Q&A: practice questions with answers

Concept: If the angles of elevation from two points at distances $d_1$ and $d_2$ from the base are complementary, then $h^2=d_1{d}_2$.

1. Let the angle at X be $\theta$; then at Y it is ${90}^{\circ }-\theta$.
2. $\tan \theta =\frac{h}{b^2}$ and $\tan ({90}^{\circ }-\theta )=\cot \theta =\frac{h}{a^2}$.
3. Multiply: $\tan \theta ·\cot \theta =1=\frac{h^2}{a^2{b}^2}$.
4. $h^2=a^2{b}^2\Rightarrow h=ab$ m.

Trick: Complementary angles → $h=\sqrt{d_1{d}_2}=\sqrt{a^2{b}^2}=ab$. (Sides of the tower don't matter.)

Answer: (D) $ab$

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

1. $(3-6)=-3$
2. $9\times (-3)=-27$
3. $\{30+(-27)\}=3$
4. $954÷3=318$

Trick: Watch the sign: $9\times (-3)$ is negative, so the curly bracket becomes just 3.

Answer: (B) 318

Concept: For the same distance, time is inversely proportional to speed: $S_1{T}_1=S_2{T}_2$.

1. Distance $=117\times 16=1872\text{km}$.
2. Nisha's time $=\frac{1872}{48}=39$ hours.

Trick: $T_2=16\times \frac{117}{48}=16\times \frac{39}{16}=39$ – 16 cancels instantly.

Answer: (A) 39 hours

Concept: CI for the 2nd year = interest of one year on the amount after the 1st year $=P(1+\frac{r}{100})\times \frac{r}{100}$.

1. $P\times \frac{105}{100}\times \frac{5}{100}=2352$
2. $P\times \frac{21}{400}=2352$
3. $P=\frac{2352\times 400}{21}=112\times 400=44800$

Trick: 2nd-year CI = $0.0525P$ at 5% → $P=\frac{2352}{0.0525}=44800$.

Answer: (C) ₹44,800

Concept: If $a^2+b^2=c^2$, the triangle is right-angled and Area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. ${18}^2+{80}^2=324+6400=6724={82}^2$ → right angle between sides 18 and 80.
2. Area $=\frac{1}{2}\times 18\times 80=720{\text{cm}}^2$.

Trick: (9, 40, 41) is a Pythagorean triplet; ×2 gives (18, 80, 82).

Answer: (D) 720

Concept: $x\%$ of $N=\frac{x}{100}\times N$; a fraction 'of' means multiply.

1. $65\%$ of $65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$
2. $\frac{1}{5}$ of $15=3$
3. Difference $=42.25-3=39.25$

Trick: ${65}^2=4225$ (numbers ending in 5: $6\times 7=42$, then append 25).

Answer: (D) 39.25

Concept: Take total work = LCM of the given days; then each person's work per day is in units.

1. Total work $=\text{LCM}(30,21,14)=210$ units.
2. Jaya $=\frac{210}{30}=7$, J+S $=\frac{210}{21}=10$, J+S+H $=\frac{210}{14}=15$ units/day.
3. Hina $=15-10=5$ units/day.
4. Jaya + Hina $=7+5=12$ units/day → time $=\frac{210}{12}=\frac{35}{2}$ days.

Trick: Hina's rate $=\frac{1}{14}-\frac{1}{21}=\frac{1}{42}$, then $\frac{1}{30}+\frac{1}{42}=\frac{2}{35}$.

Answer: (A) $\frac{35}{2}$ days

Concept: Successive discounts are applied one after another: SP $=\text{MP}\times \frac{100-d_1}{100}\times \frac{100-d_2}{100}$.

1. After 10%: $5000\times \frac{90}{100}=4500$.
2. After 5%: $4500\times \frac{95}{100}=4275$.

Trick: Net discount $=10+5-\frac{10\times 5}{100}=14.5\%$ → $5000\times 0.855=4275$.

Answer: (C) ₹4,275

Concept: Profit is shared in the ratio of (capital × time).

1. M + N = 1359 and M − N = 453 → M = 906, N = 453, so ratio $=2:1$.
2. Let N's capital $=x$; M's $=x+46000$.
3. $\frac{3(x+46000)}{4x}=\frac{2}{1}\Rightarrow 3x+138000=8x\Rightarrow x=27600$.
4. M's capital $=27600+46000=73600$.

Trick: Check options: $3\times 73600=220800$ and $4\times 27600=110400$ → exactly $2:1$.

Answer: (D) ₹73,600

Concept: Mean proportional between $a$ and $b$ $=\sqrt{ab}$.

1. $\sqrt{24\times 150}=\sqrt{3600}=60$

Trick: Check: $24:60=60:150$ since $\frac{24}{60}=\frac{60}{150}=\frac{2}{5}$.

Answer: (A) 60

Concept: Take distance MN = LCM of the journey times; then meeting time = remaining distance ÷ sum of speeds.

1. A takes 8:25 AM → 3:25 PM $=7$ h; B takes 10:25 AM → 3:25 PM $=5$ h.
2. Let $MN=35$ units: speed of A $=5$, speed of B $=7$ units/h.
3. By 10:25 AM, A has covered $2\times 5=10$ units; $25$ units remain.
4. Time to meet $=\frac{25}{5+7}=\frac{25}{12}\text{h}=2\text{h}5\text{min}$ after 10:25 AM → 12:30 PM.

Trick: Choose distance = LCM of times so that speeds become whole numbers.

Answer: (B) 12:30 PM

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Digit sum of 327087 $=3+2+7+0+8+7=27$, a multiple of 3 → divisible by 3.
2. Last digit 7 → not divisible by 5; last three digits 087 are not a multiple of 8 → not by 8.
3. Not divisible by 4 (87 is odd), so not by 12.

Trick: Odd number → straight away rule out 8 and 12.

Answer: (C) 3

Concept: Greatest number leaving the same remainder = HCF of the differences of the numbers.

1. Differences: $97-49=48$, $189-97=92$, $189-49=140$.
2. $48=2^4\times 3$, $92=2^2\times 23$, $140=2^2\times 5\times 7$.
3. HCF $=2^2=4$. Check: remainders $49,97,189$ ÷ 4 are all 1.

Trick: Same remainder → HCF of differences (not of the numbers).

Answer: (B) 4

Concept: Expand each bracket, then combine like terms.

1. $3(\frac{7}{3}{x}^2-25x+12)=7x^2-75x+36$
2. $-7(x^2+9x-13)=-7x^2-63x+91$
3. Add: $(7-7)x^2+(-75-63)x+(36+91)=-138x+127$

Trick: The $x^2$ terms cancel; watch the sign of $-7\times (-13)=+91$.

Answer: (B) $-138x+127$

Concept: $\text{Loss}\%=\frac{\text{CP}-\text{SP}}{\text{CP}}\times 100$ for the same number of articles.

1. Let CP of 1 book $=$ ₹1. SP of 20 books $=$ CP of 5 books $=$ ₹5.
2. CP of 20 books $=$ ₹20, SP $=$ ₹5 → loss $=$ ₹15.
3. $\text{Loss}\%=\frac{15}{20}\times 100=75\%$

Trick: SP of $x$ = CP of $y$ → $\%=\frac{y-x}{x}\times 100=\frac{5-20}{20}\times 100=-75\%$ (minus = loss).

Answer: (D) $75\%$ loss

Concept: One-day work = $\frac{1}{\text{days}}$; work left = $1-$ work done.

1. (A+B)'s 1-day work $=\frac{1}{12}+\frac{1}{18}=\frac{3+2}{36}=\frac{5}{36}$.
2. Work in 4 days $=4\times \frac{5}{36}=\frac{20}{36}=\frac{5}{9}$.
3. Work left $=1-\frac{5}{9}=\frac{4}{9}$.

Trick: Total work = LCM$(12,18)=36$ units; A $=3$, B $=2$ per day → 4 days $=20$ units, left $16$ of $36=\frac{4}{9}$.

Answer: (A) $\frac{4}{9}$

Concept: Angle sum of a triangle, alternate angles between parallel lines, and equal angles opposite equal sides.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{19}^{\circ }-{137}^{\circ }={24}^{\circ }$.
2. $BC\parallel AD$, so $\angle CAD=\angle BCA={24}^{\circ }$ (alternate angles).
3. $AC=CD$ → $\angle CDA=\angle CAD={24}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{24}^{\circ }-{24}^{\circ }={132}^{\circ }$.

Trick: $\angle ACD={180}^{\circ }-2\times \angle BCA$.

Answer: (C) ${132}^{\circ }$

Concept: Divisible by 11 if (sum of odd-place digits) − (sum of even-place digits) is 0 or a multiple of 11.

1. N83M92: odd places $N+3+9=N+12$; even places $8+M+2=M+10$.
2. Difference $=(N+12)-(M+10)=N-M+2$.
3. For single digits, $N-M+2=0$ → $M-N=2$ (the other case $N-M=9$ is not among options).

Trick: Check: $M=3,N=1$ → 183392 $=11\times 16672$.

Answer: (B) $M-N=2$

Concept: No profit no loss → SP = CP; $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. SP $=$ CP $=$ ₹3,500.
2. $\text{MP}\times \frac{80}{100}=3500$
3. $\text{MP}=\frac{3500\times 100}{80}=4375$ → ₹4,375.

Trick: 20% off means SP is $\frac{4}{5}$ of MP → MP $=3500\times \frac{5}{4}$.

Answer: (D) ₹4,375

Concept: Evaluate powers and roots first, then add/subtract (BODMAS).

1. ${10}^2=100$, $\sqrt{2^2}=2$, $4\sqrt{25}=4\times 5=20$.
2. $100+2-20-3=79$.

Yaad rakhein: $\sqrt{a^2}=a$ for positive $a$.

Answer: (C) 79