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

Concept: Downstream speed $=B+C$, upstream speed $=B-C$; for the same distance, time ∝ $\frac{1}{\text{speed}}$.

1. Let $B=35k$, $C=8k$: downstream $=43k$, upstream $=27k$.
2. Downstream time $=5\text{ h }10\text{ min}=310$ min.
3. Upstream time $=310\times \frac{43}{27}=\frac{13330}{27}\approx 493.7$ min.
4. $493.7$ min $\approx 8$ h $13$ min $42$ s, which is nearest to option (D).

Trick: Time ratio is the inverse of speed ratio: $27:43$ ⇒ multiply 310 min by $\frac{43}{27}$.

Answer: (D) 8 hours 13 minutes 48 seconds

Concept: If the remaining amount is 40%, the spent amount is 60% of the total.

1. Spent $=30000+45000=75000$.
2. $60\%$ of total $=75000$.
3. Total $=\frac{75000\times 100}{60}=1,25,000$.

Trick: $60\%=\frac{3}{5}$ ⇒ total $=75000\times \frac{5}{3}$.

Answer: (C) ₹1,25,000

Concept: $\frac{M_1{D}_1}{W_1}=\frac{M_2{D}_2}{W_2}$.

1. $\frac{12\times 10}{540}=\frac{M_2\times 8}{720}$.
2. $M_2=\frac{12\times 10\times 720}{540\times 8}=20$.
3. Extra people $=20-12=8$.

Trick: The question asks 'how many more' – subtract the original 12.

Answer: (A) $8$

Concept: Third side $x$ of a triangle: $|a-b|<x<a+b$.

1. $6-3.8=2.2$ and $6+3.8=9.8$.
2. So $2.2<x<9.8$.
3. 3.0, 2.4, 2.7 lie in the range; 2.2 does not (equality gives a straight line, not a triangle).

Trick: The boundary value (difference or sum) can never be a side.

Answer: (D) $2.2$

Concept: Area $=\pi r^2$; the longest chord of a circle is its diameter $=2r$.

1. $\pi r^2=256\pi \Rightarrow r^2=256\Rightarrow r=16$.
2. Longest chord $=2r=32\text{cm}$.

Yaad rakhein: Longest chord = diameter – always.

Answer: (A) $32$

Concept: When all parts are given as % of the same total, compare the percentages directly.

1. Rent + medicines + electricity $=15+12+10=37\%$.
2. Groceries + education $=20+25=45\%$.
3. Required $=\frac{37}{45}\times 100=82.22\%$.

Trick: $\frac{37}{45}$ is a little above $\frac{36}{45}=80\%$ – only 82.22% fits.

Answer: (A) $82.22\%$

Concept: Only the milk changes; water quantity stays the same.

1. In 5 L: milk $=\frac{4}{5}\times 5=4$ L, water $=1$ L.
2. Add 1 L milk: total $=6$ L, water $=1$ L.
3. Water $\%=\frac{1}{6}\times 100=\frac{100}{6}\%$.

Trick: Track the unchanged component (water) – new % = water ÷ new total.

Answer: (C) $\frac{100}{6}\%$

Concept: Read the bar values, then use $\frac{\text{part}}{\text{base}}\times 100$.

1. X in 2020 $=30$, Z in 2022 $=25$; sum $=55$ lakh tonnes.
2. W in 2021 $=25$ lakh tonnes.
3. Required $=\frac{55}{25}\times 100=220\%$.

Trick: Dividing by 25 = multiplying by 4: $55\times 4=220$.

Answer: (D) $220$

Concept: $\sin ({90}^{\circ }-\theta )=\cos \theta$, so ${\sin }^2\theta +{\sin }^2({90}^{\circ }-\theta )=1$.

1. Pairs: $({15}^{\circ },{75}^{\circ }),({25}^{\circ },{65}^{\circ }),({35}^{\circ },{55}^{\circ })$ each give 1 ⇒ $3$.
2. ${\sin }^2{45}^{\circ }=\frac{1}{2}$.
3. Total $=3+\frac{1}{2}=\frac{7}{2}$.

Trick: For $n$ terms in such a symmetric series, sum $=\frac{n}{2}$: here $\frac{7}{2}$.

Answer: (B) $\frac{7}{2}$

Concept: One-hour work of Rajiv = (together) − (Sumit alone).

1. Rajiv's 1-hour work $=\frac{1}{20}-\frac{1}{25}=\frac{5-4}{100}=\frac{1}{100}$.
2. Rajiv alone takes $100$ hours.

Trick: $\frac{ab}{b-a}=\frac{20\times 25}{25-20}=100$.

Answer: (B) $100$

Concept: Mean $=\frac{\sum fx}{\sum f}$.

1. $\sum f=2+3+4+1+5+1+2=18$.
2. $\sum fx=30+54+84+22+95+23+48=356$.
3. Mean $=\frac{356}{18}=19.78$.

Trick: Assume mean 19: deviations give $\sum fd=14$, so mean $=19+\frac{14}{18}=19.78$.

Answer: (D) $19.78$

Concept: From $\cos A=\frac{4}{5}$, use the 3-4-5 triangle: $\sin A=\frac{3}{5}$, $\tan A=\frac{3}{4}$.

1. $8+\sin A=8+\frac{3}{5}=\frac{43}{5}$.
2. $3-\tan A=3-\frac{3}{4}=\frac{9}{4}$.
3. Product $=\frac{43}{5}\times \frac{9}{4}=\frac{387}{20}$.

Trick: Remember the triplet 3-4-5: hypotenuse 5, adjacent 4, opposite 3.

Answer: (B) $\frac{387}{20}$

Concept: Average yearly production $=\frac{\text{grand total}}{\text{number of years}}$.

1. Average $=\frac{476}{6}\approx 79.33$.
2. Yearly totals: 74, 71, 75, 90, 80, 86.
3. Closest to 79.33 is 80, in 1993.

Trick: Use the 'Total' row directly – no need to add types again.

Answer: (A) 1993

Concept: Average $=\frac{\text{sum of values}}{\text{number of values}}$.

1. 2020 sales: $420+430+410+440+400+420=2520$.
2. Average $=\frac{2520}{6}=420$.

Trick: Deviations from 420: $0+10-10+20-20+0=0$ ⇒ average 420.

Answer: (C) $420$

Concept: Solve ( ), { }, [ ] from inside, converting mixed fractions to improper fractions.

1. $2\frac{1}{3}+\frac{3}{5}=\frac{7}{3}+\frac{3}{5}=\frac{44}{15}$.
2. $2\frac{1}{2}-\frac{44}{15}=\frac{75-88}{30}=-\frac{13}{30}$.
3. $3\frac{1}{5}-\frac{13}{30}=\frac{96-13}{30}=\frac{83}{30}$.
4. $55\frac{1}{3}÷\frac{83}{30}=\frac{166}{3}\times \frac{30}{83}=20$.

Trick: $166=2\times 83$ – the 83 cancels, a sign you are on track.

Answer: (C) $20$

Concept: Successive change: $a+b+\frac{ab}{100}$.

1. Volume $=lbh$; height unchanged.
2. Change $=6+8+\frac{6\times 8}{100}=14+0.48=14.48\%$.

Trick: Or $1.06\times 1.08=1.1448$ ⇒ $14.48\%$.

Answer: (D) $14.48\%$

Concept: $\text{SP}=\text{MP}\times (1-\text{d}\%)$ and $\text{CP}=\frac{\text{SP}}{1+\text{p}\%}$.

1. SP $=420\times 0.9=378$.
2. CP $=\frac{378}{1.2}=315$.

Trick: $\text{CP}=\text{MP}\times \frac{90}{120}=420\times \frac{3}{4}=315$.

Answer: (D) ₹315

Concept: ${(a-\frac{1}{a})}^3=a^3-\frac{1}{a^3}-3(a-\frac{1}{a})$.

1. $a^3=\sqrt{2}-1$, so $\frac{1}{a^3}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$.
2. Let $t=a-\frac{1}{a}$: $t^3+3t=a^3-\frac{1}{a^3}$.
3. $=(\sqrt{2}-1)-(\sqrt{2}+1)=-2$.

Trick: $\sqrt{2}-1$ and $\sqrt{2}+1$ are reciprocals – their difference is always $-2$.

Answer: (C) $-2$

Concept: A free gift is an extra cost: effective SP = selling price − gift value.

1. SP $=500\times 0.75=375$.
2. Effective receipt $=375-65=310$.
3. $1.24\times \text{CP}=310\Rightarrow \text{CP}=250$.

Trick: Check: $250\times 1.24=310$ and $310+65=375$.

Answer: (C) ₹250

Concept: Population after changes $=P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})(1-\frac{r_3}{100})$.

1. $15000\times 1.15=17250$.
2. $17250\times 1.25=21562.5$.
3. $21562.5\times 0.94=20268.75\approx 20269$.

Trick: Net factor $1.15\times 1.25\times 0.94\approx 1.35$ ⇒ about 20,250 – only (A) is close.

Answer: (A) $20269$