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

Concept: Read the required bars, add them and reduce the ratio by the HCF.

1. School A (2008, 2012, 2013): $640+900+750=2290$
2. School B (2009, 2010, 2011): $820+600+750=2170$
3. Ratio $=2290:2170=229:217$ (divide by 10)

Trick: Totals ending in 0 – just drop the zero; 229 and 217 have no common factor.

Answer: (B) $229:217$

Concept: BODMAS – "of" is done before addition; solve numerator and denominator separately.

1. Numerator: $40+\frac{3}{4}\times 32=40+24=64$
2. Denominator: $37+\frac{3}{4}\times (34-6)=37+\frac{3}{4}\times 28=37+21=58$
3. Value $=\frac{64}{58}=\frac{32}{29}=1\frac{3}{29}$

Trick: Numerator > denominator, so the value is a little more than 1 – only (A) fits.

Answer: (A) $1\frac{3}{29}$

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. Single discount $=20+20-\frac{20\times 20}{100}=36\%$
2. $36\%$ of MP $=$ ₹331.20
3. MP $=\frac{331.20\times 100}{36}=$ ₹920

Trick: $1\%=\frac{331.2}{36}=9.2$, so $100\%=920$.

Answer: (A) ₹920

Concept: When time is equal, profit is shared in the ratio of investments.

1. $13750:16250:18750=11:13:15$ (divide by 1250)
2. B's 13 parts $=$ ₹5200, so 1 part $=$ ₹400
3. Difference of A and C $=(15-11)\times 400=$ ₹1600

Trick: Find "1 part" first – any share or difference is then just parts $\times$ 400.

Answer: (A) ₹1,600

Concept: Use $xz=y^2$ to write $x^2+y^2+z^2=(x+z{)}^2-y^2$.

1. $x^2+z^2+y^2=(x+z{)}^2-2xz+y^2=(x+z{)}^2-y^2$
2. $133=(x+z-y)(x+z+y)=19(x+z-y)\Rightarrow x+z-y=7$
3. With $x+z+y=19$: $y=6,x+z=13,xz=36$
4. $x=9,z=4\Rightarrow x-z=5$

Trick: Check: $81+36+16=133$ and $9+6+4=19$ ✓.

Answer: (A) $5$

Concept: Tangents drawn from an external point to a circle are equal.

1. $AR=AQ=2.6$, $CQ=CP=2.7$, $BP=BR=3$
2. Perimeter $=2(2.6+2.7+3)$
3. $=2\times 8.3=16.6\text{cm}$

Yaad rakhein: Perimeter of a triangle with an incircle $=2\times$(sum of the three different tangent lengths).

Answer: (A) $16.6$

Concept: In the right triangle formed by wall, ground and ladder, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. Height of the top of the ladder on the wall $=24\text{m}$
2. $\tan \theta =\frac{24}{\text{base}}=\frac{12}{5}$
3. Base $=\frac{24\times 5}{12}=10\text{m}$

Trick: 12 became 24 (×2), so 5 becomes 10.

Answer: (D) $10$

Concept: Total of the middle numbers = grand total − totals of the known groups.

1. Total $=12\times 39=468$; last five $=175$; first four $=160$
2. 5th + 6th + 7th $=468-175-160=133$
3. Let 5th $=a$: 6th $=a+6$, 7th $=a-5\Rightarrow 3a+1=133\Rightarrow a=44$
4. 6th $=50$, 7th $=39$; average $=\frac{50+39}{2}=44.5$

Trick: 6th + 7th $=133-44=89$, half of it is 44.5.

Answer: (D) $44.5$

Concept: "$x\%$ less than" means $\frac{\text{difference}}{\text{base value}}\times 100$, base = the value after "than".

1. Less than 55 kg: $20+45+30=95$
2. 55 kg to 65 kg: $60+40=100$
3. Percentage less $=\frac{100-95}{100}\times 100=5\%$

Trick: When the base is 100, the difference itself is the percentage.

Answer: (B) $5\%$

Concept: Amounts of consecutive years differ by the factor $(1+\frac{r}{100})$.

1. $\frac{8469.44}{7562}=1.12\Rightarrow r=12\%$
2. CI on ₹10000 for 2 years $=10000[(1.12{)}^2-1]$
3. $=10000\times 0.2544=$ ₹2544

Trick: 2-year CI rate $=12+12+\frac{12\times 12}{100}=25.44\%$.

Answer: (A) ₹2,544

Concept: $V=\pi r^{2}h$ and CSA $=2\pi rh=\frac{2V}{r}$.

1. $\frac{22}{7}\times {14}^2\times h=6160\Rightarrow 616h=6160\Rightarrow h=10$
2. CSA $=2\times \frac{22}{7}\times 14\times 10=880{\text{cm}}^2$

Trick: CSA $=\frac{2V}{r}=\frac{2\times 6160}{14}=880$ directly.

Answer: (B) $880$

Concept: Opposite directions → relative speed is the sum; distance covered = sum of the lengths.

1. Relative speed $=74+52=126\text{km/h}=126\times \frac{5}{18}=35\text{m/s}$
2. $x+y=35\times 12=420\text{m}$
3. $y=\frac{2}{3}x\Rightarrow \frac{5x}{3}=420\Rightarrow x=252$, $y=168\text{m}$

Trick: Lengths $x:y=3:2$, so $y=\frac{2}{5}\times 420=168$.

Answer: (B) $168$

Concept: Complementary angles: $\csc ({90}^{\circ }-\theta )=\sec \theta$, $\tan ({90}^{\circ }-\theta )=\cot \theta$.

1. ${\csc }^2{57}^{\circ }={\sec }^2{33}^{\circ }\Rightarrow {\sec }^2{33}^{\circ }-{\tan }^2{33}^{\circ }=1$, so first term $=4$
2. $\cos {90}^{\circ }=0$; $\tan {66}^{\circ }\tan {24}^{\circ }=\tan {66}^{\circ }\cot {66}^{\circ }=1$
3. $4-0+y=\frac{y}{2}\Rightarrow \frac{y}{2}=-4\Rightarrow y=-8$

Yaad rakhein: If $A+B={90}^{\circ }$, then $\tan A\tan B=1$.

Answer: (D) $-8$

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

1. Total students $=15+13+10+14+12+6=70$
2. 165–175: $14+12=26$
3. $\frac{26}{70}\times 100=37.14\%\approx 37$

Trick: $\frac{26}{70}$ is a bit more than $\frac{1}{3}$ – only 37 is close.

Answer: (B) $37$

Concept: $SI=\frac{P\times R\times T}{100}$.

1. SI $=20400-12000=$ ₹8400
2. $R=\frac{8400\times 100}{12000\times 7}=\frac{840000}{84000}=10\%$

Trick: Interest is $70\%$ of principal in 7 years, so $10\%$ per year.

Answer: (D) $10\%$

Concept: Both averages are over the same 6 years, so the ratio of averages = ratio of totals.

1. School B total $=550+820+600+750+500+480=3700$
2. School A total $=640+800+500+700+900+750=4290$
3. Ratio $=3700:4290=370:429$

Trick: Same number of years – skip dividing by 6. B comes first, so the smaller number comes first.

Answer: (C) $370:429$

Concept: Number leaving the same remainder $r$ = $k\times \text{LCM}+r$; then test divisibility by 11.

1. LCM(15, 18, 36) $=180$; number $=180k+9$
2. $k=1$: 189, $k=2$: 369, … check 11: $180k+9\equiv 4k+9\left(\mathrm{mod}11\right)$
3. $4k+9\equiv 0\Rightarrow k=6$; number $=1089=11\times 99$
4. Sum of digits $=1+0+8+9=18$

Trick: 1089: alternate sum $(9+0)-(8+1)=0$ → divisible by 11.

Answer: (C) $18$

Concept: Convert averages into totals and subtract.

1. $A+B+C=3\times 65=195$
2. $B+C=2\times 61.5=123\Rightarrow A=72$
3. $A+C=2\times 68.5=137\Rightarrow C=137-72=65\text{kg}$

Trick: $C=(B+C)+(A+C)-(A+B+C)=123+137-195=65$.

Answer: (A) $65$

Concept: Efficiency ratio gives units of work per day; total work = combined efficiency × days.

1. A : B efficiency $=2:1$
2. Total work $=(2+1)\times 13=39$ units
3. A alone $=\frac{39}{2}=19\frac{1}{2}$ days

Trick: A does $\frac{2}{3}$ of the joint work rate, so A takes $13\times \frac{3}{2}=19.5$ days.

Answer: (C) $19\frac{1}{2}$

Concept: If numbers are in ratio $a:b$ (co-prime) and HCF $=h$, the numbers are $ah$ and $bh$.

1. Numbers $=7\times 28=196$ and $11\times 28=308$
2. Sum $=196+308=504$

Trick: Sum $=(7+11)\times 28=18\times 28=504$.

Answer: (A) $504$