Q&A: practice questions with answers

Concept: Salary $=\text{Expenditure}(E)+\text{Savings}(S)$.

1. Given $E=2(S-E)$ ⇒ $3E=2S$ ⇒ $E:S=2:3$.
2. Salary $=2+3=5$ parts.
3. $E\%=\frac{2}{5}\times 100=40\%$.

Trick: Convert the condition to a ratio, then $\frac{\text{part}}{\text{total}}\times 100$.

Answer: (A) $40\%$

Concept: Successive discounts $a\%,b\%$ = single $(a+b-\frac{ab}{100})\%$.

1. Option 1: $30\%$ of 8000 $=2400$.
2. Option 2: $20+15-\frac{300}{100}=32\%$ ⇒ $32\%$ of 8000 $=2560$.
3. Difference $=2560-2400=160$, Option 2 is better.

Trick: Difference $=(32-30)\%$ of $8000=2\%\times 8000=160$.

Answer: (A) Option 2 and by ₹160

Concept: Use LCM as tank capacity; emptying pipe has negative efficiency.

1. Tank $=\text{LCM}(10,15,30)=30$ units; A $=+3$, B $=+2$, C $=-1$ per hour.
2. First 6 h (all three): $4\times 6=24$ units.
3. Remaining $6$ units by B + C at $1$ unit/h ⇒ $6$ h.
4. Total time $=6+6=12$ h.

Trick: Efficiency method avoids fractions: units ÷ net rate = time.

Answer: (C) 12 hours

Concept: In a trapezium, the segment joining the midpoints of the diagonals is parallel to the bases and equals half their difference.

1. Here AB ∥ DC, so $EF=\frac{DC-AB}{2}$.
2. $EF=\frac{72-14}{2}=\frac{58}{2}=29$ cm.
3. BC and AD are extra data (not needed).

Yaad rakhein: Midpoints of sides → half the sum; midpoints of diagonals → half the difference.

Answer: (C) 29

Concept: Median of an odd number of terms = middle term after arranging in order.

1. Since $x>0$: $\frac{x}{5}<\frac{x}{4}<\frac{x}{3}<\frac{2x}{5}<\frac{2x}{3}<\frac{3x}{4}<x$.
2. 7 terms ⇒ median = 4th term $=\frac{2x}{5}$.
3. $\frac{2x}{5}=8.4$ ⇒ $x=21$.

Trick: Compare as decimals: $0.2,0.25,0.33,0.4,0.67,0.75,1$.

Answer: (D) 21

Concept: Write both situations in terms of the cost price $x$.

1. Original SP $=1.32x$.
2. New CP $=x+20$, new SP $=1.32x-42$, profit $27.5\%$.
3. $1.32x-42=1.275(x+20)=1.275x+25.5$ ⇒ $0.045x=67.5$.
4. $x=1500$.

Trick: Option check: CP 1500 ⇒ SP 1980; $1938=1.275\times 1520$ ✓.

Answer: (A) ₹1,500

Concept: Net part filled per hour = sum of filling rates − emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{20}-\frac{1}{20}=\frac{1}{4}$ per hour.
2. The 20-h filler and 20-h 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-h tap matters.

Answer: (D) 1

Concept: Solve innermost bracket first (BODMAS).

1. $26-11+3=18$.
2. $-9\times 18=-162$.
3. $30+(-162)=-132$.

Trick: Sign rule: $30-162$ ⇒ answer negative, magnitude $132$.

Answer: (A) $-132$

Concept: $24=8\times 3$ (coprime) ⇒ number must be divisible by both 8 and 3.

1. By 8: last three digits $16B$ ⇒ $160$ or $168$, so $B=0$ or $8$.
2. By 3: digit sum $=19+A+B$ must be a multiple of 3.
3. For max, take $B=8$: $27+A$ ⇒ $A=9$ is allowed.
4. Max $A+B=9+8=17$.

Trick: For a maximum, try the largest B first, then the largest valid A.

Answer: (B) 17

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a+b{)}^2={88}^2=7744$.
2. $4ab=4\times 1927=7708$.
3. $(a-b{)}^2=36$ ⇒ $a-b=6$.

Trick: Numbers are $47$ and $41$: $47\times 41=1927$ ✓.

Answer: (C) 6

Concept: For a fixed sum, the product is maximum when the two numbers are as close as possible.

1. $41$ is odd, so the closest pair is $20$ and $21$.
2. $\text{HCF}(20,21)=1$ ⇒ they are co-prime.
3. Product $=20\times 21=420$.

Yaad rakhein: Consecutive numbers are always co-prime.

Answer: (C) 420

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=10630-9248.10=1381.90$.
2. $\frac{1381.90}{10630}\times 100=13\%$.

Trick: $10\%$ of 10630 $=1063$; $3\%=318.9$; total $1381.9$ ✓.

Answer: (A) $13\%$

Concept: $V=\pi r^{2}h$; halving the radius makes volume $\frac{1}{4}$.

1. Original volume $V=\pi R^{2}H$.
2. Each small cylinder $=\pi {\left(\frac{R}{2}\right)}^{2}H=\frac{V}{4}$.
3. Three of them use $\frac{3V}{4}$.
4. Unused $=V-\frac{3V}{4}=\frac{1}{4}$ of $V$.

Trick: Volume ∝ $r^2$ (same height): $\frac{1}{2}$ radius ⇒ $\frac{1}{4}$ volume.

Answer: (B) $\frac{1}{4}$

Concept: Total value = sum of (denomination × number of notes).

1. Let ₹100 notes $=x$, ₹500 notes $=x+4$.
2. $500(x+4)+100x=24800$ ⇒ $600x=22800$.
3. $x=38$.

Trick: Check: $42\times 500+38\times 100=21000+3800=24800$ ✓.

Answer: (B) 38

Concept: Proportion of red $=\frac{\text{red}}{\text{red}+\text{blue}}$.

1. Bag A: $\frac{2}{5}=40\%$.
2. Bag B: $\frac{4}{5}=80\%$.
3. Bag C: $\frac{3}{5}=60\%$ ⇒ Bag B is highest.

Trick: All totals are 5 parts, so just compare the red parts: 2, 4, 3.

Answer: (C) Bag B

Concept: A change from loss to gain = change in SP as % of CP.

1. From $-10\%$ to $+5\%$ is a change of $15\%$ of CP.
2. $15\%$ of CP $=120$ ⇒ CP $=800$.
3. SP $=90\%$ of 800 $=720$.

Trick: Rupee difference ÷ % difference gives 1% of CP: $\frac{120}{15}=8$.

Answer: (B) ₹720

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

1. $SI=\frac{55000\times 5.5\times 5.5}{100}$.
2. $=550\times 30.25=16637.5$.

Trick: $5.5\times 5.5=30.25$ – multiply this by $\frac{P}{100}=550$.

Answer: (A) 16,637.5

Concept: Convert the fraction to a decimal: $\frac{3}{8}=0.375$.

1. Expression $=0.375+0.375-0.125$.
2. $=0.75-0.125=0.625$.

Yaad rakhein: Eighths: $\frac{1}{8}=0.125$, $\frac{3}{8}=0.375$, $\frac{5}{8}=0.625$.

Answer: (B) 0.625

Concept: $A=P{(1+\frac{R}{100})}^T$.

1. $9075=P\times (1.1{)}^2=1.21P$.
2. $P=\frac{9075}{1.21}=7500$.

Trick: $10\%$ for 2 years ⇒ $P:A=100:121$; $\frac{9075}{121}=75$, so $P=7500$.

Answer: (C) 7,500

Concept: Angle of depression = angle of elevation from the lower point (alternate angles).

1. Bottom (60°): $d=\frac{20}{\tan {60}^{\circ }}=\frac{20}{\sqrt{3}}$.
2. Top (30°): drop $=d\tan {30}^{\circ }=\frac{20}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{20}{3}$.
3. Tower $h=20-\frac{20}{3}=\frac{40}{3}\approx 13.33$ m.

Trick: For 30° & 60° depressions, tower $=\frac{2}{3}$ of building height.

Answer: (A) 13.33 m