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

Concept: Area of a square $=\frac{d^2}{2}$, so area $\propto d^2$ and $d\propto \sqrt{\text{Area}}$.

1. New area $=100\%-19\%=81\%$ of old area.
2. New diagonal $=\sqrt{0.81}=0.9$ times old diagonal.
3. Decrease $=1-0.9=0.1=10\%$.

Trick: Area factor $0.81=(0.9{)}^2$, so every length (side, diagonal) falls by $10\%$.

Answer: (A) $10\%$

Concept: If a sum of squares is zero, each square is zero. Split the expression into two perfect squares.

1. $x^2-4xy+4y^2+4y^2-12y+9=0$
2. $(x-2y{)}^2+(2y-3{)}^2=0$
3. $2y-3=0\Rightarrow y=\frac{3}{2}$; $x=2y=3$
4. $7x-8y=21-12=9$

Trick: Write $8y^2=4y^2+4y^2$: one part completes $(x-2y{)}^2$, the other $(2y-3{)}^2$.

Answer: (D) $9$

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $433.50=P\times {\left(\frac{17}{100}\right)}^2=P\times \frac{289}{10000}$
2. $P=\frac{433.50\times 10000}{289}=1.5\times 10000=15000$

Trick: $289\times 1.5=433.5$, so $P=1.5\times 10000$ directly.

Answer: (A) ₹$15000$

Concept: Add the three pair-rates: it gives twice the rate of A, B and C together.

1. Take work $=\text{LCM}(36,60,45)=180$ units. $A+B=5$, $B+C=3$, $A+C=4$ units/day.
2. $2(A+B+C)=12\Rightarrow A+B+C=6$
3. $A=6-(B+C)=6-3=3$ units/day
4. Days for A $=\frac{180}{3}=60$

Trick: One alone $=$ (all three) $-$ (the other two).

Answer: (B) $60$ days

Concept: Radius ⟂ tangent, so in quadrilateral OAPB, $\angle AOB={180}^{\circ }-\angle APB$; triangle OAB is isosceles ($OA=OB$).

1. $\angle AOB={180}^{\circ }-{100}^{\circ }={80}^{\circ }$
2. $\angle OAB=\angle OBA=\frac{{180}^{\circ }-{80}^{\circ }}{2}={50}^{\circ }$

Trick: $\angle OAB=\frac{\angle APB}{2}={50}^{\circ }$ (always half the angle between tangents).

Answer: (D) ${50}^{\circ }$

Concept: Department strength $=\text{percentage}\times \text{total}$; males $=100\%-60\%=40\%$ of it.

1. Department E $=20\%$ of $450=90$
2. Males $=40\%$ of $90=36$

Trick: Directly: $450\times 0.2\times 0.4=36$.

Answer: (B) $36$

Concept: For the same time period, profit is shared in the ratio of investments.

1. $92500:112500=37:45$
2. B's share $=\frac{45}{82}$ of profit $=9000$
3. Total profit $=\frac{9000\times 82}{45}=200\times 82=16400$

Trick: 1 ratio unit $=\frac{9000}{45}=200$; total $=82\times 200$.

Answer: (B) ₹$16400$

Concept: Area of a trapezium $=\frac{1}{2}\times (\text{sum of parallel sides})\times \text{height}$; $1{\text{m}}^2=10000{\text{cm}}^2$.

1. Area $=\frac{1}{2}\times (27+13)\times 7=140{\text{cm}}^2$
2. In ${\text{m}}^2$: $\frac{140}{10000}=0.014$

Yaad rakhein: ${\text{cm}}^2\to {\text{m}}^2$: divide by ${10}^4$ (not $100$).

Answer: (B) $0.014$

Concept: Numbers in ratio $a:b$ (co-prime) with HCF $h$ are $ha$ and $hb$; LCM $=h\times a\times b$.

1. Numbers $=8\times 7=56$ and $8\times 13=104$
2. LCM $=8\times 7\times 13=728$

Trick: Check: $\text{HCF}\times \text{LCM}=8\times 728=5824=56\times 104$.

Answer: (D) $728$

Concept: Girls $\%=\frac{\text{girls}}{\text{boys}+\text{girls}}\times 100$ for each school.

1. A: $\frac{500}{1100}\approx 45.5\%$, B: $\frac{550}{1000}=55\%$
2. C: $\frac{650}{1400}\approx 46.4\%$, D: $\frac{600}{1300}\approx 46.2\%$
3. E: $\frac{650}{1450}\approx 44.8\%$ – the lowest

Trick: Compare boys : girls; the highest boys-to-girls ratio ($\frac{800}{650}\approx 1.23$ in E) means the lowest girls share.

Answer: (C) E

Concept: Percentage more $=\frac{\text{difference}}{\text{base}}\times 100$, base = the class compared with.

1. Cars at 70–80 $\text{km/h}$ $=90$; at 50–60 $\text{km/h}$ $=70$
2. $\frac{90-70}{70}\times 100=\frac{200}{7}\approx 28.6\%$

Trick: $\frac{2}{7}=28.57\%$ – remember the sevenths.

Answer: (C) $28.6\%$

Concept: Order: Brackets, then 'of', then $÷$, $\times$, then $+$.

1. $2\frac{1}{2}\text{ of }1\frac{3}{5}=\frac{5}{2}\times \frac{8}{5}=4$
2. $3\frac{1}{3}÷4=\frac{10}{3}\times \frac{1}{4}=\frac{5}{6}$
3. Bracket: $\frac{3}{8}+\frac{1}{7}\times \frac{7}{4}=\frac{3}{8}+\frac{2}{8}=\frac{5}{8}$
4. Total $=\frac{5}{6}+\frac{5}{8}=\frac{20+15}{24}=\frac{35}{24}$

Yaad rakhein: 'of' is done before $÷$.

Answer: (B) $\frac{35}{24}$

Concept: If each of $n$ students gives ₹$n$, total $=n^2$.

1. $n^2=2209$
2. $n=\sqrt{2209}=47$ (since ${47}^2=2209$)

Trick: Unit digit 9 → root ends in 3 or 7; ${47}^2=2209$ fits.

Answer: (A) $47$

Concept: Express all in terms of C and use sum $=3\times \text{average}$.

1. $A=\frac{C}{6}$, $B=2A=\frac{C}{3}$
2. $\frac{C}{6}+\frac{C}{3}+C=90\Rightarrow \frac{3C}{2}=90\Rightarrow C=60$
3. $A=10$; $C-A=50$

Trick: Ratio $A:B:C=1:2:6$, 9 parts $=90$ → 1 part $=10$; difference $=5$ parts $=50$.

Answer: (C) $50$

Concept: Distance in one revolution $=$ circumference $=\pi d$.

1. $\pi d=\frac{22}{7}\times 42=132\text{cm}$
2. 12 revolutions $=132\times 12=1584\text{cm}=15.84\text{m}$

Yaad rakhein: Distance $=$ revolutions $\times \pi d$.

Answer: (C) $15.84$

Concept: Write all shares in terms of one share and equate to the total.

1. Let B $=x$; A $=x+500$; C $=A+300=x+800$
2. $3x+1300=8200\Rightarrow x=2300$
3. A $=2300+500=2800$

Trick: Check: $2800+2300+3100=8200$.

Answer: (C) ₹$2800$

Concept: Employees in a department $=\text{its }\%\times \text{total}$.

1. Department B $=14\%$
2. $\frac{14}{100}\times 450=63$

Trick: $1\%$ of $450=4.5$, so $14\times 4.5=63$.

Answer: (D) $63$

Concept: Bisectors of the two exterior angles at B and C meet at the excentre O, and $\angle BOC={90}^{\circ }-\frac{\angle A}{2}$.

1. $\angle BOC={90}^{\circ }-\frac{{66}^{\circ }}{2}$
2. $={90}^{\circ }-{33}^{\circ }={57}^{\circ }$

Yaad rakhein: Internal bisectors: ${90}^{\circ }+\frac{A}{2}$; external bisectors: ${90}^{\circ }-\frac{A}{2}$.

Answer: (C) ${57}^{\circ }$

Concept: $\tan \theta =\frac{\text{height}}{\text{shadow}}$; from $\cos \theta =\frac{12}{13}$, the triangle is 5–12–13.

1. $\sin \theta =\frac{5}{13}$, so $\tan \theta =\frac{5}{12}$
2. $h=18\times \frac{5}{12}=7.5\text{m}$

Trick: Triplet $(5,12,13)$: base $12k=18\Rightarrow k=1.5$, height $5k=7.5$.

Answer: (A) $7.5$

Concept: Add girls and boys over all five schools and simplify the ratio.

1. Girls $=500+550+650+600+650=2950$
2. Boys $=600+450+750+700+800=3300$
3. $2950:3300=59:66$ (divide by 50)

Trick: Order matters: girls first, so $59:66$, not $66:59$.

Answer: (C) $59:66$