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

Concept: Pick the two values carefully (Neem 2021, Birch 2022).

1. Neem 2021 $=5000$
2. Birch 2022 $=9000$
3. Ratio $=5000:9000=5:9$

Trick: Read row and column twice – $5:7$ (Birch 2021) is the trap.

Answer: (A) $5:9$

Concept: Second discount is on the price after the first discount.

1. After $18\%$: $26000\times 0.82=21320$
2. Season discount $=21320-16000=5320$
3. $\frac{5320}{21320}\times 100\approx 24.95\%$

Trick: Never take the second discount on the marked price.

Answer: (B) $24.95\%$

Concept: In a right triangle, the altitude to the hypotenuse makes $\angle QPS=\angle R$.

1. $\angle Q={180}^{\circ }-{90}^{\circ }-{47}^{\circ }={43}^{\circ }$
2. In $△PQS$: $\angle QPS={90}^{\circ }-{43}^{\circ }={47}^{\circ }$

Trick: Altitude from the right angle swaps the acute angles.

Answer: (B) ${47}^{\circ }$

Concept: Take total work = LCM $(8,12)=24$ units: Sanjay $3$/day, Rohan $2$/day.

1. Every 2 days: $5$ units → 8 days: $20$ units
2. Day 9 (Sanjay): $23$ units
3. Remaining $1$ unit by Rohan: $\frac{1}{2}$ day
4. Total $=9\frac{1}{2}$ days

Trick: Stop at the last full cycle that leaves work less than one pair.

Answer: (B) $9\frac{1}{2}$

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

1. Above poverty line in R $=76\%$ of $26000=19760$
2. Ratio $3:3$ → females $=\frac{19760}{2}=9880$

Trick: $26000\times 0.76\times \frac{1}{2}=13000\times 0.76=9880$.

Answer: (B) $9,880$

Concept: Perpendicular from the centre bisects the chord.

1. Half chord $=\sqrt{{1.7}^2-{0.8}^2}=\sqrt{2.89-0.64}=\sqrt{2.25}=1.5$
2. AB $=2\times 1.5=3\text{cm}$

Trick: $(8,15,17)$ triplet scaled by $0.1$: $0.8,1.5,1.7$.

Answer: (C) $3$

Concept: Share $\%=\frac{\text{part}}{\text{total}}\times 100$.

1. Total $=8000+2000+6000+4000=20000$
2. Clothing $=\frac{2000}{20000}\times 100=10\%$

Trick: $2000$ is $\frac{1}{10}$ of $20000$.

Answer: (A) $10\%$

Concept: Find the HCF, then solve for $m$.

1. HCF of $45$ and $55=5$
2. $55\times 5+45m=5\Rightarrow 275+45m=5$
3. $45m=-270\Rightarrow m=-6$

Trick: HCF is small, so $m$ must be negative to cancel $275$.

Answer: (A) $-6$

Concept: Savings % = $100\%-$ total spending %.

1. Spending $=40\%+25\%+15\%=80\%$
2. Monthly saving $=20\%$ of $50000=10000$
3. Yearly $=10000\times 12=120000$

Trick: Don't forget to multiply by $12$ for a year.

Answer: (C) ₹1,20,000

Concept: Overall gain factor = (price gain factor) × (weight gain factor).

1. CP $=0.88$SP → price factor $=\frac{1}{0.88}=\frac{25}{22}$
2. Needed factor $=1.25=\frac{5}{4}$
3. Weight factor $=\frac{5}{4}÷\frac{25}{22}=\frac{22}{20}=1.1$
4. So the machine shows $10\%$ more weight.

Trick: $\frac{1.25}{1.136}\approx 1.1$ → $10\%$.

Answer: (C) $10\%$

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) $9\frac{1}{11}\%$