Q&A: practice questions with answers

Concept: Use ${\sin }^2\theta +{\cos }^2\theta =1$ to reduce the equation to one ratio.

1. $10{\cos }^2\theta +7{\sin }^2\theta =7({\sin }^2\theta +{\cos }^2\theta )+3{\cos }^2\theta =7+3{\cos }^2\theta$
2. So $7+3{\cos }^2\theta =7\Rightarrow {\cos }^2\theta =0\Rightarrow \theta ={90}^{\circ }$
3. $\sin \theta +\cos \theta =1+0=1$

Trick: Write the bigger coefficient as $7+3$: the extra $3{\cos }^2\theta$ must be 0.

Answer: (A) 1

Concept: Form the equation and solve backwards.

1. $\frac{x}{3}+37=51\Rightarrow \frac{x}{3}=14$
2. $x=14\times 3=42$
3. Sum of digits = $4+2=6$

Trick: Reverse the operations: $(51-37)\times 3=42$.

Answer: (D) 6

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with T in years (9 months $=\frac{9}{12}=\frac{3}{4}$ year).

1. $\text{SI}=28260\times 9\times \frac{3}{4}÷100$
2. $=28260\times \frac{6.75}{100}=28260\times 0.0675$
3. = ₹1907.55

Trick: 9% for $\frac{3}{4}$ year = 6.75%; 1% of 28260 = 282.6, so $6.75\times 282.6=1907.55$.

Answer: (C) ₹ 1907.55

Concept: Opposite directions: relative speed = sum; distance = sum of lengths. Bridge: distance = train + bridge.

1. Relative speed = $54+72=126$ km/h $=126\times \frac{5}{18}=35$ m/s
2. Sum of lengths = $35\times 30=1050$ m $\Rightarrow 3k+2k=1050\Rightarrow k=210$; train P = 630 m
3. Speed of P = $54\times \frac{5}{18}=15$ m/s; in 90 s it covers 1350 m
4. Bridge = $1350-630=720$ m

Trick: km/h → m/s: multiply by $\frac{5}{18}$ (54 → 15, 72 → 20).

Answer: (A) 720 meters

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

1. $X:Y=12000\times 12:18000\times 6=144000:108000=4:3$
2. X's share = $25200\times \frac{4}{7}=$ ₹14,400

Trick: Cancel thousands first: $12\times 12:18\times 6=144:108=4:3$.

Answer: (C) ₹14,400

Concept: Area printed in one revolution = curved surface area = $2\pi rh$.

1. Area per revolution = $4158÷350=11.88{\text{m}}^2$
2. $2\times \frac{22}{7}\times r\times 2.1=13.2r$
3. $13.2r=11.88\Rightarrow r=0.9$ m = 90 cm

Trick: $2\pi \times 2.1=13.2$ is a neat number — divide 11.88 by it directly.

Answer: (C) 90 cm

Concept: Increase = original × rate/100.

1. Increase = $300\times \frac{5}{100}=$ ₹15
2. New price = ₹315, which is ₹15 more

Trick: $5\%=\frac{1}{20}$, so $300÷20=15$.

Answer: (A) ₹15

Concept: Net rate = filling rate − emptying rate.

1. Net per minute = $\frac{1}{20}-\frac{1}{45}=\frac{9-4}{180}=\frac{1}{36}$
2. Full tank in 36 min
3. $\frac{1}{4}$ tank $=36÷4=9$ min

Trick: Product/difference: $20\times 45÷(45-20)=\frac{900}{25}=36$ min for full tank.

Answer: (C) 9

Concept: Solve innermost brackets first, then ×/÷, then +/−.

1. $5-2=3$; $3+2\times 3=9$; $36+4\times 9=72$; $72÷(6+3)=8$
2. $40-(6+3\times 2)=40-12=28$; $8\times 2-6=10$; $28+10=38$
3. Total = $8+38=46$

Trick: Work bracket by bracket from inside out; never add before multiplying.

Answer: (B) 46

Concept: ${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$.

1. $x^4+\frac{1}{x^4}={(x^2+\frac{1}{x^2})}^2-2$
2. $={16}^2-2=256-2=254$

Trick: Square and subtract 2 — every time you go one power up.

Answer: (A) 254

Concept: Difference of percentages equals the difference of marks (pass mark is common).

1. $84\%-45\%=39\%$ of max
2. Marks gap = $62+35.5=97.5$
3. Max = $97.5\times \frac{100}{39}=250$

Trick: Above pass (+) and below pass (−) → add the two gaps.

Answer: (C) 250

Concept: For $n=p^a{q}^b$, number of divisors = $(a+1)(b+1)$.

1. $p^3\to 3+1=4$ divisors $(1,p,p^2,p^3)$
2. $p·q\to (1+1)(1+1)=4$ divisors
3. $p^2\to 2+1=3$ divisors $(1,p,p^2)$ — not four

Trick: Add 1 to each power and multiply.

Answer: (D) $p^2$ for prime p

Concept: x and y are reciprocals; use $x^2+y^2=(x+y{)}^2-2xy$.

1. $x=\frac{(\sqrt{5}+2{)}^2}{5-4}=9+4\sqrt{5}$; $y=9-4\sqrt{5}$
2. $x+y=18,xy=1$
3. $x^2+y^2={18}^2-2=324-2=322$

Trick: When $xy=1$, $x^2+y^2=(x+y{)}^2-2$.

Answer: (C) 322

Concept: Each year every person's age rises by 1; total = average × number.

1. At marriage: husband + wife = $2\times 28=56$
2. After n years: couple = $56+2n$, daughter = $n-2$
3. $56+2n+n-2=3\times 24=72\Rightarrow 3n=18\Rightarrow n=6$

Trick: Check options: $n=6\to 68+4=72$ ✓.

Answer: (B) 6

Concept: Area of regular hexagon = $6\times \left(\frac{\sqrt{3}}{4}\right)\times a^2$.

1. Square side = $\sqrt{9}=3$ cm, perimeter = 12 cm
2. Hexagon side = $12÷6=2$ cm
3. Area = $6\times \frac{\sqrt{3}}{4}\times 4=6\sqrt{3}{\text{cm}}^2$

Trick: Hexagon = 6 equilateral triangles of the same side.

Answer: (B) $6\sqrt{3}$

Concept: Check each statement against standard triangle facts.

1. (i) Angle sum = ${180}^{\circ }$ — true
2. (ii) Area = $\frac{1}{2}$ × base × altitude, not $\frac{1}{3}$ — false
3. (iii) Perimeter = sum of sides — true; (iv) Exterior angle = sum of interior opposite angles — true

Yaad rakhein: $\frac{1}{3}$ belongs to cone/pyramid volume, not triangle area.

Answer: (D) (ii)

Concept: Take total work = LCM of times; add hourly efficiencies.

1. Work = LCM(10, 15, 20) = 60 units; A = 6, B = 4, C = 3 units/h
2. A + B in 2 h = $2\times 10=20$ units; left = 40
3. A + B + C = 13 units/h ⇒ $t=\frac{40}{13}\approx 3.08\approx 3$

Trick: LCM method avoids fractions.

Answer: (D) 3

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

1. $P=\frac{720\times 100}{5\times 2}$
2. $=\frac{72000}{10}=$ ₹7200

Trick: Total rate = $5\times 2=10\%$; 10% = 720 ⇒ 100% = 7200.

Answer: (B) 7200

Concept: Mode = the value that occurs most often.

1. 55 occurs 3 times (4th, 11th, 17th values)
2. 44, 45, 52 and 54 occur twice each; all other values occur once
3. So the most frequent value is 55

Trick: Tally each value once while scanning — don't stop at the first repeated value.

Answer: (A) 55

Concept: BODMAS: brackets, then ÷ and × from left to right, then +.

1. $\frac{5}{6}+\frac{1}{3}=\frac{7}{6}$; $\frac{2}{5}+0.3=0.7=\frac{7}{10}$
2. $\frac{7}{6}÷\frac{7}{10}=\frac{10}{6}=\frac{5}{3}$
3. $\frac{5}{3}\times 0.5=\frac{5}{6}$; $\frac{5}{6}+\frac{5}{6}=\frac{10}{6}=\frac{5}{3}$

Trick: Convert decimals to fractions ($0.5=\frac{1}{2}$) so everything cancels cleanly.

Answer: (A) $\frac{5}{3}$