Q&A: practice questions with answers

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=540-405=$ ₹135.
2. $d=\frac{135}{540}\times 100=25$.

Trick: $\frac{405}{540}=\frac{3}{4}$, so SP is $75\%$ of MP; discount $=100-75=25\%$.

Answer: (D) 25

Concept: Read the highest and lowest points of the line graph and subtract.

1. Maximum production: June $=350$; minimum: January $=150$.
2. Difference $=350-150=200$.

Trick: Just scan the labelled peaks and dips – no calculation beyond one subtraction.

Answer: (C) 200

Concept: Subtract the two equations to isolate one variable.

1. $2T+3C=540$ and $2T+C=470$.
2. Subtract: $2C=70\Rightarrow C=35$, so $2T=470-35=435$.
3. Difference $=2T-2C=435-70=365$.

Trick: $2T-2C=(2T+C)-3C=470-105=$ ₹365 directly.

Answer: (C) ₹365

Concept: If price rises by $r\%$, consumption must fall by $\frac{r}{100+r}\times 100\%$ to keep expenditure same.

1. Reduction $=\frac{20}{120}\times 100$.
2. $=\frac{50}{3}=16\frac{2}{3}\%$.

Trick: $20\%=\frac{1}{5}$ increase $\Rightarrow$ $\frac{1}{5+1}=\frac{1}{6}=16\frac{2}{3}\%$ decrease.

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

Concept: A number is divisible by 9 if the sum of its digits is divisible by 9.

1. Sum $=5+k+2+1=8+k$.
2. $8+k=9\Rightarrow k=1$ (next, $18$, needs $k=10$ – not a digit).

Trick: Find the gap to the next multiple of 9: $9-8=1$.

Answer: (B) 1

Concept: Make decimal places equal, find LCM of the whole numbers, then put the decimal back.

1. $2.5=2.500$, $0.175$ → $2500$ and $175$ (3 decimal places).
2. $2500=2^2\times 5^4$, $175=5^2\times 7$; LCM $=2^2\times 5^4\times 7=17500$.
3. LCM $=\frac{17500}{1000}=17.5$.

Trick: LCM must be a multiple of both: $17.5÷2.5=7$ and $17.5÷0.175=100$ – both whole.

Answer: (C) 17.5

Concept: $A=P{(1+\frac{R}{100})}^n$, CI $=A-P$.

1. $A=2500\times (1.06{)}^2=2500\times 1.1236=2809$.
2. CI $=2809-2500=309$.

Trick: 2-year effective rate $=6+6+\frac{6\times 6}{100}=12.36\%$; $12.36\%$ of $2500=309$.

Answer: (B) ₹309

Concept: Solid hemisphere: TSA $=3\pi r^2$, CSA $=2\pi r^2$.

1. $3\pi r^2=166.32\Rightarrow \pi r^2=55.44$.
2. CSA $=2\pi r^2=2\times 55.44=110.88$ sq cm.

Trick: CSA $=\frac{2}{3}\times$ TSA $=\frac{2}{3}\times 166.32=110.88$.

Answer: (B) 110.88 sq cm

Concept: Take total work = LCM of days and use units per day.

1. Total $=\text{LCM}(24,40,20)=120$ units; Ansh $=5$, Tanvi $=3$, Shubham $=6$ units/day.
2. First 8 days: $8\times (5+3)=64$ units; left $=56$.
3. All three: $56÷14=4$ days. Total $=8+4=12$ days.

Trick: The LCM-unit method avoids fractions completely.

Answer: (C) 12 days

Concept: $\text{HCF}\times \text{LCM}=$ product of the two numbers.

1. $7\times 434=14\times x$.
2. $x=\frac{3038}{14}=217$.

Trick: $x=\frac{434}{14}\times 7=31\times 7=217$; check $\text{HCF}(14,217)=7$.

Answer: (C) 217

Concept: B’s one-day work $=$ (A+B)’s one-day work $-$ A’s one-day work.

1. B $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$, so B alone takes 4 days.
2. 20 times the work $=20\times 4=80$ days.

Trick: If A and B together take half of A’s time, B is as fast as A: $4$ days each.

Answer: (C) 80

Concept: Part of a quantity in ratio $a:b$ $=\frac{a}{a+b}\times$ total.

1. Milk $=\frac{3}{3+2}\times 20$.
2. $=12$ litres.

Trick: 5 parts $=20$ L, so 1 part $=4$ L; milk $=3\times 4=12$ L.

Answer: (D) 12 litres

Concept: $a^2-b^2=(a-b)(a+b)$.

1. Numerator $=(4.35-2.67)(4.35+2.67)=1.68\times 7.02$.
2. $\frac{1.68\times 7.02}{1.68}=7.02$.

Trick: When the denominator equals $a-b$, the answer is simply $a+b=4.35+2.67$.

Answer: (A) 7.02

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Take distance $=5$ units: $2$ at $50$ km/h, $3$ at $60$ km/h.
2. Time $=\frac{2}{50}+\frac{3}{60}=0.04+0.05=0.09$ h.
3. Average $=\frac{5}{0.09}=\frac{500}{9}=55\frac{5}{9}$ km/h.

Trick: Use distance $=300$ km: times $\frac{120}{50}+\frac{180}{60}=2.4+3=5.4$ h; $\frac{300}{5.4}=55\frac{5}{9}$.

Answer: (D) $55\frac{5}{9}$ km/h

Concept: If $a:k::k:b$, then $k^2=ab$ ($k$ is the mean proportional).

1. $k^2=144\times 36=5184$.
2. $k=\sqrt{5184}=12\times 6=72$, so $k-1=71$.

Trick: $\sqrt{144\times 36}=\sqrt{144}\times \sqrt{36}=12\times 6$.

Answer: (D) 71

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

1. $2000=\frac{10000\times R\times 2}{100}=200R$.
2. $R=10\%$.

Trick: Interest $=20\%$ of principal in 2 years $\Rightarrow$ $10\%$ per year.

Answer: (B) 10%

Concept: Required $\%=\frac{\text{cricket}}{\text{total}}\times 100$.

1. Total $=25+30+20+15+10=100$.
2. Cricket $=25$, so $\frac{25}{100}\times 100=25\%$.

Trick: When the total is exactly 100, each value is itself the percentage.

Answer: (B) 25%

Concept: Average of consecutive numbers = middle term.

1. Five numbers: $n-2,n-1,n,n+1,n+2$ (sum $5n$).
2. Next three: $n+3,n+4,n+5$ (sum $3n+12$).
3. New average $=\frac{8n+12}{8}=n+1.5$.

Trick: All 8 numbers run from $n-2$ to $n+5$; average $=\frac{(n-2)+(n+5)}{2}=n+1.5$.

Answer: (D) increase by 1.5

Concept: LCM = product of highest powers of all prime factors.

1. $36=2^2\times 3^2$, $54=2\times 3^3$, $72=2^3\times 3^2$, $96=2^5\times 3$.
2. LCM $=2^5\times 3^3=32\times 27=864$.

Trick: Check options: only $864$ is divisible by $96$ ($864÷96=9$) and by $54$.

Answer: (C) 864

Concept: Each child also ages 15 years, so the children’s total rises by 45.

1. Let children’s sum $=S$; father $=2S$.
2. After 15 years: $2S+15=S+45\Rightarrow S=30$.
3. Father now $=60$; after 15 years $=75$.

Trick: Remember: for 3 children the sum increases by $3\times 15$, not $15$.

Answer: (A) 75 years