Q&A: practice questions with answers

Concept: Fourth proportional to $a,b,c$ is $x$ with $a:b=c:x$, so $x=\frac{b\times c}{a}$. With powers of the same base, add/subtract exponents.

1. $x=\frac{2^{1.5}\times 2^{2.5}}{2^{0.5}}$
2. $=2^{1.5+2.5-0.5}=2^{3.5}$

Trick: Fourth proportional = (2nd × 3rd) ÷ 1st. For same base: multiply → add powers, divide → subtract powers.

Answer: (D) $2^{3.5}$

Concept: Apply BODMAS: division and multiplication (left to right) before addition.

1. $(-56)÷(-3)=\frac{56}{3}$, then $\frac{56}{3}\times 3=56$
2. $(-37)+56=19$

Trick: Negative ÷ negative = positive. The $÷3$ and $\times 3$ cancel, leaving $-37+56$.

Answer: (C) 19

Concept: Required percentage $=\frac{\text{part}}{\text{whole}}\times 100$; the number itself cancels out.

1. $\frac{7\%\text{ of }N}{9\%\text{ of }N}\times 100=\frac{7}{9}\times 100$
2. $=77.777\dots \%\approx 77.78\%$

Trick: Such questions never depend on the number — just divide the two percentages: $\frac{7}{9}=0.7778$.

Answer: (D) 77.78%

Concept: Write every base as a product of primes; the total number of prime factors is the sum of all exponents (counted with repetition).

1. $6^7=(2\times 3{)}^7=2^7\times 3^7$
2. $N=5^8\times 2^7\times 3^7\times {11}^4$
3. Total prime factors $=8+7+7+4=26$

Trick: Only 6 is composite here — split it first. (Distinct primes would be 4; total prime factors counts repetition.)

Answer: (A) 26

Concept: Convert each percentage relation into a ratio, then combine by making the common term (B) equal.

1. $25\%A=40\%B\Rightarrow \frac{A}{B}=\frac{40}{25}=\frac{8}{5}$
2. $30\%B=45\%C\Rightarrow \frac{B}{C}=\frac{45}{30}=\frac{3}{2}$
3. Make B equal: $A:B=24:15$, $B:C=15:10$ → $A:B:C=24:15:10$

Trick: $pA=qB$ gives $A:B=q:p$ (the numbers swap sides).

Answer: (D) 24:15:10

Concept: Sum $=$ average $\times$ number of terms.

1. Sum of three numbers $=15\times 3=45$
2. Third number $=45-(10+20)=15$

Trick: If two numbers average to the given mean (here $\frac{10+20}{2}=15$), the third must equal the mean itself.

Answer: (B) 15

Concept: Floor + ceiling $=2lb$; four walls $=2h(l+b)$. Volume $=lbh$, and $1{\text{m}}^3=1000$ litres.

1. $2\times 15\times 12=2h(15+12)\Rightarrow 360=54h\Rightarrow h=\frac{20}{3}\text{m}$
2. Volume $=15\times 12\times \frac{20}{3}=1200{\text{m}}^3$
3. $=1200\times 1000=1200\times {10}^3$ litres

Trick: Floor + ceiling = walls ⇒ $h=\frac{lb}{l+b}$. Then always convert m³ → litres by $\times 1000$.

Answer: (D) $1200\times {10}^3$ liters

Concept: Work per day $=\frac{1}{\text{days}}$. Efficiency of B $=1.5\times$ efficiency of A.

1. A's one-day work $=\frac{1}{37}-\frac{1}{74}=\frac{1}{74}$
2. B's one-day work $=\frac{3}{2}\times \frac{1}{74}=\frac{3}{148}$
3. Work done in 28 days $=\frac{28}{37}$, remaining $=\frac{9}{37}$
4. B alone: $\frac{9}{37}÷\frac{3}{148}=\frac{9}{37}\times \frac{148}{3}=12$ days

Trick: Take total work $=148$ units: group does 4/day, A does 2/day, B does 3/day. Remaining $148-28\times 4=36$ units $\Rightarrow 36÷3=12$ days.

Answer: (C) 12 days

Concept: Division and multiplication first (left to right), then subtraction.

1. $12÷12=1$, then $1\times 8=8$
2. $11-8=3$

Trick: Never subtract first: $11-12$ is the classic trap. Resolve the $÷$ and $\times$ chain before touching $-$.

Answer: (A) 3

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

1. Times: $\frac{30}{20}=1.5$ h, $\frac{50}{25}=2$ h, $\frac{20}{10}=2$ h → total $5.5$ h
2. Average speed $=\frac{100}{5.5}=18.18\approx 18\text{km/h}$

Trick: Different distances at different speeds → compute each time separately; the slow 10 km/h stretch pulls the average well below 20.

Answer: (A) 18 km/h

Concept: Units produced $\propto$ workers $\times$ hours $\times$ days.

1. Hours and days unchanged; workers doubled ($5\to 10$)
2. Units $=2400\times \frac{10}{5}=4800$

Trick: $\frac{M_1{H}_1{D}_1}{W_1}=\frac{M_2{H}_2{D}_2}{W_2}$ — change only the quantity that changed.

Answer: (B) 4800 units

Concept: Amount $=P{(1+\frac{r}{100})}^n$. Equal amounts after different years ⇒ principals are in inverse ratio of the extra growth.

1. $V(1.08{)}^{18}=N(1.08{)}^{19}\Rightarrow V=1.08N$
2. $V:N=108:100=27:25$
3. Nitin's share $=\frac{25}{52}\times 1352=25\times 26=650$

Trick: One extra year of CI at $r\%$ ⇒ shares are $(100+r):100$. At 8%: $27:25$; $1352÷52=26$.

Answer: (D) 650

Concept: Expenditure $=$ Income $-$ Savings. Set incomes as $6x,8x$ and use the expenditure ratio.

1. $\frac{6x-51000}{8x-51000}=\frac{2}{4}=\frac{1}{2}$
2. $12x-102000=8x-51000\Rightarrow 4x=51000\Rightarrow x=12750$
3. Mohan's income $=6\times 12750=76,500$

Trick: Equal savings: income difference $=$ expenditure difference. Incomes differ by $2x$, expenditures by $y$ (ratio 1:2) ⇒ $y=2x$, $6x-y=51000\Rightarrow x=12750$.

Answer: (A) 76,500

Concept: Speed $=\frac{\text{distance}}{\text{time}}$; convert km to metres first.

1. $1.45\text{km}=1450\text{m}$
2. Speed $=\frac{1450}{5}=290\text{m/s}$

Trick: Match units before dividing: km → m ($\times 1000$). $1450÷5=290$.

Answer: (A) 290 m/s

Concept: Successive changes multiply: price $\times (1-\text{discount})\times (1+\text{tax})$.

1. After 25% discount: $45000\times \frac{3}{4}=33750$
2. After 10% tax: $33750\times 1.1=37125$

Trick: $-25\%$ then $+10\%$ is a net $-17.5\%$ (by $a+b+\frac{ab}{100}$): $45000\times 0.825=37125$.

Answer: (B) ₹37,125

Concept: Selling at a 5% 'loss' means SP $=0.95\times$ CP of 1 kg. Actual gain 40% means true cost of goods given $=\frac{\text{SP}}{1.4}$.

1. SP of '1 kg' $=0.95C$ (C = cost of a real kg)
2. True cost of goods handed over $=\frac{0.95C}{1.4}=0.6786C$
3. Weight given $=0.6786\times 1000\approx 678.6$ g

Trick: False-weight formula: $\text{weight used}=1000\times \frac{100-\text{loss}\%}{100+\text{gain}\%}=1000\times \frac{95}{140}$.

Answer: (C) 678.6 gm

Concept: Number of shots $=\frac{\text{volume of cuboid}}{\text{volume of one sphere}}$, sphere volume $=\frac{4}{3}\pi r^3$.

1. Cuboid $=54\times 44\times 30=71280{\text{cm}}^3$
2. One shot $=\frac{4}{3}\times \frac{22}{7}\times 27=\frac{792}{7}{\text{cm}}^3$
3. $N=71280\times \frac{7}{792}=90\times 7=630$

Trick: Cancel before multiplying: $71280÷792=90$ (since $792\times 90=71280$), then $\times 7$.

Answer: (C) 630

Concept: Compare each ratio by its decimal value, or by how far it is below 1.

1. $\frac{6}{7}=0.857$, $\frac{31}{35}=0.886$, $\frac{21}{23}=0.913$, $\frac{11}{12}=0.917$, $\frac{43}{45}=0.956$
2. Ascending: $6:7<31:35<21:23<11:12<43:45$

Trick: Each ratio is $1-\frac{k}{d}$: deficits $\frac{1}{7},\frac{4}{35},\frac{2}{23},\frac{1}{12},\frac{2}{45}$. Larger deficit ⇒ smaller ratio. Watch $21:23$ vs $11:12$: $\frac{2}{23}=0.087>\frac{1}{12}=0.083$.

Answer: (A) 6:7, 31:35, 21:23, 11:12, 43:45

Concept: Hemisphere: volume $=\frac{2}{3}\pi r^3$, total surface area $=3\pi r^2$ (curved $2\pi r^2$ + flat top $\pi r^2$).

1. $\frac{2}{3}\times 3.14\times r^3=7.065\Rightarrow r^3=\frac{7.065\times 3}{6.28}=3.375$
2. $r=1.5\text{m}$
3. TSA $=3\times 3.14\times (1.5{)}^2=3\times 3.14\times 2.25=21.195{\text{m}}^2$

Trick: $3.375={1.5}^3$ is worth memorising (like $0.125={0.5}^3$). Check with options: TSA of a hemisphere $=3\pi r^2$, not $2\pi r^2$.

Answer: (A) 21.195 m²

Concept: Change in total $=$ change in average $\times$ number of students.

1. Excess entered $=68-59=9$
2. $n\times 1.8=9\Rightarrow n=5$

Trick: Wrong-entry questions: $n=\frac{\text{difference of entries}}{\text{change in average}}$.

Answer: (D) 5