Q&A: practice questions with answers

Concept: Work done by a pipe = (time open) $\times$ (its 1-minute rate). Pipe A runs the full 30 min; B runs only $t$ min.

1. A's rate $=\frac{1}{37.5}=\frac{2}{75}$, B's rate $=\frac{1}{45}$.
2. A in 30 min fills $30\times \frac{2}{75}=\frac{4}{5}$ of the cistern.
3. Remaining $1-\frac{4}{5}=\frac{1}{5}$ is filled by B: $\frac{t}{45}=\frac{1}{5}\Rightarrow t=9$ min.

Trick: Take LCM capacity 225 units: A $=6$ units/min, B $=5$ units/min. A gives $6\times 30=180$; B must give 45 units $\Rightarrow 45÷5=9$ min.

Answer: (A) 9 min

Concept: If numbers are $5x$ and $3x$, their difference equals $(5-3)x$.

1. $5x-3x=46\Rightarrow 2x=46\Rightarrow x=23$.
2. Smaller number $=3x=3\times 23=69$ (larger $=115$).

Trick: Difference of ratio terms $=2$ parts $=46$, so 1 part $=23$; smaller $=3$ parts $=69$.

Answer: (D) 69

Concept: With two candidates and all votes valid, the winner gets $100\%-38\%=62\%$. The losing margin is the difference of their percentages.

1. Margin $=62\%-38\%=24\%$ of total votes.
2. $24\%$ of total $=7200\Rightarrow \text{Total}=\frac{7200\times 100}{24}=30000$.

Trick: $1\%=\frac{7200}{24}=300$, so $100\%=30000$.

Answer: (D) 30,000

Concept: Write every term with the same base 2, then equate the powers: $a^m\times a^n=a^{m+n}$.

1. $4^{12}=(2^2{)}^{12}=2^{24}$ and ${16}^{11}=(2^4{)}^{11}=2^{44}$.
2. $2^K\times 2^{24}=2^{44}\Rightarrow K+24=44$.
3. $K=20$.

Yaad rakhein: $(a^m{)}^n=a^{mn}$ — always convert $4,8,16,32$ into powers of 2 first.

Answer: (A) 20

Concept: Wages are shared in the ratio of work done. In 3 days the whole work is done, so find C's 1-day work first.

1. C's 1-day work $=\frac{1}{3}-\frac{1}{6}-\frac{1}{8}=\frac{8-4-3}{24}=\frac{1}{24}$.
2. C's work in 3 days $=3\times \frac{1}{24}=\frac{1}{8}$.
3. C's share $=\frac{1}{8}\times 3848=481$.

Trick: Work done in 3 days: A $=\frac{3}{6}=\frac{1}{2}$, B $=\frac{3}{8}$, C $=$ rest $=1-\frac{1}{2}-\frac{3}{8}=\frac{1}{8}$; ratio $4:3:1$.

Answer: (A) ₹481

Concept: Convert everything into grams: $1\text{ kg}=1000\text{ g}$. Empty box = full box $-$ dry fruits.

1. Full box $=12.650\text{ kg}=12650\text{ g}$.
2. Dry fruits $=9\text{ kg }800\text{ g}=9800\text{ g}$.
3. Empty box $=12650-9800=2850\text{ g}$.

Yaad rakhein: Three digits after the decimal in kg $=$ grams: $12.650\text{ kg}=12\text{ kg }650\text{ g}$.

Answer: (B) 2850 g

Concept: If $\frac{x}{7}=\frac{y}{9}$, then $x:y=7:9$. Divide 96 in this ratio.

1. Total parts $=7+9=16$; one part $=\frac{96}{16}=6$.
2. First part $=7\times 6=42$, second part $=9\times 6=54$.
3. Check: $\frac{42}{7}=6=\frac{54}{9}$ ✓. Larger part $=54$.

Trick: $\frac{x}{a}=\frac{y}{b}\Rightarrow x:y=a:b$ directly.

Answer: (D) 54

Concept: Divisible by 2, 3 and 5 means divisible by $2\times 3\times 5=30$: the number must end in 0 and its digit sum must be divisible by 3.

1. Ending in 0: only 2240 and 3150.
2. Digit sum of 2240 $=8$ (not divisible by 3) ✗.
3. Digit sum of 3150 $=9$ ✓, so $3150÷30=105$.

Trick: Divisible by 10 (last digit 0) + digit sum multiple of 3 $\Rightarrow$ divisible by 30.

Answer: (C) 3150

Concept: A tunnel is open at both ends, so only its curved surface is needed: $\text{CSA}=2\pi rh$.

1. Diameter 2 m $\Rightarrow r=1$ m; length $h=40$ m.
2. $\text{CSA}=2\pi \times 1\times 40=80\pi {\mathrm{m}}^2$.

Yaad rakhein: Tunnel/pipe = curved surface only (no top and bottom circles).

Answer: (A) $80\pi$

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

1. Interest for 2 years $=2\times 12.5\%=25\%$ of $P$.
2. $A=1.25P=17500\Rightarrow P=14000$.
3. $\text{SI}=17500-14000=3500$.

Trick: $25\%=\frac{1}{4}$: if principal $=4$ parts, amount $=5$ parts. $5=17500\Rightarrow$ SI $=1$ part $=3500$.

Answer: (D) ₹3,500

Concept: If $x$ is subtracted from $a$ to get $b$: $a-x=b\Rightarrow x=a-b$.

1. $-\frac{5}{6}-x=1$.
2. $x=-\frac{5}{6}-1=-\frac{5}{6}-\frac{6}{6}=-\frac{11}{6}$.

Trick: Check: $-\frac{5}{6}-(-\frac{11}{6})=\frac{6}{6}=1$ ✓.

Answer: (D) $\frac{-11}{6}$

Concept: Total surface = outer curved surface + bottom circle + top ring + curved wall of the hole + floor of the hole.

1. Outer CSA $=2\pi \times 5\times 7=70\pi$; bottom $=\pi \times 5^2=25\pi$.
2. Top ring $=\pi (5^2-3^2)=16\pi$.
3. Hole wall $=2\pi \times 3\times 2=12\pi$; hole floor $=\pi \times 3^2=9\pi$.
4. Total $=70\pi +25\pi +16\pi +12\pi +9\pi =132\pi$.

Trick: Ring + hole floor $=$ full top circle $25\pi$. So TSA $=$ original TSA $(2\pi \times 5\times 12=120\pi )$ + hole wall $12\pi =132\pi$.

Answer: (D) $132\pi$

Concept: $\text{Gain}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. A: gain $=576-384=192$; $\text{Gain}\%=\frac{192}{384}\times 100=50\%$.
2. B: gain $=1672-1254=418$; $\text{Gain}\%=\frac{418}{1254}\times 100=33\frac{1}{3}\%$.
3. Ratio $=50:\frac{100}{3}=\frac{3}{2}$.

Trick: $\frac{576}{384}=\frac{3}{2}$ (gain $\frac{1}{2}$) and $\frac{1672}{1254}=\frac{4}{3}$ (gain $\frac{1}{3}$); ratio $\frac{1}{2}:\frac{1}{3}=3:2$.

Answer: (A) $\frac{3}{2}$

Concept: $cosec\theta =\sec ({90}^{\circ }-\theta )$. So if $cosecA=\sec B$, then $A+B={90}^{\circ }$.

1. $cosec2\theta =\sec (\theta -{30}^{\circ })$.
2. $2\theta +(\theta -{30}^{\circ })={90}^{\circ }\Rightarrow 3\theta ={120}^{\circ }$.
3. $\theta ={40}^{\circ }$.

Trick: Check: $cosec{80}^{\circ }=\sec {10}^{\circ }$ ✓ because ${80}^{\circ }+{10}^{\circ }={90}^{\circ }$.

Answer: (A) 40

Concept: $a^{-n}=\frac{1}{a^n}$, so ${\left(\frac{1}{a}\right)}^{-n}=a^n$; and $a^{\frac{1}{3}}=\sqrt[3]{a}$.

1. ${\left(\frac{1}{64}\right)}^{-\frac{1}{3}}={64}^{\frac{1}{3}}$.
2. ${64}^{\frac{1}{3}}=\sqrt[3]{64}=4$.

Trick: Negative power flips the fraction; then take the cube root: $4^3=64$.

Answer: (D) 4

Concept: Use $(a-b)(a+b)=a^2-b^2$ with $a=10000$, $b=6$.

1. $9994\times 10006=(10000-6)(10000+6)$.
2. $={10000}^2-6^2=100000000-36$.
3. $=99999964$.

Trick: Numbers equidistant from a round base: product $={\text{base}}^2-{\text{gap}}^2$.

Answer: (C) 99999964

Concept: Moving towards each other, the gap closes at the sum of speeds: $12+13=25$ km/h.

1. To be 6 km apart (before meeting), the gap must reduce by $56-6=50$ km.
2. Time $=\frac{50}{25}=2$ hours.

Yaad rakhein: After meeting they would be 6 km apart again at $\frac{62}{25}$ h; only 2 h appears in the options.

Answer: (B) 2 hours

Concept: Take the youngest as $x$ and express the others in terms of $x$.

1. Tanu $=x$, Kanu $=2x$, Anu $=2x+2$.
2. $x+2x+2x+2=32\Rightarrow 5x=30\Rightarrow x=6$.
3. Kanu $=2\times 6=12$ years.

Trick: Check options: Kanu $=12\Rightarrow$ Tanu $=6$, Anu $=14$; $6+12+14=32$ ✓.

Answer: (C) 12 years

Concept: Total marks of class = (boys' total) + (girls' total).

1. Let girls' average $=g$; boys' average $=g-5$.
2. $30(g-5)+20g=50\times 25=1250\Rightarrow 50g=1400\Rightarrow g=28$.
3. Boys' average $=23$; ratio $=23:28$.

Trick: Class average 25 is a weighted mean: $25=\frac{30b+20(b+5)}{50}\Rightarrow b=23$.

Answer: (B) 23 : 28

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

1. $33\frac{1}{3}\%=\frac{1}{3}$: B $=3$, A $=4$.
2. B is less by $\frac{1}{4}\times 100=25\%$.

Trick: Fraction method: $\frac{1}{3}$ more $\Rightarrow \frac{1}{3+1}=\frac{1}{4}$ less $=25\%$.

Answer: (A) 25%