Q&A: practice questions with answers

Concept: Take total work = LCM of times and work in units/hour.

1. LCM(10,20,12) $=60$ units. A+B $=6$, B+C $=3$, C+A $=5$ units/h
2. Adding: $2(A+B+C)=14\Rightarrow A+B+C=7$
3. B $=7-(C+A)=7-5=2$ units/h
4. Time for B $=\frac{60}{2}=30$ hours

Trick: Single pipe = (sum of all three pairs)$÷2$ − (pair without it).

Answer: (D) 30 hours

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=\frac{10}{2}=5$ m
2. Area $=4\times 3.14\times 25=314{\text{m}}^2$
3. Cost $=314\times 5=$ ₹1,570

Trick: $4\pi r^2=\pi d^2$; with $d=10$, area $=100\pi =314$ directly.

Answer: (C) ₹1570

Concept: New member = new total − old total.

1. Old total $=6\times 24000=1,44,000$
2. New total $=7\times 25500=1,78,500$
3. New employee $=178500-144000=$ ₹34,500

Trick: New income = new average + (rise in average × old count) $=25500+1500\times 6=34500$.

Answer: (D) ₹34,500

Concept: Proportion of science students $=\frac{\text{science}}{\text{total}}$; compare the fractions.

1. A: $\frac{5}{13}\approx 0.385$
2. B: $\frac{7}{18}\approx 0.389$
3. C: $\frac{3}{8}=0.375$
4. Highest is Batch B.

Trick: Compare $\frac{5}{13}$ and $\frac{7}{18}$ by cross-multiplying: $5\times 18=90<7\times 13=91$, so $\frac{7}{18}$ is bigger.

Answer: (A) Batch B

Concept: Third proportional to $a$ and $b$ is $\frac{b^2}{a}$.

1. $m=\frac{{20}^2}{4}=100$
2. $n=\frac{{50}^2}{100}=25$

Trick: $a:b::b:c\Rightarrow c=\frac{b^2}{a}$ – the middle term is squared.

Answer: (D) 25

Concept: Angles opposite equal sides are equal; angle sum of a triangle $={180}^{\circ }$.

1. $XY=XZ\Rightarrow \angle Z=\angle Y$
2. $\angle Y+\angle Z={180}^{\circ }-{40}^{\circ }={140}^{\circ }$
3. $\angle Z=\frac{{140}^{\circ }}{2}={70}^{\circ }$

Trick: Base angle of an isosceles triangle $=\frac{{180}^{\circ }-\text{vertex angle}}{2}$.

Answer: (B) ${70}^{\circ }$

Concept: Mean $=\frac{\sum fx}{\sum f}$.

1. $\frac{10\times 5+15p}{5+p}=12.5$
2. $50+15p=62.5+12.5p\Rightarrow 2.5p=12.5$
3. $p=5$

Trick: Mean 12.5 is exactly midway between 10 and 15, so the frequencies must be equal: $p=5$.

Answer: (B) 5

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (never the average of speeds).

1. Times: $\frac{100}{50}=2$ h, $\frac{150}{60}=2.5$ h, $\frac{50}{40}=1.25$ h
2. Total: 300 km in $5.75$ h
3. Average speed $=\frac{300}{5.75}=52.17\text{km/h}$

Trick: $\frac{300}{5.75}=\frac{1200}{23}\approx 52.17$ – only option B lies between 50 and 60 close to this value.

Answer: (B) 52.17 km/h

Concept: Successive changes multiply: $P\times (1+10\%)\times (1-15\%)$.

1. $P\times 1.1\times 0.85=0.935P$
2. $0.935P=187000\Rightarrow P=\frac{187000}{0.935}=2,00,000$

Trick: Net change $=10-15-\frac{10\times 15}{100}=-6.5\%$, so $93.5\%$ of $P=187000$.

Answer: (A) 2,00,000

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum at even places) is 0 or a multiple of 11.

1. 3762: $(3+6)-(7+2)=0$ ✓ divisible
2. 6184: $(6+8)-(1+4)=9$ ✗; 4915: $(4+1)-(9+5)=-9$ ✗; 7393: $(7+9)-(3+3)=10$ ✗

Trick: $3762=11\times 342$ – alternate-digit difference zero means divisible.

Answer: (A) 3,762

Concept: Dividend $=\text{Divisor}\times \text{Quotient}+\text{Remainder}$.

1. Divisor $=5\times 56=280$
2. Quotient $=\frac{280}{14}=20$
3. Dividend $=280\times 20+56=5656$

Trick: Find the divisor from the remainder first – it links both given relations.

Answer: (B) 5656

Concept: Net change for $+a\%$ then $-b\%$ $=a-b-\frac{ab}{100}$.

1. Net $=12-10-\frac{120}{100}=+0.8\%$
2. $0.8\%$ of salary $=1584\Rightarrow$ salary $=\frac{1584\times 100}{0.8}=$ ₹1,98,000

Trick: $1.12\times 0.9=1.008$; the extra $0.008$ part equals ₹1,584, so salary $=1584÷0.008$.

Answer: (D) 1,98,000

Concept: Speed $=\frac{\text{Distance}}{\text{Time}}$.

1. Original speed $=\frac{300}{5}=60\text{km/h}$
2. New speed $=\frac{300}{6}=50\text{km/h}$
3. Change $=60-50=10\text{km/h}$ decrease

Trick: More time for the same distance always means lower speed – so only "decrease" options qualify.

Answer: (D) 10 km/h decrease

Concept: Multiply inside each bracket, then divide.

1. $0.25\times 1.6=0.4$
2. $0.04\times 0.5=0.02$
3. $\frac{0.4}{0.02}=\frac{40}{2}=20$

Trick: Remove decimals: $\frac{25\times 16}{4\times 5}\times \frac{{10}^{-3}}{{10}^{-4}}=20\times 1=20$… or simply note $\frac{0.25}{0.5}=0.5$ and $\frac{1.6}{0.04}=40$, product 20.

Answer: (A) 20

Concept: Total work = LCM of days; efficiency = work per day.

1. LCM(12,39) $=156$ units; A $=13$, B $=4$ units/day
2. 4 days together: $4\times 17=68$ units; remaining $=88$
3. B alone: $\frac{88}{4}=22$ days

Trick: In 4 days A does $\frac{4}{12}=\frac{1}{3}$ of the work; B must do the rest alone, so B's remaining time $=39\times (1-\frac{1}{3}-\frac{4}{39})=22$.

Answer: (D) 22

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

1. $SI=\frac{8000\times 7.5\times 3}{100}=1800$
2. Amount $=8000+1800=$ ₹9,800

Trick: $7.5\%\times 3=22.5\%$ of 8000 $=1800$ in one step.

Answer: (A) ₹9800

Concept: $SP=CP\times (1+\text{profit}\%)$. Let marked price $=M$.

1. $CP=0.8M$, $SP=M+10$
2. $M+10=1.4\times 0.8M=1.12M$
3. $0.12M=10\Rightarrow M=\frac{10}{0.12}=83.33$

Trick: $1.4\times 0.8=1.12$, so the ₹10 above MP is exactly $12\%$ of MP.

Answer: (A) ₹83.33

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $23+(-14)+(-9)=0$
2. Sum $=3\times 23\times (-14)\times (-9)=3\times 23\times 126$
3. $=69\times 126=8694$

Trick: Whenever three cubes are added and the bases sum to zero, skip cubing – just use $3abc$.

Answer: (B) 8694

Concept: Translate words into an equation: $A=2B+5$.

1. $B=15\Rightarrow 2B=30$
2. $A=30+5=35$ years

Trick: "5 more than twice" = double first, then add 5.

Answer: (A) 35

Concept: Overall $\%=\frac{\text{net gain}}{\text{total CP}}\times 100$.

1. One-third: CP ₹800, profit $20\%=+160$
2. One-fourth: CP ₹600, loss $10\%=-60$
3. Rest at cost: $0$. Net gain $=100$
4. $\frac{100}{2400}\times 100=\frac{25}{6}=4\frac{1}{6}\%$ profit

Trick: Weighted profit $=\frac{1}{3}(20)+\frac{1}{4}(-10)+0=\frac{20}{3}-\frac{5}{2}=\frac{25}{6}\%$ – no need to use ₹2400 at all.

Answer: (A) $4\frac{1}{6}\%$ profit