Q&A: practice questions with answers

Concept: In a partnership, profit is shared in the ratio of (capital $\times$ time).

1. Total profit = ₹1,494 and M gets ₹498 more, so M $=\frac{1494+498}{2}=996$, N $=498$; ratio M : N $=2:1$.
2. Let N's capital $=x$; M's $=x+17000$. Then $\frac{6(x+17000)}{5x}=\frac{2}{1}$.
3. $6x+102000=10x\Rightarrow x=25500$.
4. M's capital $=25500+17000=42500$.

Trick: Check options: M = 42,500 → N = 25,500; $\frac{42500\times 6}{25500\times 5}=\frac{255000}{127500}=2$ ✓.

Answer: (A) ₹42,500

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with R and T in the same time unit.

1. Here $P=1200$, $R=4\%$ per month, $T=10$ months.
2. $\text{SI}=\frac{1200\times 4\times 10}{100}=480$.
3. (If the rate were 4% per annum, $T=\frac{10}{12}$ year would give $\frac{1200\times 4\times 10}{1200}=40$ – but the question says per month.)

Trick: 4% per month for 10 months = 40% of 1200 = 480.

Answer: (C) ₹480

Concept: Expand each bracket, then combine like terms.

1. $x(7x-5)=7x^2-5x$
2. $3(x^2-2)=3x^2-6$
3. Sum: $7x^2-5x+3x^2-6+14=10x^2-5x+8$

Trick: Put $x=1$: $1(2)+3(-1)+14=13$; only $10-5+8=13$ ✓.

Answer: (C) $10x^2-5x+8$

Concept: Find each pipe's full-tank time, take LCM as tank capacity, and add efficiencies (emptying pipe negative).

1. P fills $\frac{4}{5}$ in 12 h → full in $12\times \frac{5}{4}=15$ h; Q fills $\frac{2}{7}$ in 30 h → full in $30\times \frac{7}{2}=105$ h.
2. Tank $=105$ units: P $=7$, Q $=1$ unit/h. In 5 h: $5\times 8=40$ units.
3. R empties 40 units in 8 h → R $=-5$ unit/h.
4. P + Q + R $=7+1-5=3$ unit/h → time $=\frac{105}{3}=35$ h.

Yaad rakhein: Filling pipe +, emptying pipe −; time $=\frac{\text{capacity}}{\text{net efficiency}}$.

Answer: (B) 35 hours

Concept: First $n$ even numbers: $2,4,\dots ,2n$; their sum $=n(n+1)$, so average $=n+1$.

1. Here $n=163$; last term $=2\times 163=326$.
2. Average $=\frac{2+326}{2}=164$.
3. Or directly: $n+1=163+1=164$.

Trick: Average of first $n$ even numbers $=n+1$; of first $n$ odd numbers $=n$.

Answer: (D) 164

Concept: Simplify the inner (continued) fraction first, then add/subtract using the LCM.

1. $1+\frac{4}{9}=\frac{13}{9}$, so $\frac{1}{1+\frac{4}{9}}=\frac{9}{13}$.
2. Expression $=\frac{5}{3}+\frac{9}{13}-\frac{2}{5}$; LCM of 3, 13, 5 $=195$.
3. $=\frac{325+135-78}{195}=\frac{382}{195}$.

Trick: Denominator must be the LCM 195 – only option B has it.

Answer: (B) $\frac{382}{195}$

Concept: Consecutive natural numbers are $n$ and $n+1$.

1. $n(n+1)=110\Rightarrow n^2+n-110=0$.
2. $(n+11)(n-10)=0\Rightarrow n=10$ (natural).
3. Numbers: 10 and 11; greater $=11$.

Trick: $\sqrt{110}\approx 10.5$ → the numbers lie on either side: $10\times 11=110$.

Answer: (C) 11

Concept: For a decimal, take the square root of the digits and halve the number of decimal places.

1. $\sqrt{121}=11$
2. $\sqrt{0.0225}=0.15$ (since ${15}^2=225$)
3. $\sqrt{5.76}=2.4$ (since ${24}^2=576$)
4. $11+0.15-2.4=8.75$

Trick: Decimal places in the square = 2 × decimal places in the root: 0.0225 (4) → 0.15 (2).

Answer: (C) 8.75

Concept: Use $1+{\tan }^2\beta ={\sec }^2\beta$ and $\tan \beta \text{cosec}\beta =\frac{\sin \beta }{\cos \beta }·\frac{1}{\sin \beta }=\sec \beta$.

1. Numerator: ${\tan }^2\beta {\text{cosec}}^2\beta ={\sec }^2\beta$.
2. Denominator: $1+{\tan }^2\beta ={\sec }^2\beta$.
3. $\frac{{\sec }^2\beta }{{\sec }^2\beta }=1$.

Trick: Put $\beta ={45}^{\circ }$: $\frac{1\times 2}{1+1}=1$.

Answer: (C) 1

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $72\times 432=144\times x$
2. $x=\frac{72\times 432}{144}=\frac{432}{2}=216$.
3. Check: HCF(144, 216) $=72$, LCM $=432$ ✓.

Trick: $\frac{144}{72}=2$, so $x=\frac{432}{2}=216$.

Answer: (A) 216

Concept: An exterior angle of a triangle equals the sum of the two opposite interior angles.

1. In right $△BDC$: $\angle BDC={90}^{\circ }$, so $\angle C={180}^{\circ }-{90}^{\circ }-{58}^{\circ }={32}^{\circ }$.
2. In $△AEC$, $\angle AEB$ is the exterior angle at E (since B, E, C are collinear).
3. $\angle AEB=\angle EAC+\angle ECA={16}^{\circ }+{32}^{\circ }={48}^{\circ }$.

Trick: In a right triangle the two acute angles add to ${90}^{\circ }$, so $\angle C={90}^{\circ }-{58}^{\circ }$ directly.

Answer: (D) ${48}^{\circ }$

Concept: Volume of a cylinder $V=\pi r^{2}h$.

1. $r=3.5\text{m}$, so $r^2=12.25$.
2. $V=\pi \times 12.25\times 5.2=63.7\pi {\text{m}}^3$.

Trick: ${3.5}^2\times 5.2=12.25\times 5.2$: $12\times 5.2=62.4$ and $0.25\times 5.2=1.3$, total $63.7$.

Answer: (C) $63.7\pi$ cubic metres

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $96\times 54=5184$.
2. $\sqrt{5184}=72$.

Trick: $96=6\times 16$, $54=6\times 9$, so $\sqrt{96\times 54}=6\times 4\times 3=72$.

Answer: (D) 72

Concept: Evaluate powers and roots first, then add/subtract left to right.

1. $7^2=49$, $\sqrt{8^2}=8$, $5\sqrt{16}=5\times 4=20$.
2. $49+8-20-14=23$.

Trick: $\sqrt{a^2}=a$ for positive $a$ – no need to square 8.

Answer: (C) 23

Concept: Work per hour adds up: combined rate $=\frac{1}{11}+\frac{1}{24}+\frac{1}{19}$.

1. LCM $=11\times 24\times 19=5016$.
2. Rate $=\frac{456+209+264}{5016}=\frac{929}{5016}$ tank per hour.
3. Time $=\frac{5016}{929}=5\frac{371}{929}$ hours (since $929\times 5=4645$, remainder $371$).

Trick: Estimate: rate $\approx 0.091+0.042+0.053=0.186$, so time $\approx 5.4$ h – only option (C) starts with 5.

Answer: (C) $5\frac{371}{929}$ hours

Concept: In 'one free with one', the customer pays the marked price of 1 article but gets 2.

1. Let CP of one set $=100$; he gives 2 sets, so CP $=200$.
2. Profit $20\%$ ⇒ SP $=200\times 1.2=240$.
3. This SP is the marked price of one set: $240$, i.e. $140\%$ above the CP of $100$.

Trick: Mark-up $=\frac{2\times 120}{100}-1=140\%$.

Answer: (C) $140\%$

Concept: Divisibility by 13 (also 7, 11): split into 3-digit groups from the right and take the alternating sum; if it is divisible by 13, so is the number.

1. Groups: 716 | 710 | 501.
2. Alternating sum $=501-710+716=507=13\times 39$ ⇒ divisible by 13.
3. Ends in 1 ⇒ not divisible by 20 or 16; $716710501÷19$ leaves remainder 6.

Trick: An odd number can never be divisible by an even number like 16 or 20 – eliminate them at once.

Answer: (D) 13

Concept: A prime has no divisor other than 1 and itself; test primes up to $\sqrt{381}\approx 19.5$.

1. Even numbers 374, 376, 378, 380 and 375 (ends in 5) are out.
2. $377=13\times 29$, $381=3\times 127$ – not prime.
3. 379 is not divisible by 2, 3, 5, 7, 11, 13, 17, 19 ⇒ prime. Count $=1$.

Trick: Digit sum $3+8+1=12$ shows 381 is divisible by 3 instantly.

Answer: (D) 1

Concept: Find each quantity and subtract.

1. $65\%$ of $75=\frac{65\times 75}{100}=48.75$.
2. $\frac{3}{5}$ of $45=27$.
3. Difference $=48.75-27=21.75$.

Trick: $65\%$ of 75 $=75\%$ of 65 $=\frac{3}{4}\times 65=48.75$.

Answer: (D) 21.75

Concept: Express CP in terms of the actual SP $S$.

1. $\frac{5}{14}S$ gives $25\%$ loss ⇒ $\frac{5}{14}S=0.75\text{CP}$ ⇒ $\text{CP}=\frac{10}{21}S$.
2. New SP $=0.58S$.
3. $\text{Profit}\%=(\frac{0.58S}{\frac{10}{21}S}-1)\times 100=(1.218-1)\times 100=21.8\%$.

Trick: Take $S=210$: CP $=100$, new SP $=0.58\times 210=121.8$ ⇒ $21.8\%$.

Answer: (D) $21.8\%$