Q&A: practice questions with answers

Concept: Translate the statement into an equation: 20% of N = 20% of 620 + 200.

1. 20% of 620 $=124$.
2. $0.2N=124+200=324$.
3. $N=\frac{324}{0.2}=1620$.

Trick: 20% of (N − 620) = 200 ⇒ $N-620=1000\Rightarrow N=1620$.

Answer: (C) 1620

Concept: In a right triangle with a 30° angle, the side opposite 30° is half the hypotenuse and the other leg is $\sqrt{3}$ times it.

1. ∠B = 90°, ∠C = 30° → AB is opposite 30°: $AB=5$.
2. Hypotenuse $AC=2\times AB=10$; $BC=AB\sqrt{3}=5\sqrt{3}$ (or $\tan {30}^{\circ }=\frac{AB}{BC}$).
3. $AC+BC=10+5\sqrt{3}$.

Trick: 30-60-90 sides are in ratio $1:\sqrt{3}:2$ (opposite 30°, 60°, 90°). Multiply by 5.

Answer: (C) $10+5\sqrt{3}$

Concept: When equal time is spent at different speeds, average speed = simple average of the speeds.

1. Speeds: stage 1 $=x$, stage 2 $=1.5x$, stage 3 $=0.8\times 1.5x=1.2x$.
2. Average $=\frac{x+1.5x+1.2x}{3}=\frac{3.7x}{3}=48$.
3. $x=\frac{144}{3.7}\approx 38.92\approx 38.9$.

Trick: Equal time → arithmetic mean of speeds; equal distance → harmonic mean. The 120 km is not needed here.

Answer: (A) 38.9

Concept: If A is $p\%$ more than B, B is less than A by $\frac{p}{100+p}\times 100\%$.

1. Let B $=100$; then A $=230$.
2. B is less than A by $230-100=130$, on base A.
3. $\frac{130}{230}\times 100=56.52\%$.

Trick: $\frac{130}{230}=\frac{13}{23}$; $13÷23=0.5652$. Always use the larger value as the base when asking "less than".

Answer: (B) 56.52%

Concept: Use the simplest equation to substitute one variable and reduce to two equations.

1. From $x+y-z=0$: $z=x+y$.
2. Eq. 1: $x+4y+2(x+y)=12\Rightarrow 3x+6y=12\Rightarrow x+2y=4$.
3. Eq. 2: $2x-y+(x+y)=3\Rightarrow 3x=3\Rightarrow x=1$.
4. $y=\frac{4-1}{2}=\frac{3}{2}$, $z=1+\frac{3}{2}=\frac{5}{2}$.

Trick: Options can be tested: only $x=1,y=\frac{3}{2},z=\frac{5}{2}$ satisfies $x+y-z=0$ and $2x-y+z=3$ together.

Answer: (C) $x=1$, $y=\frac{3}{2}$, $z=\frac{5}{2}$

Concept: Mode $=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $f_1$ = modal-class frequency, $f_0$, $f_2$ = frequencies of the classes before and after.

1. Mode 36 lies in 30–40 → $l=30$, $h=10$, $f_1=9$, $f_2=5$, $f_0=x$ (20–30).
2. $36=30+\frac{9-x}{18-x-5}\times 10\Rightarrow \frac{9-x}{13-x}=0.6$.
3. $9-x=7.8-0.6x\Rightarrow 0.4x=1.2\Rightarrow x=3$.

Trick: Check: $\frac{9-3}{13-3}=\frac{6}{10}=0.6$ → $30+6=36$ ✓.

Answer: (A) 3

Concept: Angle at the centre = 2 × angle at any point of the remaining (major) arc; angles in opposite segments are supplementary.

1. ∠AOB = 120° → for a point M on the major arc, ∠AMB $=\frac{{120}^{\circ }}{2}={60}^{\circ }$.
2. S lies on the minor arc, so ASBM is a cyclic quadrilateral: ∠ASB $={180}^{\circ }-{60}^{\circ }={120}^{\circ }$.

Trick: Point on the minor arc sees the chord at ${180}^{\circ }-\frac{\text{central angle}}{2}$; here ${180}^{\circ }-{60}^{\circ }={120}^{\circ }$. 60° (option D) is the trap for a point on the major arc.

Answer: (A) 120°

Concept: Take total capacity = LCM of the times; add filling rates, subtract the emptying rate.

1. Capacity $=\text{LCM}(24,36,18)=72$ units.
2. Rates: $\frac{72}{24}=3$, $\frac{72}{36}=2$, outlet $\frac{72}{18}=-4$ units/min.
3. Net $=3+2-4=1$ unit/min → time $=72$ minutes.

Trick: With LCM units the net rate is often 1 or 2 – here net 1 unit/min means time = capacity = 72 min.

Answer: (D) 72 minutes

Concept: Total provision (man-days) is constant: $M_1{D}_1=M_2{D}_2$.

1. Man-days $=100\times 26=2600$.
2. Remaining men $=100-48=52$.
3. $D_2=\frac{2600}{52}=50$ days.

Trick: Fewer men → more days (inverse). $\frac{100}{52}\times 26=\frac{2600}{52}=50$.

Answer: (C) 50

Concept: Write present ages as multiples of the ratio and form an equation for the future ratio.

1. Let P $=3x$, Q $=4x$.
2. After 6 years: $\frac{3x+6}{4x+6}=\frac{4}{5}\Rightarrow 15x+30=16x+24\Rightarrow x=6$.
3. P $=3\times 6=18$ years.

Trick: Difference of ages is constant: $x$ (present) = 1 unit (future) and the units differ by 6 years → $x=6$.

Answer: (A) 18 years

Concept: Compare powers of each prime. LCM takes the maximum power, HCF the minimum.

1. $120=2^3·3·5$, $360=2^3·3^2·5$, $900=2^2·3^2·5^2$, $90=2·3^2·5$. Let $N=2^a{3}^b{5}^c$ (no other primes, since N | 360).
2. Prime 2: $min(2,a)=1\Rightarrow a=1$. Prime 3: $max(1,b)=2$ and $min(2,b)=2\Rightarrow b=2$. Prime 5: $min(2,c)=1\Rightarrow c=1$.
3. $N=2·9·5=90$ only. Check: LCM(120, 90) = 360 ✓, HCF(900, 90) = 90 ✓.

Trick: N must be a multiple of HCF 90 and a divisor of LCM 360: candidates 90, 180, 360; only 90 keeps HCF(900, N) = 90.

Answer: (C) 1

Concept: If a sum becomes $n$ times under SI, the interest is $(n-1)$ times the principal: $R=\frac{(n-1)\times 100}{T}$.

1. Five times → SI $=4P$.
2. $4P=\frac{P\times R\times 10}{100}\Rightarrow R=40\%$.

Trick: $R=\frac{(5-1)\times 100}{10}=40\%$ per annum.

Answer: (C) 40%

Concept: Average $=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=528$, count $=12$.
2. Average $=\frac{528}{12}=44$.

Trick: $12\times 44=528$ ($12\times 40=480$, $12\times 4=48$).

Answer: (C) 44

Concept: Put $t=\sqrt{m}$ to turn the equation into a quadratic in $t$.

1. $2m=5\sqrt{m}-2\Rightarrow 2t^2-5t+2=0$.
2. $(2t-1)(t-2)=0\Rightarrow t=2$ or $t=\frac{1}{2}$.
3. $m=t^2=4$ or $\frac{1}{4}$.
4. Difference $=4-\frac{1}{4}=\frac{15}{4}$.

Trick: Whenever $m$ and $\sqrt{m}$ appear together, substitute $t=\sqrt{m}$ – it is just a quadratic.

Answer: (B) $\frac{15}{4}$

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$.

1. Total $=21\times 465+11\times 410+17\times 466=9765+4510+7922=22197$.
2. Members $=21+11+17=49$.
3. Average $=\frac{22197}{49}=453$.

Trick: Take 450 as base: deviations $21\times 15+11\times (-40)+17\times 16=315-440+272=147$; $\frac{147}{49}=3$, so average $=453$.

Answer: (B) 453

Concept: Number of coins $8x:4x:3x$; value = number × denomination.

1. Value $=8x\times 1+4x\times 2+3x\times 5=8x+8x+15x=31x$.
2. $31x=930\Rightarrow x=30$.
3. Total coins $=(8+4+3)x=15\times 30=450$.

Trick: Multiply each ratio term by its denomination, add, and divide the total value by that sum to get $x$.

Answer: (B) 450

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (not the average of speeds).

1. Distance 1 $=20\times 90=1800\text{m}$; distance 2 $=65\times 60=3900\text{m}$.
2. Total $=5700\text{m}$ in $150$ min.
3. Average speed $=\frac{5700}{150}=38\text{m/min}$.

Trick: With times in ratio $3:2$, weighted mean $=\frac{3\times 20+2\times 65}{5}=\frac{190}{5}=38$.

Answer: (C) 38

Concept: Let the numbers be $3x,2x,3x$; total cost = sum of (number × price).

1. Cost $=3x\times 280+2x\times 180+3x\times 120=840x+360x+360x=1560x$.
2. $1560x=6240\Rightarrow x=4$.
3. Articles of type Q $=2x=8$.

Trick: $1560\times 4=6240$ – spotting that $6240$ is $4\times 1560$ gives $x$ instantly.

Answer: (B) 8

Concept: Divide by $x^2$ to get $x^2+\frac{1}{x^2}$, then use ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$ and $x^3-\frac{1}{x^3}=a^3+3a$ where $a=x-\frac{1}{x}$.

1. $x^4-83x^2+1=0\Rightarrow x^2+\frac{1}{x^2}=83$.
2. ${(x-\frac{1}{x})}^2=83-2=81\Rightarrow x-\frac{1}{x}=9$ (taking the value that keeps the root real).
3. $x^3-x^{-3}=9^3+3\times 9=729+27=756$.
4. $\sqrt{756+28}=\sqrt{784}=28$.

Trick: $x^2+\frac{1}{x^2}=83\Rightarrow x-\frac{1}{x}=\sqrt{81}=9$; then $a^3+3a$ – remember ‘minus’ identity has ‘plus $3a$’.

Answer: (C) 28

Concept: For similar triangles, $\frac{A_1}{A_2}={\left(\frac{h_1}{h_2}\right)}^2$ (ratio of areas = square of ratio of corresponding altitudes/sides).

1. $\frac{h_1}{h_2}=\sqrt{\frac{841}{676}}$.
2. $\sqrt{841}=29$, $\sqrt{676}=26$.
3. Ratio $=29:26$.

Trick: Areas → take square root for any length ratio (side, altitude, median, perimeter).

Answer: (B) 29 : 26