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

Concept: Use $a^2-b^2=(a-b)(a+b)$, then divide and multiply left to right.

1. $7\frac{1}{4}=7.25$, $5\frac{4}{5}=5.8$
2. ${7.25}^2-{5.8}^2=1.45\times 13.05=18.9225$
3. $18.9225÷5.8=3.2625$
4. $3.2625\times \frac{16}{9}=5.8$

Trick: $1.45\times 13.05÷5.8\times \frac{16}{9}=\frac{1}{4}\times 13.05\times \frac{16}{9}=5.8$ since $\frac{1.45}{5.8}=\frac{1}{4}$.

Answer: (C) $5.8$

Concept: $1+{\tan }^{2}A={\sec }^{2}A$.

1. ${\tan }^{4}A+{\tan }^{2}A={\tan }^{2}A({\tan }^{2}A+1)$
2. $=({\sec }^{2}A-1){\sec }^{2}A$
3. $={\sec }^{4}A-{\sec }^{2}A$

Trick: Put $A={45}^{\circ }$: LHS $=2$; ${\sec }^4-{\sec }^2=4-2=2$ ✔

Answer: (A) ${\sec }^{4}A-{\sec }^{2}A$

Concept: Downstream speed $=b+c$, upstream speed $=b-c$.

1. Downstream speed $=\frac{16.9}{52/60}=19.5\text{km/h}$
2. Boat speed $=19.5-3=16.5$; upstream $=16.5-3=13.5\text{km/h}$
3. Time $=\frac{84}{13.5}\approx 6.22\text{h}$

Trick: Upstream $=$ downstream $-2c=19.5-6$.

Answer: (A) $6.22$

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Total time $=\frac{100}{40}=2.5\text{h}=150$ min
2. Used $40+50=90$ min, so last $20\text{km}$ takes $60$ min
3. Speed $=\frac{20}{1}=20\text{km/h}$

Trick: Work in minutes: $150-90=60$ min for $20\text{km}$.

Answer: (D) $20$

Concept: Apply BODMAS to the known terms and solve the linear equation.

1. $6÷3=2$; $45÷5\times 4=36$
2. $7m+4-2-7+36+49=7m+80$
3. $7m+80=87\Rightarrow m=1$

Trick: Option check: $m=1$ gives $7+80=87$ ✔

Answer: (C) $1$

Concept: Find each student's total, then $\frac{\text{Ronak}}{\text{others}}\times 100$.

1. Ronak $=75+89+79+82=325$
2. Shivam $=360$, Sonu $=355$, Arnav $=352$; together $1067$
3. $\frac{325}{1067}\times 100\approx 30.46\%$

Trick: Ronak $\approx$ one third of $1067$ minus a bit → just above $30\%$.

Answer: (D) $30.46\%$

Concept: Take total work = LCM of days and find work done in each 2-day cycle.

1. Work $=120$ units; Akash $3$/day, Vikash $2$/day
2. One 2-day cycle $=5$ units
3. $\frac{120}{5}=24$ cycles, finishing exactly → $48$ days

Trick: Work divides exactly into cycles, so days $=2\times 24$.

Answer: (C) $48$

Concept: 'of' is solved before $÷$ and $\times$; then left to right.

1. $2\times 8÷4=4$; $5\text{ of }3=15$
2. $6÷3\times 2=4$; $6\text{ of }3=18$
3. $4-15+4-18+5=-20$

Trick: Sum positives $13$, negatives $33$: $13-33=-20$.

Answer: (A) $-20$

Concept: $(x+y)(x^2-xy+y^2)=x^3+y^3$ with $x=4a$, $y=3b$.

1. Product $=64a^3+27b^3$
2. $=64\times 8+27\times 27=512+729$
3. $=1241$

Trick: Spot the cube identity: $(4a{)}^3+(3b{)}^3=8^3+9^3$.

Answer: (C) $1241$

Concept: Similar rectangles have the same length : breadth ratio.

1. $224:160=7:5$
2. $168:120=7:5$, $273:195=7:5$, $210:150=7:5$
3. $343:255$: $343\times 5=1715\ne 255\times 7=1785$ – not $7:5$

Trick: Breadth must be a multiple of $5$ and length the matching multiple of $7$: $255=5\times 51$ needs length $357$.

Answer: (B) Length = 343 m; Breadth = 255 m

Concept: Turn the percentage equation into a ratio, then use the new ratio.

1. $42R=56B\Rightarrow R:B=4:3$; let $R=4k$, $B=3k$
2. $\frac{4k}{3k+200}=\frac{5}{4}\Rightarrow 16k=15k+1000$
3. $k=1000\Rightarrow R=4000$

Trick: Option check: $R=4000\Rightarrow B=3000$; $4000:3200=5:4$ ✔

Answer: (B) ₹4000

Concept: Change from $16\%$ loss to $15\%$ gain $=31\%$ of CP.

1. $31\%$ of CP $=192.20\Rightarrow$ CP $=620$
2. New SP $=1.15\times 620=713$

Trick: $1\%$ of CP $=6.2$; new SP $=115\times 6.2$.

Answer: (D) ₹713

Concept: Brackets first; inside, 'of' before $÷$/$\times$.

1. Bracket: $6+6\times 6÷(6\text{ of }6)-6÷6\times 6=6+1-6=1$
2. Outside: $4\times 5÷(4\text{ of }5)\times 4=20÷20\times 4=4$
3. $4÷1=4$

Trick: 'of' $=$ multiplication done before ordinary $÷$.

Answer: (B) $4$

Concept: $a^2+b^2=2ab⟺(a-b{)}^2=0$.

1. $({\sin }^2\theta -{\cos }^2\theta {)}^2=0\Rightarrow {\sin }^2\theta ={\cos }^2\theta$
2. $\tan \theta =1\Rightarrow \theta ={45}^{\circ }$
3. $\cot \theta =1$

Trick: Equality in AM–GM holds only when both terms are equal.

Answer: (C) $1$

Concept: Number of candles $=\frac{\text{volume of wax block}}{\text{volume of one candle}}$.

1. Wax $=70\times 44\times 20=61600{\text{cm}}^3$
2. Candle $=\frac{22}{7}\times 7^2\times 20=3080{\text{cm}}^3$
3. $\frac{61600}{3080}=20$

Trick: Heights are both $20$ – cancel: $\frac{70\times 44}{154}=20$.

Answer: (A) $20$

Concept: Same direction: relative speed $=$ difference; distance $=$ sum of lengths.

1. Relative speed $=78-60=18\text{km/h}=5\text{m/s}$
2. Distance $=140+120=260\text{m}$
3. Time $=\frac{260}{5}=52\text{s}$

Trick: $18\text{km/h}\times \frac{5}{18}=5\text{m/s}$.

Answer: (D) $52$ seconds

Concept: Area $=\frac{1}{2}ab$; hypotenuse by Pythagoras.

1. $\frac{1}{2}x(x+2)=24\Rightarrow x^2+2x-48=0\Rightarrow x=6$
2. Sides $6$, $8$; hypotenuse $=\sqrt{36+64}=10$
3. Perimeter $=6+8+10=24\text{cm}$

Trick: Product $=48$ with difference $2$ → $6,8$ → triplet $(6,8,10)$.

Answer: (D) $24$

Concept: Opposite direction: relative speed $=$ sum; distance $=$ sum of lengths.

1. Relative speed $=44+56.8=100.8\text{km/h}=28\text{m/s}$
2. Distance $=320+240=560\text{m}$
3. $y=\frac{560}{28}=20$

Trick: $100.8\times \frac{5}{18}=28$.

Answer: (D) $20$

Concept: Marks $=\text{percentage}\times \text{maximum marks}$.

1. Sohan: $90\%$ of $200=180$
2. Shyam: $70\%$ of $200=140$
3. $180-140=40$ more

Trick: Same maximum: $(90-70)\%$ of $200=40$.

Answer: (A) 40 more

Concept: Percentage $=\frac{\text{part}}{\text{row total}}\times 100$.

1. School U total $=150+160+162+160+161+140=933$
2. $\frac{162}{933}\times 100\approx 17.4\%$
3. Nearest option $17\%$

Trick: $6$ classes of similar size → each $\approx \frac{100}{6}\approx 16.7\%$.

Answer: (D) $17\%$