Q&A: practice questions with answers

Concept: Average speed = total distance ÷ total time (not the average of speeds).

1. Total distance $=50+75=125\text{km}$
2. Total time $=40+40=80$ min $=\frac{4}{3}$ h
3. Average speed $=125\times \frac{3}{4}=93.75=93\frac{3}{4}\text{km/h}$

Trick: Equal times → average speed = mean of speeds: $\frac{75+112.5}{2}=93.75$.

Answer: (C) $93\frac{3}{4}$

Concept: Yearly average = sum of yearly totals ÷ number of years; then match it with a year's total.

1. Sum $=115+108+149+102+101=575$ thousand
2. Average $=\frac{575}{5}=115$ thousand
3. Total in 1985 $=115$ → equal to the average.

Trick: Use deviations from 115: $0-7+34-13-14=0$, so 115 is exactly the average.

Answer: (C) $1985$

Concept: When a solid is drawn into a wire, volume remains the same: $\frac{4}{3}\pi R^3=\pi r^{2}L$. Keep units the same (cm).

1. Sphere $R=6\text{cm}$; wire $r=2\text{mm}=0.2\text{cm}$
2. $\frac{4}{3}\pi \times 216=288\pi$
3. $\pi \times 0.04\times L=288\pi \Rightarrow L=\frac{288}{0.04}=7200\text{cm}$

Trick: $L=\frac{4R^3}{3r^2}=\frac{4\times 216}{3\times 0.04}=7200$ ($\pi$ cancels, so its value is not needed).

Answer: (D) $7200$

Concept: Take total work as LCM of the days and find work done in each 2-day cycle.

1. Total work $=\text{LCM}(13,26)=26$ units; A $=2$/day, B $=1$/day.
2. One cycle (A + B, 2 days) $=3$ units; 8 cycles $=16$ days $=24$ units.
3. Remaining $2$ units: A on day 17 does $2$ units → done.
4. Total $=17$ days.

Trick: Stop the cycles just before the end and let the starter finish – check the leftover each time.

Answer: (B) $17$

Concept: Area of a sector $=\frac{1}{2}\times l\times r$, where $l$ is arc length.

1. $2\pi r=88\Rightarrow r=\frac{88\times 7}{44}=14\text{cm}$
2. Area $=\frac{1}{2}\times 22\times 14=154{\text{cm}}^2$

Trick: Arc is $\frac{22}{88}=\frac{1}{4}$ of the circle, so area $=\frac{1}{4}\times \pi \times {14}^2=\frac{616}{4}=154$.

Answer: (B) $154{\text{cm}}^2$

Concept: To cross a platform, the train covers its own length + the platform length.

1. Distance $=255+745=1000\text{m}$
2. Speed $=72\times \frac{5}{18}=20\text{m/s}$
3. Time $=\frac{1000}{20}=50$ seconds

Trick: km/h → m/s: multiply by $\frac{5}{18}$.

Answer: (A) 50 seconds

Concept: For the same distance, time is inversely proportional to speed; 4 times the time means one-fourth the speed.

1. Downstream $=7.5+s$, upstream $=7.5-s$
2. $7.5+s=4(7.5-s)\Rightarrow 7.5+s=30-4s$
3. $5s=22.5\Rightarrow s=4.5\text{km/h}$

Trick: Speed ratio $4:1$ → boat : stream $=\frac{4+1}{2}:\frac{4-1}{2}=5:3$, so stream $=7.5\times \frac{3}{5}=4.5$.

Answer: (B) $4.5\text{km/h}$

Concept: Simplify from the bottom of the fraction upwards.

1. $\frac{7}{1-7}=-\frac{7}{6}$; $1+(-\frac{7}{6})=-\frac{1}{6}$
2. $\frac{1}{-\frac{1}{6}}=-6$; $1-(-6)=7$
3. $\frac{7}{7}=1$; $1-1=0$

Trick: Watch the signs at each level – the negative turns $1-(\dots )$ into $1+6$.

Answer: (D) $0$

Concept: When only tin is added, the copper quantity stays constant; use it to find the new total.

1. Tin $=25\%$ of $750=187.5\text{kg}$; copper $=562.5\text{kg}$
2. Finally tin $70\%$ → copper $30\%$: new total $=\frac{562.5}{0.3}=1875\text{kg}$
3. Tin added $=1875-750=1125\text{kg}$

Trick: Fix the unchanged component: copper $75\%\to 30\%$, so total becomes $750\times \frac{75}{30}=1875$.

Answer: (B) 1125 kg

Concept: $\cos \alpha ·\csc \alpha =\frac{\cos \alpha }{\sin \alpha }=\cot \alpha$, so the expression is ${\cot }^2\alpha$.

1. $\sin \alpha =\frac{5}{13}$ → triplet $5,12,13$ → $\cos \alpha =\frac{12}{13}$
2. $\cot \alpha =\frac{12}{5}$
3. ${\cot }^2\alpha =\frac{144}{25}$

Trick: Spot Pythagorean triplet $(5,12,13)$ instantly.

Answer: (C) $\frac{144}{25}$

Concept: Do × and ÷ left to right, then addition.

1. $\frac{1}{4}\times \frac{4}{3}=\frac{1}{3}$
2. $\frac{1}{3}÷\frac{5}{3}=\frac{1}{3}\times \frac{3}{5}=\frac{1}{5}$
3. $\frac{1}{5}+\frac{1}{2}=\frac{7}{10}=0.7$

Trick: Division by a fraction = multiply by its reciprocal.

Answer: (A) $0.7$

Concept: Females below poverty line = population × BPL% × female share of the BPL ratio.

1. State A: BPL $=36\%$ of $18000=6480$; females $=\frac{3}{8}\times 6480=2430$
2. State D: BPL $=32\%$ of $18000=5760$; females $=\frac{2}{3}\times 5760=3840$
3. Total $=2430+3840=6270$

Trick: Use only the 'Below poverty line' ratio column; the 'Above' column is a trap.

Answer: (B) $6270$

Concept: Aarav's gain = money received on selling − money paid to buy it back, measured on the original worth.

1. Aarav sells at $10000\times 1.15=₹11500$
2. Bhavin sells back at $11500\times 0.85=₹9775$
3. Gain $=11500-9775=₹1725$; $\frac{1725}{10000}\times 100=17.25\%$ profit

Trick: He gets the watch back and keeps the difference: $11500\times 15\%=1725$, i.e. $1.15\times 15=17.25\%$.

Answer: (D) 17.25% profit

Concept: Successive changes multiply: increase → ×1.15, decrease → ×0.85.

1. 2012 increase, 2013 decrease, 2014 increase (three changes by the start of 2015).
2. Factor $=1.15\times 0.85\times 1.15=0.9775\times 1.15=1.124125$
3. Change $=+12.41\%$ (increase)

Trick: $+15$ and $-15$ give $-2.25\%$; then $-2.25+15-\frac{2.25\times 15}{100}=12.41\%$.

Answer: (B) Increased by 12.41 %

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

1. $180÷30\times 6=6\times 6=36$; $60÷3=20$
2. $30+36-15-16-20-15$
3. $=66-66=0$

Trick: Add all positives ($66$) and all negatives ($66$) separately.

Answer: (C) $0$

Concept: Convert mixed fractions to improper fractions and solve the innermost bracket first.

1. $3\frac{3}{5}÷4\frac{1}{2}=\frac{18}{5}\times \frac{2}{9}=\frac{4}{5}$; $5\frac{1}{4}\times \frac{4}{5}=\frac{21}{5}$
2. $1\frac{3}{4}÷\frac{21}{5}=\frac{7}{4}\times \frac{5}{21}=\frac{5}{12}$; $3\frac{2}{3}÷\frac{5}{12}=\frac{11}{3}\times \frac{12}{5}=\frac{44}{5}$
3. Numerator $=12+\frac{44}{5}=\frac{104}{5}$; denominator $=\frac{4}{3}\times \frac{26}{5}=\frac{104}{15}$
4. Value $=\frac{104}{5}\times \frac{15}{104}=3$

Trick: Same $104$ in numerator and denominator – a sign the answer is a neat integer.

Answer: (B) $3$

Concept: Count the subjects (columns) in which at least one student scored 100.

1. Math: A and C scored 100 ✓
2. Chemistry: A scored 100 ✓
3. Information Technology: A and C scored 100 ✓
4. English and Physics: no 100 → $3$ subjects.

Trick: The question counts subjects, not the number of 100s (which is 5).

Answer: (C) $3$

Concept: Total = average × number; divide the total in the given ratio.

1. Total age $=25\times 3=75$ years
2. Ratio sum $=3+5+7=15$
3. Youngest $=\frac{3}{15}\times 75=15$ years

Trick: One part $=\frac{75}{15}=5$; youngest $=3\times 5$.

Answer: (C) 15 years

Concept: Amounts are proportional to the central angles.

1. Savings ${60}^{\circ }=₹8000$
2. Education ${30}^{\circ }=\frac{30}{60}\times 8000=₹4000$

Trick: Education angle is half of savings angle → half the amount.

Answer: (B) $4000$

Concept: $a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c{)}^2$.

1. Here $a=25,b=9,c=6$
2. Value $=(25+9+6{)}^2={40}^2=1600$

Yaad rakhein: Three squares + all doubled cross-products = square of the sum.

Answer: (C) $1600$