Q&A: practice questions with answers

Concept: Convert the decimal fraction to a whole-number fraction by multiplying numerator and denominator by $10$.

1. $\frac{1.2}{0.3}=\frac{12}{3}=4$
2. $\frac{2}{5}=0.4$
3. $4-0.4=3.6$

Trick: Same number of decimal places in numerator and denominator → just drop the decimal points.

Answer: (A) $3.6$

Concept: Two events sound together again after every $\text{LCM}$ of their intervals.

1. $\text{LCM}(12,18)=36$ minutes
2. Together at 9:36, 10:12, 10:48 AM
3. Next would be 11:24 AM — after 11:00
4. So they sound together $3$ times

Trick: Window $=120$ min; $\lfloor \frac{120}{36}\rfloor =3$ (11:00 itself is not a meeting time).

Answer: (A) $3$

Concept: In BODMAS, 'of' is done before $÷$ and $\times$; convert mixed fractions first.

1. $2\frac{1}{4}=\frac{9}{4}$, $2\frac{2}{5}=\frac{12}{5}$, $3\frac{1}{5}=\frac{16}{5}$
2. $\frac{4}{3}\text{ of }\frac{9}{4}=3$ and $\frac{9}{4}\text{ of }\frac{16}{5}=\frac{36}{5}=7.2$
3. $3÷\frac{12}{5}=\frac{5}{4}=1.25$
4. $3+1.25-7.2=-2.95$

Trick: Order inside the bracket: of → $÷$ → $+$ → $-$.

Answer: (A) $-2.95$

Concept: Find maximum marks from the given percentage condition, then convert the friend's marks to a percentage.

1. $46\%$ of maximum $=448+81=529$
2. Maximum $=\frac{529}{0.46}=1150$
3. Friend's $\%=\frac{805}{1150}\times 100=70\%$

Trick: $529={23}^2$ and $46=2\times 23$, so maximum $=\frac{529\times 100}{46}=1150$ quickly.

Answer: (B) $70\%$

Concept: $\text{Sum}=\text{Average}\times \text{Number of terms}$.

1. Total of 4 numbers $=15\times 4=60$
2. Known sum $=6.5+10.8+20=37.3$
3. $a=60-37.3=22.7$

Trick: Deviation method: $-8.5-4.2+5+(a-15)=0\Rightarrow a=22.7$.

Answer: (C) $22.7$

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$; take a convenient distance (LCM of speeds).

1. Let $AB=36\text{km}$
2. Onward: $\frac{12}{12}+\frac{24}{18}=1+\frac{4}{3}=\frac{7}{3}\text{h}$
3. Return: $\frac{18}{9}+\frac{18}{18}=2+1=3\text{h}$
4. Average $=\frac{72}{\frac{7}{3}+3}=\frac{72}{\frac{16}{3}}=13.5\text{km/h}$

Trick: Never average the speeds directly; always total distance ÷ total time.

Answer: (B) $13.5\text{km/hr}$

Concept: Same direction: relative speed $=$ difference of speeds; distance to cross $=$ sum of lengths.

1. Relative speed $=60-40=20\text{km/h}=20\times \frac{5}{18}=\frac{50}{9}\text{m/s}$
2. Distance $=120+180=300\text{m}$
3. Time $=300÷\frac{50}{9}=54\text{s}$

Trick: Time (s) $=\frac{\text{total length}\times 18}{\text{relative speed}\times 5}=\frac{300\times 18}{100}=54$.

Answer: (A) $54$ seconds

Concept: Number $=10a+b$; reversed $=10b+a$; difference $=9(b-a)$.

1. $a+b=11$
2. $9(b-a)=27\Rightarrow b-a=3$
3. $b=7$, $a=4$ → number $=47$

Trick: Reversed is larger by 27 → units digit exceeds tens digit by $\frac{27}{9}=3$; only 47 fits ($4+7=11$).

Answer: (B) $47$

Concept: Angle sum of a triangle $={180}^{\circ }$; an isosceles triangle has two equal angles. An angle $>{90}^{\circ }$ can only be the vertex angle.

1. ${110}^{\circ }$ must be the unequal angle (two angles of ${110}^{\circ }$ would exceed ${180}^{\circ }$)
2. Remaining $={180}^{\circ }-{110}^{\circ }={70}^{\circ }$
3. Each base angle $=\frac{{70}^{\circ }}{2}={35}^{\circ }$

Trick: Base angle $=\frac{{180}^{\circ }-\text{vertex}}{2}$.

Answer: (B) ${35}^{\circ }$ and ${35}^{\circ }$

Concept: Compare the effective price paid for 7 toys under each scheme (take marked price $=100$ per toy).

1. A: $0.95\times 0.79=0.7505$ → pays $75.05\%$
2. B: $0.62\times 0.67\times 0.79\approx 0.328$ → pays about $32.8\%$
3. C: $6\times 92+1\times 73=625$ out of $700$ → pays about $89.3\%$
4. D: 3 free on 7 → 10 toys for the price of 7 → pays $70\%$ per toy
5. Highest payment = least benefit → scheme C

Trick: Least beneficial = largest effective price; successive discounts multiply, never add.

Answer: (B) C

Concept: Volume of a sphere $V=\frac{4}{3}\pi r^3$.

1. $V=\frac{4}{3}\times \frac{22}{7}\times 2^3=\frac{704}{21}\approx 33.52{\text{m}}^3$
2. Water $=80\%$ of $33.52=0.8\times 33.52\approx 26.8$
3. Nearest whole number $=27{\text{m}}^3$

Trick: Option 33 is the full tank — a common trap; remember to take $80\%$.

Answer: (D) $27$

Concept: Take total work $=\text{LCM}$ of days; efficiency $=\frac{\text{work}}{\text{days}}$.

1. Total work $=12$ units; A $=1$, B $=1$ unit/day
2. In 3 days together: $3\times 2=6$ units
3. Remaining $=12-6=6$ units
4. B alone: $\frac{6}{1}=6$ days

Trick: Equal workers → in 3 days they finish half; B needs half of 12 days $=6$.

Answer: (A) $6$

Concept: Subtract all deductions from the total; the remainder is divided in the given ratio.

1. Total deductions $=39+47+40=126$
2. Remaining $=2288-126=2162$
3. Ratio sum $=17+15+14=46$; one part $=\frac{2162}{46}=47$
4. Meera after deduction $=15\times 47=705$

Trick: Question asks share after deduction — don't add back ₹47 (that gives ₹752).

Answer: (C) ₹$705$

Concept: Follow BODMAS: brackets first, then 'of', then $÷$/$\times$, then $+$.

1. $\frac{12/24}{12/6}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$
2. Bracket: $\frac{2}{9}\times \frac{18}{6}+\frac{4}{8}=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$; so $\frac{1}{4}÷\frac{7}{6}=\frac{3}{14}$
3. $\frac{28}{11}\text{ of }\frac{11}{9}=\frac{28}{9}$; $\frac{6}{9}÷\frac{28}{9}=\frac{6}{28}=\frac{3}{14}$
4. Total $=\frac{3}{14}+\frac{3}{14}=\frac{3}{7}$

Trick: Do 'of' before $÷$: $\frac{6}{9}÷(\frac{28}{11}\times \frac{11}{9})$.

Answer: (C) $\frac{3}{7}$

Concept: $A=P{(1+\frac{R}{100})}^n$; $CI=A-P$.

1. $A=3000\times (1.2{)}^2=3000\times 1.44=4320$
2. $CI=4320-3000=1320$

Trick: 2-year CI rate $=20+20+\frac{20\times 20}{100}=44\%$; $44\%$ of $3000=1320$.

Answer: (D) ₹$1,320$

Concept: For $n$ (odd) observations in ascending order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 38, 39, 41, 42, 44, 45, 46, 47, 48, 49, 50, 52, 53, 55, 60
2. $n=15$ → $\frac{15+1}{2}=8^{\text{th}}$ value
3. Median $=47$

Trick: Always sort first — the middle of the unsorted list (39) is wrong.

Answer: (A) $47$ minutes

Concept: Amounts of consecutive years differ by the factor $(1+\frac{R}{100})$.

1. $\frac{8651.50}{7865}=1.1$ → $R=10\%$
2. $P\times (1.1{)}^2=7865$
3. $P=\frac{7865}{1.21}=6500$

Trick: Interest in 3rd year $=786.5$ is $10\%$ of $7865$ → rate $10\%$.

Answer: (A) ₹$6,500$

Concept: Two successive discounts $a\%$, $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. Single equivalent of 20% and 12%: $20+12-\frac{240}{100}=29.6\%$
2. Difference in discount $=30\%-29.6\%=0.4\%$
3. $0.4\%$ of $80000=320$

Trick: Find only the gap in discount %, then apply it once to the marked price.

Answer: (A) ₹$320$

Concept: Overall $\%=\frac{\text{Total SP}-\text{Total CP}}{\text{Total CP}}\times 100$.

1. SP of first $=120$, SP of second $=90$
2. Total SP $=210$, Total CP $=200$
3. Gain $=\frac{10}{200}\times 100=5\%$

Trick: Equal CPs → overall $\%=\frac{20+(-10)}{2}=5\%$ gain.

Answer: (D) $5\%$ gain

Concept: Successive % changes multiply: $P\times (1-\frac{10}{100})(1+\frac{20}{100})$.

1. After year 1: $5000\times 0.9=4500$
2. After year 2: $4500\times 1.2=5400$

Trick: Net change $=-10+20-\frac{200}{100}=8\%$ → $5000\times 1.08=5400$.

Answer: (D) $5400$