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

Concept: Equal distances at speeds $x$ and $y$: average speed $=\frac{2xy}{x+y}$.

1. $x=30,y=70$
2. Average $=\frac{2\times 30\times 70}{30+70}=\frac{4200}{100}$
3. $=42\text{km/h}$

Trick: Average speed for equal distances is always less than the simple mean (50 here).

Answer: (A) $42\text{km/h}$

Concept: If A is $r\%$ more than B, then B is less than A by $\frac{r}{100+r}\times 100\%$.

1. Let Madhu $=100$, then Rupu $=120$
2. Madhu is less by $\frac{20}{120}\times 100$
3. $=\frac{50}{3}=16\frac{2}{3}\%$

Trick: $20\%=\frac{1}{5}$; answer $=\frac{1}{5+1}=\frac{1}{6}=16\frac{2}{3}\%$.

Answer: (D) $16\frac{2}{3}\%$

Concept: Under SI, each instalment $x$ earns interest till the due date; $\text{Debt}=\sum x(1+\frac{rt}{100})$.

1. 1st instalment earns 2 years, 2nd earns 1 year, 3rd earns 0
2. $x(1+0.245)+x(1+0.1225)+x=3.3675x$
3. $3.3675x=9429\Rightarrow x=2800$

Trick: Formula: $x=\frac{100A}{100n+\frac{n(n-1)}{2}r}=\frac{942900}{300+36.75}=2800$.

Answer: (B) 2800

Concept: Compare the column totals of all schools.

1. Totals: V $=917$, VI $=924$, VII $=904$, VIII $=919$, IX $=903$, X $=910$
2. Smallest total $=903$ (class IX)

Trick: VII (904) and IX (903) are very close — compare the exact totals.

Answer: (B) IX

Concept: Least number divisible by all = LCM (highest power of every prime).

1. $20=2^2\times 5,28=2^2\times 7,34=2\times 17,60=2^2\times 3\times 5,75=3\times 5^2$
2. LCM $=2^2\times 3\times 5^2\times 7\times 17$
3. $=300\times 119=35700$

Trick: The LCM must be divisible by 17 and 7, i.e. by 119: only $35700=119\times 300$ works.

Answer: (D) 35700

Concept: Mean proportional of $a$ and $b$ $=\sqrt{ab}$.

1. $ab=0.03\times 0.0003=0.000009$
2. $\sqrt{0.000009}=0.003$

Trick: Decimal places in the square root = half of 6, i.e. 3: $\sqrt{9}=3\to 0.003$.

Answer: (C) 0.003

Concept: $a^3+b^3+c^3-3abc=(a+b+c)[(a+b+c{)}^2-3(ab+bc+ca)]$.

1. $a+b+c=5,ab+bc+ca=7$
2. $(a+b+c{)}^2-3(ab+bc+ca)=25-21=4$
3. Value $=5\times 4=20$

Trick: Remember the short form $S(S^2-3P)$ with $S=a+b+c$, $P=ab+bc+ca$.

Answer: (B) 20

Concept: Central angle $=\frac{\text{percentage}}{100}\times {360}^{\circ }$; the total surveyed is not needed.

1. Football $=30\%$
2. Angle $=\frac{30}{100}\times {360}^{\circ }={108}^{\circ }$

Trick: $1\%={3.6}^{\circ }$, so $30\times {3.6}^{\circ }={108}^{\circ }$.

Answer: (A) ${108}^{\circ }$

Concept: Net effect of $+a\%$ then $-b\%$: $a-b-\frac{ab}{100}$.

1. Let CP $=100$; MP $=150$
2. SP $=150\times 0.6=90$
3. Loss $=100-90=10\Rightarrow 10\%$

Trick: $+50-40-\frac{50\times 40}{100}=-10\%$.

Answer: (C) $10\%$

Concept: $\text{SP}=\text{MP}\times (1-\frac{d}{100})$.

1. $z\times 0.88=6622$
2. $z=\frac{6622}{0.88}=7525$

Trick: Check: 12% of 7525 $=903$; $7525-903=6622$.

Answer: (C) ₹7,525

Concept: Required $\%=\frac{\text{Hemanth Mid-I}}{\text{total Mid-I}}\times 100$.

1. Mid-I marks: Sunil 33, Hemanth 30, Rehman 36, Mohan 21
2. Total $=33+30+36+21=120$
3. $\frac{30}{120}\times 100=25\%$

Trick: $\frac{30}{120}=\frac{1}{4}=25\%$.

Answer: (D) $25\%$

Concept: Find the maximum and minimum values and subtract.

1. Maximum: Sales $=115$ lakh
2. Minimum: Communication $=60$ lakh
3. Difference $=115-60=55$ lakh

Trick: Scan the row once, marking the highest and lowest.

Answer: (D) 55

Concept: Half-yearly: rate $=\frac{r}{2}$, periods $=2t$; $\text{CI}=P[{(1+\frac{R}{100})}^n-1]$.

1. Rate $=10\%$ per half-year, $n=2$
2. Amount $=8000\times {1.1}^2=9680$
3. CI $=9680-8000=1680$

Trick: Two periods at 10%: effective rate $=10+10+\frac{10\times 10}{100}=21\%$; $21\%$ of 8000 $=1680$.

Answer: (D) ₹1,680

Concept: Factorise both expressions and take the common factor.

1. $x^6+1=(x^2{)}^3+1=(x^2+1)(x^4-x^2+1)$
2. $x^4-1=(x^2-1)(x^2+1)$
3. Common factor $=x^2+1$

Trick: $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=x^2,b=1$.

Answer: (D) $(1+x^2)$

Concept: Convert mixed fractions, then do $÷$ and $\times$ left to right before $+$ and $-$.

1. Numerator: $\frac{3}{2}÷\frac{13}{4}=\frac{6}{13}$, $\frac{1}{2}÷\frac{13}{14}=\frac{7}{13}$; sum $=1+\frac{1}{5}=\frac{6}{5}$
2. Denominator: $\frac{1}{5}\times \frac{7}{2}=\frac{7}{10}$; $\frac{1}{3}÷\frac{7}{4}\times \frac{7}{2}=\frac{4}{21}\times \frac{7}{2}=\frac{2}{3}$
3. $\frac{7}{10}-\frac{2}{3}=\frac{1}{30}$
4. Value $=\frac{6}{5}÷\frac{1}{30}=36$

Trick: $\frac{6}{13}+\frac{7}{13}=1$ — spotting the same denominator saves time.

Answer: (B) 36

Concept: Convert everyone to one unit using $2\text{M}=4\text{W}$, i.e. $1\text{M}=2\text{W}$.

1. Work $=4\text{W}\times 34=136$ woman-days
2. $6\text{M}+5\text{W}=12\text{W}+5\text{W}=17\text{W}$
3. Days $=\frac{136}{17}=8$

Trick: $M_1{D}_1=M_2{D}_2$: $4\times 34=17\times D\Rightarrow D=8$.

Answer: (B) 8

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

1. $78=2\times 3\times 13,84=2^2\times 3\times 7$
2. $90=2\times 3^2\times 5,112=2^4\times 7$
3. Only 2 is common to all $\Rightarrow$ HCF $=2$

Trick: 112 is not divisible by 3 and 78 not by 7, so nothing beyond 2 can divide all.

Answer: (B) 2

Concept: Amount $=\frac{\text{percentage}}{100}\times \text{total}$.

1. Laptop $=30\%$, TV $=15\%$
2. Difference $=30\%-15\%=15\%$
3. $15\%$ of 300000 $=45000$

Trick: Subtract percentages first, then convert once.

Answer: (D) ₹45,000

Concept: $\sin ({180}^{\circ }-A)=\sin A$.

1. $\cos A=\frac{1}{2}\Rightarrow A={60}^{\circ }$
2. $\sin ({180}^{\circ }-{60}^{\circ })=\sin {60}^{\circ }$
3. $=\frac{\sqrt{3}}{2}$

Yaad rakhein: ${180}^{\circ }-A$ lies in the 2nd quadrant, where only sin (and cosec) is positive.

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

Concept: Profit $\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$, where CP is the cost of the quantity actually given.

1. Let CP $=1$ per gm; he charges for 50 gm at 24% loss: SP $=50\times 0.76=38$
2. He gives only 19 gm: actual CP $=19$
3. Profit $=38-19=19\Rightarrow \frac{19}{19}\times 100=100\%$

Trick: $\frac{\text{SP}}{\text{CP}}=\frac{76}{100}\times \frac{50}{19}=2$, i.e. SP is double the CP.

Answer: (D) $100\%$ profit