Q&A: practice questions with answers

Concept: Write both selling prices in terms of the cost price $x$ and compare their difference with ₹126.

1. Original: SP $=1.15x$.
2. New CP $=0.9x$, gain $20\%$ ⟹ new SP $=1.2\times 0.9x=1.08x$.
3. Difference: $1.15x-1.08x=0.07x=126$ ⟹ $x=1800$.

Trick: Successive % on CP: $0.9\times 1.2=1.08$, so the gap is $(115-108)\%=7\%$ of CP $=$ ₹126.

Answer: (B) ₹1,800

Concept: If numbers are $4k,13k,9k$ with $k$ their HCF, then LCM $=k\times \text{LCM}(4,13,9)$.

1. $\text{LCM}(4,13,9)=468$ (all co-prime).
2. $468k=6084$ ⟹ $k=13$.
3. HCF $=k=13$.

Trick: For co-prime ratio terms, HCF $={\displaystyle \frac{\text{LCM}}{\text{product of ratio terms}}}$.

Answer: (B) 13

Concept: Form linear equations with son's present age $s$.

1. Man $=2s+4$.
2. After 6 years: $(2s+4+6)+(s+6)=58$ ⟹ $3s+16=58$.
3. $s=14$ years.

Trick: Present sum $=58-12=46$; then $3s+4=46$ ⟹ $s=14$.

Answer: (C) 14 years

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

1. $1320={\displaystyle \frac{P\times 20\times 3}{100}}={\displaystyle \frac{60P}{100}}$.
2. $P={\displaystyle \frac{1320\times 100}{60}}=2200$.

Trick: $20\%\times 3=60\%$ of sum $=1320$ ⟹ sum $=1320÷0.6=2200$.

Answer: (B) 2,200

Concept: For a fixed perimeter, the triangle of maximum area is equilateral.

1. Each side $={\displaystyle \frac{18}{3}}=6$ cm.
2. Area $={\displaystyle \frac{\sqrt{3}}{4}}{a}^2={\displaystyle \frac{\sqrt{3}}{4}}\times 36=9\sqrt{3}$ cm².

Yaad rakhein: Fixed perimeter ⟹ most "regular" shape has the greatest area (equilateral triangle, square, circle).

Answer: (C) $9\sqrt{3}$

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