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

Concept: Write $\tan A=\frac{\sin A}{\cos A}$, then rationalise using $(\cos A-\sin A)$.

1. $\frac{1-\tan A}{1+\tan A}=\frac{\cos A-\sin A}{\cos A+\sin A}$
2. Multiply up and down by $(\cos A-\sin A)$: $\frac{(\cos A-\sin A{)}^2}{{\cos }^{2}A-{\sin }^{2}A}$
3. Numerator $=1-2\sin A\cos A=1-\sin 2A$; denominator $=\cos 2A$
4. So the value $=\sqrt{\frac{1-\sin 2A}{\cos 2A}}$

Trick: Check with $A={15}^{\circ }$: $\frac{1-\tan {15}^{\circ }}{1+\tan {15}^{\circ }}=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$ and $\frac{1-\sin {30}^{\circ }}{\cos {30}^{\circ }}=\frac{1}{\sqrt{3}}$ ✓.

Answer: (B) $\sqrt{\frac{1-\sin 2A}{\cos 2A}}$

Concept: Volume of a sphere $V=\frac{4}{3}\pi r^3$, so $V\propto d^3$.

1. New diameter $=\frac{d}{2}$, so new volume $={\left(\frac{1}{2}\right)}^{3}V=\frac{V}{8}$
2. The volume becomes $\frac{1}{8}$ of the former volume, i.e. it is reduced (to one-eighth).
3. Options (A) and (C) say "increased", which is impossible when the size shrinks.

Trick: Length scale $k$ → volume scale $k^3$: half the diameter → $\frac{1}{8}$ the volume.

Answer: (D) reduced by $\frac{1}{8}$ of the former volume (new volume $=\frac{1}{8}$ of old)

Concept: HCF = product of common prime factors with the lowest powers.

1. $3888=2^4\times 3^5$
2. $3969={63}^2=3^4\times 7^2$
3. Common part $=3^4=81$

Trick: Difference $3969-3888=81$; HCF must divide 81, and 81 divides both ($81\times 48=3888$, $81\times 49=3969$).

Answer: (A) 81

Concept: Smallest number exactly divisible by all = LCM (highest powers of all primes).

1. $96=2^5\times 3$, $108=2^2\times 3^3$, $144=2^4\times 3^2$
2. LCM $=2^5\times 3^3=32\times 27=864$

Trick: Check the smallest option first: $864÷96=9$, $864÷108=8$, $864÷144=6$ ✓.

Answer: (A) 864

Concept: HCF of several numbers = largest number dividing each of them.

1. $240=2^4\times 3\times 5$, $280=2^3\times 5\times 7$, $560=2^4\times 5\times 7$
2. Common: $2^3\times 5=40$

Trick: 560 is a multiple of 280, so HCF(240, 280) is enough: $280-240=40$ and 40 divides both.

Answer: (D) 40

Concept: Read the two required cells from the table and subtract.

1. Store Q in 2013 = 198
2. Store S in 2015 = 205
3. Difference $=205-198=7$

Trick: Locate row first, then column – avoid picking S 2013 (243) by mistake.

Answer: (B) 7

Concept: Compare P and Q year by year and pick the years where Q > P.

1. 2001: $96<100$; 2003: $100<108$; 2004: $180<190$
2. 2002: $130>120$ ✓
3. 2005: $275>240$ ✓

Trick: Mark only rows where the right column is bigger – 2002 and 2005.

Answer: (D) 2002, 2005

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$, so total distance $=73\times 5$.

1. Total distance $=73\times 5=365\text{km}$
2. 75 min at 85: $\frac{5}{4}\times 85=106.25\text{km}$; 3 h at 70: $210\text{km}$
3. First part $=365-106.25-210=48.75\text{km}$
4. Speed $=\frac{48.75}{0.75}=65\text{km/h}$

Trick: Use deviations from 73: $1.25(85-73)+3(70-73)=15-9=6$, so first part must give $-6$: $0.75(x-73)=-6\Rightarrow x=65$.

Answer: (A) 65

Concept: SI $=\frac{PRT}{100}$; CI for 3 years at 10% $=33.1\%$ of principal.

1. Let Peter invest $2x$, Rachel $x$.
2. Peter SI $=2x\times \frac{10\times 2}{100}=0.2\times 2x=0.4x$
3. Rachel CI $=x[(1.1{)}^3-1]=0.331x$
4. $0.4x-0.331x=0.069x=897\Rightarrow x=13000$

Trick: 10% CI for 3 years = 33.1% – remember $10,21,33.1$ for 1, 2, 3 years.

Answer: (A) ₹13,000

Concept: Expenditure = Price × Consumption. If price rises by $r\%$, cut consumption by $\frac{r}{100+r}\times 100\%$.

1. $r=10$
2. Decrease $=\frac{10}{110}\times 100=\frac{100}{11}=9\frac{1}{11}\%$

Trick: Price $\frac{11}{10}$ → consumption $\frac{10}{11}$: fall $=\frac{1}{11}=9\frac{1}{11}\%$.

Answer: (A) $9\frac{1}{11}\%$

Concept: Find the cost price from the marked price, then add the new profit.

1. $\text{CP}=\frac{5670}{1.35}=4200$
2. New SP $=4200\times 1.12=4704$

Trick: $5670\times \frac{112}{135}=42\times 112=4704$ in one line.

Answer: (C) 4,704

Concept: For the same distance, speed $\propto \frac{1}{\text{time}}$.

1. Speeds $=\frac{1}{12}:\frac{1}{15}:\frac{1}{10}$
2. Multiply by LCM 60: $5:4:6$

Trick: Invert the time ratio and clear fractions with the LCM of the terms.

Answer: (C) 5 : 4 : 6

Concept: Area $=\frac{1}{2}ab\sin C$ when two sides and the included angle are given.

1. $a=4$, $b=5$, $C={45}^{\circ }$
2. Area $=\frac{1}{2}\times 4\times 5\times \frac{1}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}{\text{cm}}^2$

Trick: $\frac{10}{\sqrt{2}}=5\sqrt{2}$ – rationalise quickly.

Answer: (C) $5\sqrt{2}{\text{cm}}^2$

Concept: $\text{MP}\times (1-\text{discount})=\text{CP}\times (1+\text{profit})$.

1. SP $=450\times 1.3=585$
2. MP $=\frac{585}{0.9}=650$

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{130}{90}=\frac{13}{9}$, so MP $=450\times \frac{13}{9}=650$.

Answer: (B) ₹650

Concept: Fourth proportional $d$ to $a,b,c$: $a:b=c:d\Rightarrow d=\frac{bc}{a}$.

1. $d=\frac{0.8\times 0.108}{0.3}$
2. $=\frac{0.0864}{0.3}=0.288$

Trick: $\frac{0.8}{0.3}=\frac{8}{3}$, and $0.108\times \frac{8}{3}=0.036\times 8=0.288$.

Answer: (C) 0.288

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

1. $a=3.254$, $b=0.746$
2. Value $=a-b=3.254-0.746=2.508$

Trick: Cubes minus over "squares plus product" → just subtract the bases.

Answer: (D) 2.508

Concept: Successive profits multiply: $1.2\times 1.25\times (1+x)=1.65$.

1. $1.2\times 1.25=1.5$
2. $1+x=\frac{1.65}{1.5}=1.1$
3. $x=10\%$

Trick: Take price 100 → 120 → 150 → 165; third gain $=\frac{15}{150}=10\%$.

Answer: (A) 10%

Concept: Percentage $=\frac{\text{Rock total}}{\text{Grand total}}\times 100$.

1. Rock $=9+7+13=29$
2. Total sample $=28+32+46=106$
3. $\frac{29}{106}\times 100=27.358\dots \approx 27.36\%$

Trick: 29 is a bit more than a quarter of 106 (26.5) → answer just above 25%: only 27.36% fits.

Answer: (A) 27.36%

Concept: Above poverty line = $100\%-$ below; then split it in the given male : female ratio.

1. Town C: below = 20%, so above $=80\%$ of 8000 $=6400$
2. Above-line ratio M : F $=5:3$
3. Females $=6400\times \frac{3}{8}=2400$

Yaad rakhein: Use the "Above poverty line" column, not the "Below" one (that would give a different answer).

Answer: (C) 2400

Concept: $\frac{a^2+ab+b^2}{a^3-b^3}=\frac{1}{a-b}$ since $a^3-b^3=(a-b)(a^2+ab+b^2)$.

1. $a=157$, $b=133$
2. Value $=\frac{1}{157-133}=\frac{1}{24}$

Trick: When the difference of cubes is in the denominator, the answer is simply $\frac{1}{a-b}$.

Answer: (A) $\frac{1}{24}$