ halved.

Answer: (D) $43:45$

Concept: Solve the innermost bracket first (BODMAS), separately for numerator and denominator.

1. Numerator: $(\frac{15}{7}+\frac{43}{7})÷\frac{9}{7}=\frac{58}{7}\times \frac{7}{9}=\frac{58}{9}$
2. Inner: $\frac{4}{9}÷\frac{4}{3}=\frac{1}{3}$; $\frac{7}{9}+\frac{3}{9}=\frac{10}{9}$
3. Denominator: $\frac{10}{9}÷\frac{10}{9}=1$
4. Value $=\frac{58}{9}$

Trick: Denominator becomes $1$, so the answer is just the numerator.

Answer: (C) $\frac{58}{9}$

Concept: $\text{Required}\%=\frac{\text{A total}}{\text{D total}}\times 100$.

1. A total $=29+37+38+42=146$
2. D total $=37+35+42+45=159$
3. $\frac{146}{159}\times 100\approx 91.82\%$

Trick: A is a little below D, so the answer is just above $90\%$ – only $91.82\%$ fits.

Answer: (A) $91.82\%$

Concept: If boys are $20\%$ more than girls, take girls $=100$.

1. Girls $=100$, boys $=120$
2. Boys : Girls $=120:100=6:5$

Trick: $20\%=\frac{1}{5}$, so girls $5$ parts and boys $6$ parts.

Answer: (A) $6:5$

Concept: Angle sum of a triangle is ${180}^{\circ }$.

1. Let $\angle P>\angle Q+\angle R$
2. Then $2\angle P>\angle P+\angle Q+\angle R={180}^{\circ }$
3. So $\angle P>{90}^{\circ }$ – the triangle is obtuse-angled

Yaad rakhein: If one angle equals the sum of the other two, it is right-angled; if greater, obtuse; if smaller (for every angle), acute.

Answer: (C) obtuse angle

Concept: Mean price $=\frac{\text{total cost}}{\text{total quantity}}$.

1. Take $3\text{kg}$ at ₹25 and $2\text{kg}$ at ₹40
2. Cost $=75+80=155$ for $5\text{kg}$
3. Price per kg $=\frac{155}{5}=31$

Trick: Alligation check: $(40-31):(31-25)=9:6=3:2$ ✔

Answer: (A) ₹31

Concept: Equal group sizes ($10$ each), so compare the totals.

1. Boys total $=56+75+34+49+89+99+94+59+82+79=716$, average $71.6$
2. Girls total $=35+88+95+99+67+82+97+63+79+81=786$, average $78.6$
3. Girls' average is higher

Trick: Same count – just compare the column sums.

Answer: (C) The average achievement of the girls is more than that of the boys.

Concept: Three successive $10\%$ discounts leave ${0.9}^3$ of the price.

1. ${0.9}^3=0.729$, so equivalent discount $=27.1\%$
2. Difference $=28\%-27.1\%=0.9\%$
3. $0.9\%$ of $2000=18$

Trick: Compare rates first, then apply to ₹2000 once.

Answer: (C) ₹18

Concept: Work rate = $\frac{1}{\text{days}}$. Amita's rate equals Bina + Sita.

1. $A=B+S$ and $A+B=\frac{1}{25}$, so $2B+S=\frac{1}{25}$
2. $S=\frac{1}{35}$: $2B=\frac{1}{25}-\frac{1}{35}=\frac{2}{175}$
3. $B=\frac{1}{175}$ → $175$ days

Trick: LCM $175$: $A+B=7$ units, $S=5$; $2B=2\Rightarrow B=1$ unit/day.

Answer: (D) $175$ days

Concept: Total profit $\%$ is on the total CP of both articles.

1. CP of each $=x$; SP₁ $=1.12x$, SP₂ $=1.12x+3600$
2. Total profit $=0.24x+3600$
3. $15\frac{27}{51}\%=\frac{792}{51}\%$, so $0.24x+3600=\frac{792}{5100}\times 2x$
4. $3600=\frac{360}{5100}x\Rightarrow x=51000$

Trick: Second article gives extra $\frac{3600}{x}$; net $\%$ $=12+\frac{3600\times 100}{2x}=15\frac{27}{51}$, so $\frac{180000}{x}=3\frac{27}{51}=\frac{180}{51}$.

Answer: (C) ₹51,000