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

Concept: Convert every scheme into an effective discount %.

1. A: $24+27-\frac{24\times 27}{100}=44.52\%$
2. B: pay for 5, get 8 ⇒ $\frac{3}{8}\times 100=37.5\%$
3. C: pay for 6, get 10 ⇒ $\frac{4}{10}\times 100=40\%$
4. Minimum is scheme B.

Trick: Free-item offers: discount $=\frac{\text{free}}{\text{total received}}\times 100$.

Answer: (A) B

Concept: Volume of sphere $=\frac{4}{3}\pi r^3$.

1. $r^3={2.1}^3=9.261$
2. $V=\frac{4}{3}\times \frac{22}{7}\times 9.261=\frac{88}{21}\times 9.261$
3. $=88\times 0.441=38.808{\text{cm}}^3$

Trick: $9.261÷21=0.441$ — dividing first keeps the arithmetic short.

Answer: (C) $38.808{\text{cm}}^3$

Concept: Count symbols and multiply by the key (1 symbol $=15$ scouts).

1. VI: $8$ symbols, VII: $4$, VIII: $3$
2. Total symbols $=15$
3. Scouts $=15\times 15=225$

Trick: Add symbols first, multiply once.

Answer: (A) $225$

Concept: $\cot \theta =\frac{\text{base}}{\text{perpendicular}}$; find hypotenuse by Pythagoras.

1. Base $=28$, perpendicular $=45$, hypotenuse $=\sqrt{{28}^2+{45}^2}=\sqrt{2809}=53$
2. $\sin \theta =\frac{45}{53}$, $\cos \theta =\frac{28}{53}$
3. $\sin \theta -\cos \theta =\frac{17}{53}$

Yaad rakhein: Triplet $(28,45,53)$ is a standard Pythagorean triple.

Answer: (B) $\frac{17}{53}$

Concept: Successive discounts multiply the remaining fractions.

1. Price left $=0.9\times 0.8\times 0.8=0.576$
2. Single discount $=1-0.576=0.424=42.4\%$

Trick: Two at a time: $20,20\to 36\%$; then $36,10\to 36+10-3.6=42.4\%$.

Answer: (B) $42.4\%$

Concept: Use LCM units and the work done in one 3-day cycle.

1. Total $=200$ units; A $=1$, B $=2$ units/day
2. Cycle: $3+1+2=6$ units in $3$ days
3. $33$ cycles $=99$ days ⇒ $198$ units; left $2$ units
4. Day 100 (A + B, $3$ units/day): $\frac{2}{3}$ day ⇒ total $99\frac{2}{3}$ days

Trick: Stop the full cycles just before the total and finish the rest in the cycle's own order.

Answer: (B) $99\frac{2}{3}$

Concept: Adding water keeps the quantity of acid unchanged.

1. Acid $=30\%$ of $12=3.6$ L
2. New total with $20\%$ acid: $\frac{3.6}{0.2}=18$ L
3. Water to add $=18-12=6$ L

Trick: Concentration falls $30\to 20$ ⇒ volume rises in ratio $20:30$, i.e. $12\to 18$.

Answer: (B) $6$

Concept: Solve the bracket; 'of' means multiply, done before ÷ and +.

1. $(18-11)=7$; $7$ of $4=28$
2. $18÷3+3+28=6+3+28=37$
3. $52-37=15$

Yaad rakhein: Order: Bracket → Of → Division → Multiplication → Addition → Subtraction.

Answer: (B) $15$

Concept: Percentage increase $=\frac{\text{new}-\text{old}}{\text{old}}\times 100$.

1. 2019 $=20$ lakh, 2022 $=35$ lakh
2. $\frac{35-20}{20}\times 100=75\%$

Trick: The base is always the earlier year (2019), not 2022.

Answer: (B) $75\%$

Concept: Turn the condition into a ratio $x:y$, then substitute.

1. $x+y=4(x-y)$ ⇒ $3x=5y$ ⇒ $x=5k$, $y=3k$
2. $5xy=75k^2$; $3(x^2-y^2)=3(25k^2-9k^2)=48k^2$
3. Value $=\frac{75}{48}=\frac{25}{16}$

Trick: Put $x=5$, $y=3$ directly: $\frac{75}{3\times 16}=\frac{25}{16}$.

Answer: (A) $\frac{25}{16}$

Concept: $x^3+y^3=(x+y{)}^3-3xy(x+y)$.

1. $(x+y{)}^3={12}^3=1728$
2. $3xy(x+y)=3\times 27\times 12=972$
3. $x^3+y^3=1728-972=756$

Trick: Here $x,y=9,3$: $729+27=756$.

Answer: (C) $756$

Concept: Cut-off 60 ⇒ read the '60 and above' column.

1. Science, 60 and above $=43$
2. Aggregate, 60 and above $=39$
3. Difference $=43-39=4$

Trick: In a cumulative 'and above' table, the column itself gives the pass count.

Answer: (D) $4$

Concept: TSA of cone $=\pi R(L+R)$ with $R=\frac{r}{4}$, $L=4l$.

1. $\text{TSA}=\pi ·\frac{r}{4}(4l+\frac{r}{4})$
2. $=\pi r(l+\frac{r}{16})$

Trick: Take the $4$ out of the bracket: $\frac{r}{4}\times 4=r$, and $\frac{r}{4}÷4=\frac{r}{16}$.

Answer: (C) $\pi r(l+\frac{r}{16})$

Concept: Cloth area $=$ curved surface area $=\pi rl$; length $=\frac{\text{area}}{\text{width}}$.

1. $l=\sqrt{{21}^2+{28}^2}=\sqrt{1225}=35\text{m}$
2. CSA $=\frac{22}{7}\times 21\times 35=2310{\text{m}}^2$
3. Length $=\frac{2310}{5}=462\text{m}$

Trick: $21:28:35=3:4:5$ triangle — slant height instantly.

Answer: (B) $462$

Concept: Bracket first; inside it, × before + and −.

1. $3-8\times 3+27=3-24+27=6$
2. $24÷6=4$
3. $4+3=7$

Trick: Inside the bracket, do $8\times 3$ first — not $(3-8)$.

Answer: (B) $7$

Concept: $\tan t+\cot t=\frac{1}{\sin t\cos t}$ and $(\sin t+\cos t{)}^2=1+2\sin t\cos t$.

1. $\frac{1}{\sin t\cos t}=1$ ⇒ $\sin t\cos t=1$
2. $(\sin t+\cos t{)}^2=1+2=3$ ⇒ $\sin t+\cos t=\pm \sqrt{3}$
3. $\frac{1}{\sin t+\cos t}=\pm \frac{1}{\sqrt{3}}$; one value $=\frac{1}{\sqrt{3}}$

Trick: Whenever $\sin t\cos t$ is known, square $(\sin t+\cos t)$.

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

Concept: Under CI, the multiplying factor repeats every equal period.

1. $\times 2$ in $4$ years
2. $8=2^3$ ⇒ three doublings
3. Time $=3\times 4=12$ years

Yaad rakhein: If a sum becomes $n$ times in $T$ years, it becomes $n^k$ times in $kT$ years.

Answer: (C) $12$

Concept: Chord $=$ radius ⇒ triangle with the centre is equilateral ⇒ central angle ${60}^{\circ }$.

1. $\theta ={60}^{\circ }$
2. Arc $=\frac{60}{360}\times 2\pi r=\frac{1}{6}\times 2\times \frac{22}{7}\times 21$
3. $=\frac{132}{6}=22$ units

Trick: Arc for ${60}^{\circ }$ $=\frac{\pi r}{3}$: $\frac{22\times 3}{3}=22$.

Answer: (C) $22$

Concept: Let total voters $=N$; valid votes $=$ votes cast $-$ blank.

1. Cast $=0.88N$; valid $=0.88N-84$
2. Ramagya $=0.54N$; Shravan $=0.88N-84-0.54N=0.34N-84$
3. Difference: $0.20N+84=1456$ ⇒ $0.2N=1372$ ⇒ $N=6860$

Trick: Option check: $20\%$ of $6860=1372$, $1372+84=1456$ ✓.

Answer: (B) $6860$

Concept: Convert mixed fractions; ÷ and × left to right.

1. $\frac{11}{3}÷\frac{11}{5}=\frac{5}{3}$; $\frac{5}{3}\times \frac{1}{4}=\frac{5}{12}$
2. $1+\frac{3}{1-\frac{2}{5}}=1+\frac{3}{\frac{3}{5}}=1+5=6$
3. $\frac{5}{12}\times 6=\frac{5}{2}=2\frac{1}{2}$

Trick: Solve the complex fraction from the bottom up: $1-\frac{2}{5}$ first.

Answer: (D) $2\frac{1}{2}$