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

Concept: $\frac{4\pi r^2}{\frac{4}{3}\pi r^3}=\frac{3}{r}$.

1. $\frac{3}{r}=\frac{2}{7}$.
2. $r=\frac{21}{2}=10.5$ cm.

Yaad rakhein: For a sphere, S : V $=3:r$.

Answer: (B) 10.5 cm

Concept: $\frac{a^3+b^3}{a^2-ab+b^2}=a+b$.

1. Let $a=427,b=325$; denominator $=\frac{a^2+b^2-ab}{100}$ since $42.7=\frac{a}{10}$, $32.5=\frac{b}{10}$.
2. Value $=100(a+b)=100\times 752=75200$.

Trick: Count the decimal shift: denominator is 100 times smaller.

Answer: (C) 75200

Concept: For incentre I: $\angle BIC={90}^{\circ }+\frac{\angle A}{2}$.

1. $\angle BIC={90}^{\circ }+\frac{{54}^{\circ }}{2}={90}^{\circ }+{27}^{\circ }={117}^{\circ }$.

Yaad rakhein: Incentre: $90+\frac{A}{2}$; excentre: $90-\frac{A}{2}$.

Answer: (D) ${117}^{\circ }$

Concept: $\tan \theta =\frac{\text{height}}{\text{distance}}$; both towers are at the same distance d from the midpoint.

1. $h_1=d\tan {30}^{\circ }=\frac{d}{\sqrt{3}}$.
2. $h_2=d\tan {45}^{\circ }=d$.
3. $h_1:h_2=\frac{1}{\sqrt{3}}:1=1:\sqrt{3}$.

Trick: Equal distances ⇒ heights ratio = ratio of tangents.

Answer: (B) $1:\sqrt{3}$

Concept: $\%\text{ less}=\frac{\text{difference}}{\text{base}}\times 100$; percentages of the same total can be compared directly.

1. Education = 6%, Defence = 12%.
2. $\%\text{ less}=\frac{12-6}{12}\times 100=50\%$.

Trick: Education is exactly half of Defence ⇒ 50% less.

Answer: (B) $50\%$

Concept: Equal volumes: $r_1^2{h}_1=r_2^2{h}_2$, so $h\propto \frac{1}{r^2}$.

1. $r_1:r_2=1:4$.
2. $h_1:h_2=4^2:1^2=16:1$.

Trick: Square the radius ratio and invert it.

Answer: (A) 16 : 1

Concept: Let usual time be t hours; speed $=\frac{400}{t}$.

1. $\frac{400}{t-2}-\frac{400}{t}=10\Rightarrow \frac{800}{t(t-2)}=10$.
2. $t(t-2)=80\Rightarrow t=10$.
3. Check: $40\text{km/h}$ in 10 h; $50\text{km/h}$ in 8 h ✓.

Trick: Find two numbers 2 apart with product 80: 10 and 8.

Answer: (D) 10

Concept: Amount $=\frac{\text{percentage}}{100}\times \text{total}$.

1. Proceeds to states = 34%.
2. $\frac{34}{100}\times 1680=571.2$ billion rupees.

Trick: $1\%$ of 1680 = 16.8; $\times 34=571.2$.

Answer: (A) 571.20

Concept: Order: 'of' first, then ÷ and × left to right, then + and −.

1. $\frac{3}{10}\text{ of }\frac{4}{9}=\frac{2}{15}$; $\frac{2}{3}÷\frac{2}{15}=5$.
2. $\frac{4}{5}\times \frac{10}{9}=\frac{8}{9}$; $\frac{8}{9}÷\frac{8}{15}=\frac{5}{3}$.
3. $\frac{3}{4}÷\frac{1}{2}=\frac{3}{2}$.
4. $5-\frac{5}{3}+\frac{3}{2}=\frac{30-10+9}{6}=\frac{29}{6}$.

Trick: 'of' binds tighter than ÷ — never divide before resolving 'of'.

Answer: (A) $\frac{29}{6}$

Concept: Net rate = fill rate − leak rate.

1. Fill rate $=1$ tank/h; net rate $=\frac{1}{4/3}=\frac{3}{4}$.
2. Leak rate $=1-\frac{3}{4}=\frac{1}{4}$ ⇒ empties in 4 hours.

Trick: Leak time $=\frac{ab}{b-a}=\frac{1\times \frac{4}{3}}{\frac{1}{3}}=4$.

Answer: (D) 4

Concept: Build the smallest number backwards from the last quotient.

1. Last quotient 0: $7\times 0+5=5$; $4\times 5+3=23$; $3\times 23+2=71$.
2. General number $=84k+71$ (since $3\times 4\times 7=84$).
3. $84$ is a multiple of 42, so remainder $=71-42=29$.

Trick: Remainder by a factor of 84 is just the remainder of 71.

Answer: (C) 29

Concept: Change from loss to profit is a percentage of CP.

1. Difference $=13.5\%+9.5\%=23\%$ of CP.
2. $23\%\text{ of CP}=1104\Rightarrow \text{CP}=\frac{1104\times 100}{23}=4800$.

Trick: $1104÷23=48$, so CP = 4800.

Answer: (B) ₹4,800

Concept: A number divisible by 5, 6 and 8 is a multiple of their LCM.

1. $\text{LCM}(5,6,8)=120$.
2. Multiples between 300 and 700: 360, 480, 600.
3. Count = 3.

Trick: $\lfloor \frac{700}{120}\rfloor -\lfloor \frac{300}{120}\rfloor =5-2=3$.

Answer: (A) 3

Concept: $P=\frac{SI\times 100}{RT}$; $CI=P[(1+\frac{r}{100}{)}^n-1]$.

1. $P=\frac{8100\times 100}{10\times 2}=40500$.
2. $CI=40500\times ({1.15}^2-1)=40500\times 0.3225=13061.25$.
3. Nearest rupee: 13,061.

Trick: Two-year CI rate at 15% $=15+15+\frac{15\times 15}{100}=32.25\%$.

Answer: (D) ₹13,061

Concept: Compare the single equivalent discounts.

1. Successive 20% and 15%: $20+15-\frac{20\times 15}{100}=32\%$.
2. Difference $=35\%-32\%=3\%$ of MP $=504$.
3. MP $=\frac{504\times 100}{3}=16800$.

Trick: Equivalent discount: $a+b-\frac{ab}{100}$.

Answer: (B) ₹16,800

Concept: Exterior angle bisectors at Q and R meet at the excentre U: $\angle QUR={90}^{\circ }-\frac{\angle P}{2}$.

1. ${59}^{\circ }={90}^{\circ }-\frac{\angle P}{2}$.
2. $\frac{\angle P}{2}={31}^{\circ }\Rightarrow \angle P={62}^{\circ }$.

Yaad rakhein: Two external bisectors: $90-\frac{P}{2}$; two internal: $90+\frac{P}{2}$.

Answer: (B) ${62}^{\circ }$

Concept: Average = sum of Arts students ÷ 4.

1. Arts: $30+50+20+20=120$.
2. Average $=\frac{120}{4}=30$.

Trick: Read only the Arts bars (orange in the chart).

Answer: (B) 30

Concept: If $a+b+c=0$ then $a^3+b^3+c^3=3abc$.

1. $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}$.
2. $=\frac{3abc}{abc}=3$.

Trick: Check with a = 1, b = 1, c = −2: $-\frac{1}{2}-\frac{1}{2}+4=3$.

Answer: (B) 3

Concept: Express both groups below 14 years as fractions of total N.

1. Boys $=\frac{7N}{12}$; below 14: $\frac{4}{7}\times \frac{7N}{12}=\frac{4N}{12}$.
2. Girls $=\frac{5N}{12}$; below 14: $\frac{3}{5}\times \frac{5N}{12}=\frac{3N}{12}$.
3. $\frac{7N}{12}=1120\Rightarrow N=1920$.

Trick: Girls below 14 is $1-\frac{2}{5}=\frac{3}{5}$ of girls — don't use $\frac{2}{5}$.

Answer: (D) 1920

Concept: Square side $=\frac{\text{diagonal}}{\sqrt{2}}$; triangle area $=\frac{1}{2}bh$.

1. Side $=\frac{9\sqrt{2}}{\sqrt{2}}=9$ ⇒ perimeter (base) $=36$ cm.
2. Height $=\sqrt{144}=12$ cm.
3. Area $=\frac{1}{2}\times 36\times 12=216{\text{cm}}^2$.

Trick: Diagonal $a\sqrt{2}$ ⇒ side a directly.

Answer: (A) 216