Q&A: practice questions with answers

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

1. ${(1+\frac{R}{100})}^3=\frac{1092727}{1000000}=1.092727$
2. Test $R=3$: ${1.03}^3=1.0609\times 1.03=1.092727$ ✔
3. So $R=3\%$ per annum.

Trick: Last digits 727 → cube ending in 7 → ${103}^3=1092727$, so $R=3\%$.

Answer: (D) $3\%$

Concept: $\text{Sum}=\text{Average}\times \text{Number of terms}$

1. Sum of first two $=2\times 79=158$; third $=135$.
2. Sum of last two $=655-158-135=362$.
3. Their average $=\frac{362}{2}=181$

Trick: Subtract known parts from the total, then divide.

Answer: (B) $181$

Concept: Margin of victory $=(52\%-48\%)$ of valid votes $=4\%$ of valid votes.

1. Let registered voters $=x$. Votes polled $=0.8x$.
2. Valid votes $=98\%$ of $0.8x=0.784x$.
3. $4\%$ of $0.784x=0.03136x=1568$
4. $x=\frac{1568}{0.03136}=50000$

Trick: Check options: $50000\times 0.8\times 0.98\times 0.04=1568$ ✔

Answer: (C) $50000$

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum at even places) is $0$ or a multiple of $11$.

1. Digits of N37M01 from left: odd places N, 7, 0; even places 3, M, 1.
2. Difference $=(N+7+0)-(3+M+1)=N-M+3$.
3. Take $N-M+3=0\Rightarrow M-N=3$ (e.g., N = 1, M = 4 gives 137401 $=11\times 12491$).

Trick: Alternate digits; differences 0, 11, 22… only.

Answer: (C) $M-N=3$

Concept: ${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$

1. $\sec \theta +\tan \theta =\frac{3}{2}\Rightarrow \sec \theta -\tan \theta =\frac{2}{3}$
2. Adding: $2\sec \theta =\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\Rightarrow \sec \theta =\frac{13}{12}$
3. $\cos \theta =\frac{1}{\sec \theta }=\frac{12}{13}$

Trick: If $\sec \theta +\tan \theta =\frac{a}{b}$, then $\cos \theta =\frac{2ab}{a^2+b^2}=\frac{12}{13}$.

Answer: (A) $\frac{12}{13}$

Concept: $\text{LCM}\times \text{HCF}=\text{Product of the two numbers}$

1. So $ab=150$ and $a-b=5$.
2. $(a+b{)}^2=(a-b{)}^2+4ab=25+600=625\Rightarrow a+b=25$
3. $a=15,b=10$ (check: $15\times 10=150$, $15-10=5$ ✔)

Trick: Option check: only 15, 10 has product 150 and difference 5.

Answer: (A) $15,10$

Concept: Increasing a number by $70\%$ multiplies it by $1.7$.

1. $1.7x=3570$
2. $x=\frac{3570}{1.7}=2100$

Trick: $170\%\to 3570\Rightarrow 1\%\to 21\Rightarrow 100\%\to 2100$

Answer: (A) $2100$

Concept: Form two linear equations with son's age $S$ and father's age $F$.

1. $2S+F=42$ …(i) and $S+2.2F=89$ …(ii)
2. From (i): $F=42-2S$. Put in (ii): $S+92.4-4.4S=89$
3. $3.4S=3.4\Rightarrow S=1$, so $F=42-2=40$

Trick: Option check: $F=40\Rightarrow S=1$ from (i); (ii): $1+88=89$ ✔

Answer: (B) $40$

Concept: Equal simple interest for the same time means $P_1\times r_1=P_2\times r_2$.

1. Let the sum at 14% be $x$; the rest is $21600-x$ at 4%.
2. Equal interest: $\frac{x\times 14\times 6}{100}=\frac{(21600-x)\times 4\times 6}{100}$
3. $14x=4(21600-x)\Rightarrow 18x=86400\Rightarrow x=4800$

Trick: Sums are in the inverse ratio of rates: $x:(21600-x)=4:14=2:7$, so $x=\frac{2}{9}\times 21600=4800$.

Answer: (B) ₹4,800

Concept: Filling pipes add rates, an emptying pipe subtracts its rate: $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{1}{t}$.

1. $\frac{1}{33}+\frac{1}{22}=\frac{2+3}{66}=\frac{5}{66}$
2. Emptying rate $=\frac{5}{66}-\frac{1}{18}=\frac{15-11}{198}=\frac{4}{198}=\frac{2}{99}$
3. Time to empty $=\frac{99}{2}=49.5$ minutes

Trick: Take capacity = LCM(33, 22, 18) = 198 units: rates 6 + 9 − 11 = 4 units/min for the outlet → $\frac{198}{4}=49.5$.

Answer: (B) 49.5 minutes

Concept: Expand both brackets; the $x^2$ terms cancel, leaving a linear equation.

1. $2(5x^2-5)=10x^2-10$
2. $5(2x^2+9x-5)=10x^2+45x-25$
3. Difference: $10x^2-10-10x^2-45x+25=-45x+15=18$
4. $-45x=3\Rightarrow x=-\frac{3}{45}=-\frac{1}{15}$

Trick: Check: $-45\times (-\frac{1}{15})+15=3+15=18$ ✓.

Answer: (D) $-\frac{1}{15}$

Concept: BODMAS: innermost bracket first, then multiplication, then addition, then the division outside.

1. $(4-5)=-1$
2. $22+6\times (-1)=22-6=16$
3. $640÷16=40$

Trick: Solve brackets from inside out: ( ) → { } → [ ].

Answer: (D) 40

Concept: $SP=CP\times \frac{100+\text{markup}}{100}\times \frac{100-\text{discount}}{100}$

1. Let $CP=100$; $MP=140$
2. $SP=140\times \frac{55}{100}=77$
3. Loss $=100-77=23$, i.e. $23\%$

Trick: Net effect: $40-45-\frac{40\times 45}{100}=-23\%$ (negative = loss).

Answer: (B) 23%

Concept: Write both selling prices in terms of the original CP and equate.

1. Let $CP=x$; $SP=1.54x$
2. New CP $=x+75$, new SP $=1.54x-66$
3. $1.54x-66=1.408(x+75)=1.408x+105.6$
4. $0.132x=171.6\Rightarrow x=1300$

Trick: Check option: $CP=1300$ → $SP=2002$; new $SP=1936$, new $CP=1375$; $\frac{1936}{1375}=1.408$ ✓.

Answer: (D) ₹1,300

Concept: If $p\%$ of $A=\frac{m}{n}$ of $B$, then $\frac{A}{B}=\frac{m/n}{p/100}$.

1. $\frac{69}{100}A=\frac{4}{5}B$
2. $\frac{A}{B}=\frac{4}{5}\times \frac{100}{69}=\frac{80}{69}$

Trick: $\frac{4}{5}=80\%$, so $69\%A=80\%B\Rightarrow A:B=80:69$.

Answer: (A) 80 : 69

Concept: Walk one way + drive one way is given; driving both ways is 4 h less. Find each one-way time.

1. Walk + Drive $=4\frac{1}{4}$ h
2. 2 × Drive $=4\frac{1}{4}-4=\frac{1}{4}$ h, so Drive $=\frac{1}{8}$ h
3. Walk $=4\frac{1}{4}-\frac{1}{8}=4\frac{1}{8}$ h
4. Walk both ways $=2\times 4\frac{1}{8}=8\frac{1}{4}$ h = 8 h 15 min

Trick: Walk both = 2 × (W + D) − (D + D) = $2\times 4.25-0.25=8.25$ h.

Answer: (C) 8 hours 15 minutes

Concept: Average of the first $n$ even numbers $=n+1$.

1. Sum $=n(n+1)$, so average $=\frac{n(n+1)}{n}=n+1$
2. $n=168\Rightarrow$ average $=169$

Yaad rakhein: First $n$ even → $n+1$; first $n$ odd → $n$; first $n$ natural → $\frac{n+1}{2}$.

Answer: (C) 169

Concept: The altitude of an isosceles triangle bisects the base; use Pythagoras for the height.

1. Half base $=36$ cm
2. $h=\sqrt{{85}^2-{36}^2}=\sqrt{7225-1296}=\sqrt{5929}=77$ cm
3. Area $=\frac{1}{2}\times 72\times 77=2772{\text{cm}}^2$

Trick: 36, 77, 85 is a Pythagorean triplet; or use $\frac{b}{4}\sqrt{4a^2-b^2}$.

Answer: (A) 2772

Concept: $SP=MP\times (1-d)=CP\times (1+p)$, so $\frac{MP}{CP}=\frac{100+p}{100-d}$.

1. $0.8MP=1.28CP$
2. $\frac{MP}{CP}=\frac{128}{80}=\frac{8}{5}$

Trick: $\frac{MP}{CP}=\frac{100+\text{profit}\%}{100-\text{discount}\%}$ directly.

Answer: (A) 8 : 5

Concept: Perimeter of a rectangle $=2(l+b)$.

1. $l+b=8+5=13$ cm
2. $P=2\times 13=26$ cm

Trick: 13 cm (option D) is only half the perimeter — don't stop at $l+b$.

Answer: (B) 26 cm