Q&A: practice questions with answers

Concept: Apply the changes one after another as multipliers.

1. After $10\%$ increase: $600\times 1.1=660$.
2. After $15\%$ discount: $660\times 0.85=561$.

Trick: Net change $=10-15-{\displaystyle \frac{10\times 15}{100}}=-6.5\%$; $600\times 0.935=561$.

Answer: (C) ₹561

Concept: To cross a bridge the train covers (its own length + bridge length).

1. Speed $=72\times {\displaystyle \frac{5}{18}}=20$ m/s.
2. Distance $=180+350=530$ m.
3. Time $={\displaystyle \frac{530}{20}}=26.5$ s.

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

Answer: (A) 26.5 seconds

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a-b{)}^2={27}^2-4\times 170=729-680=49$.
2. $a-b=7$.

Trick: Check: $17\times 10=170$, $17+10=27$, difference $7$.

Answer: (A) 7

Concept: Savings $=$ discount $\%$ of the yearly monthly-payment total.

1. Monthly for a year: $150\times 12=1800$.
2. Savings $=15\%$ of $1800=270$.

Trick: $10\%$ of 1800 $=180$, $5\%=90$; total $270$.

Answer: (D) 270

Concept: CI for 2 years at $10\%$ $=21\%$ of principal (since ${1.1}^2=1.21$).

1. $0.21y=462$ ⟹ $y=2200$.
2. Amount $=2200+462=2662$.

Trick: $21\%$ → 462, so $1\%$ → 22 and $121\%$ → $2662$.

Answer: (C) ₹2,662

Concept: Surface area of sphere $=4\pi r^2$.

1. $4\times {\displaystyle \frac{22}{7}}\times r^2=28962$ ⟹ $r^2={\displaystyle \frac{28962\times 7}{88}}\approx 2303.8$.
2. $r\approx 48$ cm (since ${48}^2=2304$).
3. Diameter $=2r=96$ cm.

Trick: ${\displaystyle \frac{28962}{88}}\approx 329$, $\times 7\approx 2304={48}^2$. Don't stop at $r$ – the question asks the diameter.

Answer: (D) 96 cm

Concept: Two successive changes of $a\%$ and $b\%$ give net $a+b+{\displaystyle \frac{ab}{100}}$.

1. $-10-10+{\displaystyle \frac{(-10)(-10)}{100}}=-20+1=-19\%$.
2. Check: $100\to 90\to 81$, fall $=19\%$.

Trick: Depreciation is less than $20\%$ because the second $10\%$ acts on a smaller value.

Answer: (C) 19%

Concept: Compare what fraction of the marked price (MP) of 10 toys the customer actually pays under each scheme.

1. (A) $0.84\times 0.71=0.5964$ → pays $59.64\%$.
2. (B) $0.96\times 0.66\times 0.63=0.3992$ → pays $39.92\%$ (discount $\approx 60.1\%$).
3. (C) ${\displaystyle \frac{4\times 0.88+6\times 0.73}{10}}=0.79$ → pays $79\%$.
4. (D) Buying 10 toys she pays full price for 10 (the 4 free toys are extra, not a price cut); effectively ${\displaystyle \frac{10}{14}}\approx 71.4\%$ per toy.
5. Scheme B stands apart from the other three – it is the only one with a discount above $50\%$.

Trick: Multiply the "paid" fractions $(1-d)$; never add the discount percentages.

Answer: (B) B

Concept: Work per hour $={\displaystyle \frac{1}{\text{time}}}$; split the tank into the part filled with Z and without Z.

1. X+Y+Z fill ${\displaystyle \frac{1}{8}}$ per hour ⟹ in 3 h: ${\displaystyle \frac{3}{8}}$; remaining ${\displaystyle \frac{5}{8}}$.
2. X+Y fill ${\displaystyle \frac{5}{8}}$ in 10 h ⟹ ${\displaystyle \frac{5}{80}}={\displaystyle \frac{1}{16}}$ per hour.
3. Z $={\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}={\displaystyle \frac{1}{16}}$ ⟹ Z alone takes 16 h.

Trick: Take tank $=16$ units: all three $=2$/h, X+Y $=10÷10=1$/h, so Z $=1$/h → 16 h.

Answer: (B) 16

Concept: Bells ring together again after LCM of the intervals.

1. $1764=2^2\times 3^2\times 7^2$, $2268=2^2\times 3^4\times 7$.
2. LCM $=2^2\times 3^4\times 7^2=4\times 81\times 49=15876$ s.
3. $15876=14400+1476$ ⟹ 4 h $+$ 24 min 36 s.

Trick: HCF $=252$; LCM $={\displaystyle \frac{1764\times 2268}{252}}=7\times 2268=15876$.

Answer: (A) 4 hours 24 minutes 36 seconds

Concept: Net rate $=$ sum of filling rates $-$ emptying rate.

1. Net per hour $={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{37}}-{\displaystyle \frac{1}{37}}={\displaystyle \frac{1}{2}}$.
2. Half cistern $={\displaystyle \frac{1}{2}}÷{\displaystyle \frac{1}{2}}=1$ hour.

Trick: The 37-hour filler and the 37-hour emptier cancel each other – only the 2-hour tap matters.

Answer: (B) 1

Concept: Let the numbers be $3x$ and $5x$ and use the new ratio.

1. ${\displaystyle \frac{3x-6}{5x-6}}={\displaystyle \frac{1}{2}}$ ⟹ $6x-12=5x-6$ ⟹ $x=6$.
2. Numbers: 18 and 30; difference $=12$.

Trick: Difference $=2x$; once $x=6$, answer $=12$ directly.

Answer: (D) 12

Concept: Number $=10a+b$; reversing gives $10b+a$, and the difference is $9(b-a)$.

1. $9(b-a)=27$ ⟹ $b-a=3$.
2. $a+b=9$ ⟹ $b=6,a=3$; number $=36$.
3. Unit digit $=6$ (check: $36+27=63$).

Trick: Adding $9k$ reverses digits ⟹ digits differ by $k$ (here $k=3$).

Answer: (D) 6

Concept: Bracket first, then multiplication, then addition/subtraction left to right.

1. ${\displaystyle \frac{5}{6}}\times {\displaystyle \frac{9}{10}}={\displaystyle \frac{45}{60}}={\displaystyle \frac{3}{4}}$.
2. ${\displaystyle \frac{3}{4}}+{\displaystyle \frac{3}{4}}={\displaystyle \frac{3}{2}}$.
3. ${\displaystyle \frac{3}{2}}-{\displaystyle \frac{2}{3}}={\displaystyle \frac{9-4}{6}}={\displaystyle \frac{5}{6}}$.

Trick: Cancel before multiplying: ${\displaystyle \frac{5}{6}}\times {\displaystyle \frac{9}{10}}={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{2}}$.

Answer: (D) $\frac{5}{6}$

Concept: Excluded number $=$ old total $-$ new total.

1. Old total $=15\times 82=1230$.
2. New mean $=76$, new total $=14\times 76=1064$.
3. Excluded $=1230-1064=166$.

Trick: Removed number $=$ new mean $+$ (fall in mean $\times$ original count) $=76+6\times 15=166$.

Answer: (C) 166

Concept: Diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs = half-diagonals.

1. Side $={\displaystyle \frac{100}{4}}=25$ m; half of given diagonal $=7$ m.
2. Half of other diagonal $=\sqrt{{25}^2-7^2}=\sqrt{576}=24$ m.
3. Other diagonal $=2\times 24=48$ m.

Yaad rakhein: $(7,24,25)$ is a Pythagorean triplet.

Answer: (A) 48 meters

Concept: Convert each ratio to a decimal and compare.

1. ${\displaystyle \frac{5}{8}}=0.625$, ${\displaystyle \frac{3}{5}}=0.6$, ${\displaystyle \frac{7}{11}}\approx 0.636$, ${\displaystyle \frac{4}{7}}\approx 0.571$.
2. Greatest $={\displaystyle \frac{7}{11}}$.

Trick: Compare ${\displaystyle \frac{5}{8}}$ and ${\displaystyle \frac{7}{11}}$ by cross-multiplying: $5\times 11=55<7\times 8=56$, so ${\displaystyle \frac{7}{11}}$ is bigger.

Answer: (C) 7 : 11

Concept: Multiply by $\sin \theta$ to get a perfect square in $\sin \theta$.

1. $4\sin \theta +{\displaystyle \frac{1}{\sin \theta }}=4$ ⟹ $4{\sin }^2\theta -4\sin \theta +1=0$ ⟹ $(2\sin \theta -1{)}^2=0$.
2. $\sin \theta ={\displaystyle \frac{1}{2}}$ ⟹ $\theta ={30}^{\circ }$.
3. $\sin {90}^{\circ }+\cos {90}^{\circ }=1+0=1$.

Trick: $x+{\displaystyle \frac{1}{x}}=2$ type (here $2\sin \theta +{\displaystyle \frac{1}{2\sin \theta }}=2$) forces $x=1$, i.e. $2\sin \theta =1$.

Answer: (C) 1

Concept: Combined average $={\displaystyle \frac{\text{total of all scores}}{\text{total students}}}$.

1. Boys total $=8\times 56=448$; girls total $=4\times 80=320$.
2. Average $={\displaystyle \frac{768}{12}}=64$.

Trick: Ratio $8:4=2:1$, so the average lies ${\displaystyle \frac{1}{3}}$ of the way from 56 to 80: $56+8=64$.

Answer: (D) 64

Concept: Divisible by $24$ ⟺ divisible by both $8$ and $3$.

1. By 8: last three digits $84B$ must be divisible by 8 ⟹ $B=0$ (840) or $B=8$ (848).
2. By 3: digit sum $=4+6+A+4+8+4+B=26+A+B$ must be a multiple of 3.
3. With $B=0$: $A=1$ works ($27$) ⟹ $A+B=1$, the minimum.

Trick: Try the smallest $B$ first; the smallest $A$ that fixes the digit sum gives the minimum.

Answer: (C) 1