Q&A: practice questions with answers

Concept: Net filling rate = sum of filling rates − emptying rate.

1. Rates: $\frac{1}{4}+\frac{1}{19}-\frac{1}{19}=\frac{1}{4}$ per hour
2. The 19-hour filler and 19-hour emptier cancel out
3. Time for $\frac{1}{4}$ tank $=\frac{1/4}{1/4}=1$ hour

Trick: Equal fill and empty times cancel — only the 4-hour tap matters.

Answer: (B) $1$ hour

Concept: In a trapezium, the segment joining mid-points of the diagonals is parallel to the bases and $EF=\frac{AB-DC}{2}$.

1. Parallel sides: $AB=21$, $DC=11$
2. $EF=\frac{21-11}{2}=5\text{cm}$
3. BC and AD are extra data (not needed)

Yaad rakhein: Mid-points of diagonals → half the difference of parallel sides; mid-points of legs → half the sum.

Answer: (D) $5$

Concept: Number of small solids $=\frac{\text{volume of big}}{\text{volume of one small}}$.

1. $n=\frac{\pi \times {35}^2\times 50}{\pi \times {3.5}^2\times 15}$
2. $\frac{{35}^2}{{3.5}^2}=100$, so $n=\frac{100\times 50}{15}=333.33$
3. Complete cylinders $=333$

Trick: Radius ratio $10\Rightarrow$ area ratio $100$; then multiply by height ratio $\frac{50}{15}$.

Answer: (C) $333$

Concept: Profit is shared in the ratio of investments.

1. Total parts $=5+7=12$
2. Raghav's share $=\frac{7}{12}\times 48000=28000$

Trick: One part $=\frac{48000}{12}=4000$; Raghav $=7\times 4000$.

Answer: (B) ₹$28,000$

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

1. $1920=\frac{P\times 8\times 3}{100}$
2. $P=\frac{1920\times 100}{24}=8000$

Trick: 3 years at 8% $=24\%$ of P $=1920\Rightarrow P=8000$.

Answer: (A) ₹$8,000$

Concept: He gets 1200 g for the price of 1000 g and gives 840 g for the price of 1000 g.

1. Let price of 1 kg $=$ ₹$1000$. He buys 1200 g for ₹$1000$ and sells 840 g for ₹$1000$.
2. Goods bought for ₹$1000$ (1200 g) are sold as $\frac{1200}{840}$ lots of ₹$1000$
3. $\text{Profit}\%=\frac{1200-840}{840}\times 100=\frac{360}{840}\times 100=42.86\%$

Trick: $\text{Profit}\%=\frac{\text{goods taken}-\text{goods given}}{\text{goods given}}\times 100$.

Answer: (A) $42.86\%$

Concept: Make the common term (female students) equal in both ratios.

1. $M:F=4:3=32:24$ (×8)
2. $F:T=8:1=24:3$ (×3)
3. $M:F:T=32:24:3$, so $(M+F):T=56:3$

Trick: Common term: LCM of 3 and 8 is 24.

Answer: (D) $56:3$

Concept: Combined average $=\frac{\sum n_i{\overline{x}}_i}{\sum n_i}$.

1. Totals: $60\times 65=3900$, $40\times 72=2880$, $50\times 60=3000$
2. Sum $=9780$
3. Average $=\frac{9780}{150}=65.2$

Trick: Deviation from 65: $0+40\times 7-50\times 5=30$; $65+\frac{30}{150}=65.2$.

Answer: (B) $65.2$

Concept: Mode is the value that occurs most often.

1. 53 occurs 3 times
2. 43, 47, 54, 55 occur 2 times each; all others once
3. Highest frequency → 53

Trick: Sort or tally quickly; only one value repeats 3 times.

Answer: (C) $53$

Concept: Let larger $=x$, smaller $=y$ and form two equations.

1. $x-y=8$ and $2x=3y+4$
2. $2(y+8)=3y+4\Rightarrow y=12$
3. $x=12+8=20$

Trick: Option check: $x=20,y=12$: $40=36+4$ ✓.

Answer: (C) $20$

Concept: $\text{Increase}\%=\frac{\text{increase}}{\text{original}}\times 100$.

1. Increase $=220-200=20$
2. $\frac{20}{200}\times 100=10\%$

Yaad rakhein: Always divide by the original (old) value.

Answer: (B) $10\%$

Concept: Divide both sides by $3^{-24}$: $a^m÷a^n=a^{m-n}$.

1. $k=3^{-22+24}+3^{-19+24}+3^{-17+24}+3^{-16+24}$
2. $k=3^2+3^5+3^7+3^8$
3. $k=9+243+2187+6561=9000$

Trick: Multiply each power by $3^{24}$ — just add 24 to every exponent.

Answer: (A) $9000$

Concept: Take tank $=60$ units (LCM of 20, 30, 60).

1. Rates: A $=3$, B $=2$, C $=-1$ unit/min; together $=4$
2. In 10 min: $40$ units filled
3. Left $20$ units at $3+2=5$ per min → $4$ min
4. Total $=10+4=14$ min

Trick: LCM-units method avoids fractions.

Answer: (A) $14$ minutes

Concept: Crossing a pole: length $=vt$; opposite trains: $L_A+L_B=(v_A+v_B)t$.

1. $v_B=54\times \frac{5}{18}=15\text{m/s}$; $L_A=24v$
2. $24v+150=20(v+15)\Rightarrow 4v=150\Rightarrow v=37.5\text{m/s}$
3. $v=37.5\times \frac{18}{5}=135\text{km/h}$, $L_A=24\times 37.5=900\text{m}$

Trick: Only option (D) has 135 km/h with $L=24\times 37.5=900$.

Answer: (D) 900 m and 135 km/h

Concept: Effective price per item $=\frac{\text{total paid}}{\text{items received}}$.

1. Pay for 4 full: $4\times 47=188$
2. 5th pen at 50% off: $23.50$
3. Total $=211.50$ for 5 pens → $\frac{211.5}{5}=42.30$

Trick: Discount $=\frac{0.5}{5}=10\%$; $47\times 0.9=42.30$.

Answer: (A) ₹$42.30$

Concept: Volume of material $=\pi h(R^2-r^2)$.

1. $R^2-r^2=64-49=15$
2. $V=3.14\times 20\times 15=942{\text{cm}}^3$

Trick: $R^2-r^2=(R+r)(R-r)=15\times 1$.

Answer: (D) $942$ cubic cm

Concept: Fourth proportional $x$ of $a,b,c$: $a:b=c:x\Rightarrow x=\frac{bc}{a}$.

1. $x=\frac{25\times 888}{37}$
2. $\frac{888}{37}=24$, so $x=25\times 24=600$

Yaad rakhein: $888=37\times 24$ — spot the factor first.

Answer: (D) $600$

Concept: Same distance: difference of times = delay.

1. $\frac{d}{30}-\frac{d}{40}=\frac{20}{60}=\frac{1}{3}$
2. $\frac{d}{120}=\frac{1}{3}\Rightarrow d=40\text{km}$

Trick: $d=\frac{S_1{S}_2}{S_1-S_2}\times t=\frac{40\times 30}{10}\times \frac{1}{3}=40$.

Answer: (A) $40\text{km}$

Concept: ${\cos }^2\theta +{\sin }^2\theta =1$ and $1+{\tan }^2\theta ={\sec }^2\theta$.

1. (D): $1+{\tan }^2\theta +1-{\sec }^2\theta ={\sec }^2\theta +1-{\sec }^2\theta =1$ ✓
2. (A) gives $2{\sec }^2\theta -1$, not 1
3. (B), (C) are always greater than 1

Yaad rakhein: ${\sec }^2\theta -{\tan }^2\theta =1$, ${\csc }^2\theta -{\cot }^2\theta =1$.

Answer: (D) $1+{\tan }^2\theta +(1-{\sec }^2\theta )$

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

1. $R\times T=12\times 4=48\%$
2. $P=\frac{1320\times 100}{48}=2750$

Trick: $48\%$ of P $=1320$ → $1\%=27.5$ → $P=2750$.

Answer: (D) ₹$2,750$