Competitive Exams : Quantitative Aptitude - Percentage (Set 2) | Concept of Maths

Question 1

1. The population of a town increases by $15\%$ in the first year and decreases by $10\%$ in the second year. If the current population is $20,700$, what was the population two years ago?

A) $18,000$

B) $20,000$

C) $21,500$

D) $19,000$

Answer

Correct Answer: B

Solution: Let the population two years ago be $P$. After 1st year: $P \times (1 + \frac{15}{100}) = P \times 1.15$ After 2nd year: $(P \times 1.15) \times (1 - \frac{10}{100}) = 20,700$ $P \times 1.15 \times 0.9 = 20,700$ $P \times 1.035 = 20,700$ $P = \frac{20,700}{1.035} = 20,000$.

Question 2

2. The price of sugar rises by $25\%$. By what percentage should a household reduce its consumption of sugar so that their expenditure on sugar increases by only $10\%$?

A) $15\%$

B) $10\%$

C) $12\%$

D) $18\%$

Answer

Correct Answer: C

Solution: Let initial Price be $100$ and initial Consumption be $100$. Initial Expenditure = $10,000$. New Price = $125$. Target Expenditure = $10,000 \times 1.10 = 11,000$. New Consumption = \frac{11,000}{125} = 88. Reduction in consumption = $100 - 88 = 12\%$.

Question 3

3. In an election between two candidates, $10\%$ of the voters did not cast their votes. $10\%$ of the votes cast were found invalid. The winning candidate got $54\%$ of the valid votes and won by a majority of $1,620$ votes. Find the total number of voters enrolled?

A) $25,000$

B) $20,000$

Question 4

4. A man spends $75\%$ of his income. His income increases by $20\%$ and his expenditure increases by $10\%$. By what percentage do his savings increase?

A) $30\%$

B) $40\%$

C) $25\%$

D) $50\%$

Answer

Correct Answer: D

Solution: Let Income = $100$, Expenditure = $75$, Savings = $25$. New Income = $120$. New Expenditure = $75 \times 1.10 = 82.5$. New Savings = $120 - 82.5 = 37.5$. Increase in savings = $37.5 - 25 = 12.5$. $\%$ increase = $(\frac{12.5}{25}) \times 100 = 50\%$.

Question 5

5. In an examination, a candidate scoring $30\%$ marks fails by $15$ marks, while another candidate who scores $40\%$ marks gets $35$ marks more than the minimum pass marks. What is the passing percentage?

A) $35\%$

B) $42\%$

C) $33\%$

D) $38\%$

Answer

Correct Answer: C

Solution: Let total marks be $M$. $0.30M + 15 = 0.40M - 35$ $0.10M = 50 \Rightarrow M = 500$. Pass marks = $0.30(500) + 15 = 150 + 15 = 165$. Pass percentage = $(\frac{165}{500}) \times 100 = 33\%$.

Question 6

6. A vessel contains $40$ liters of a mixture of milk and water, containing $10\%$ water. How much water should be added to make it $20\%$ water in the new mixture?

A) $5$ liters

B) $4$ liters

C) $8$ liters

D) $10$ liters

Answer

Correct Answer: A

Solution: Initial water = $10\%$ of $40 = 4$ liters. Milk = $36$ liters. In the new mixture, milk ($36$ liters) will be $80\%$ (since water is $20\%$). $80\%$ of New Total = $36$. New Total = \frac{36}{0.8} = 45$ liters. Water added = $45 - 40 = 5$ liters.

Question 7

7. The sales tax on an item is reduced by $20\%$, and its consumption increases by $15\%$. What is the overall percentage effect on the revenue collected?

A) $5\%$ increase

B) $8\%$ decrease

C) $10\%$ decrease

D) $4\%$ increase

Answer

Correct Answer: B

Solution: Revenue = Tax Rate $\times$ Consumption. Let initial Revenue = $100 \times 100 = 10,000$. New Tax Rate = $80$. New Consumption = $115$. New Revenue = $80 \times 115 = 9,200$. Decrease = $10,000 - 9,200 = 800$. $\%$ change = $(\frac{-800}{10,000}) \times 100 = -8\%$.

Question 8

8. If $A$ is $20\%$ more than $B$, and $B$ is $25\%$ less than $C$, then by what percentage is $A$ less than $C$?

A) $5\%$

B) $12\%$

C) $10\%$

D) $15\%$

Answer

Correct Answer: C

Solution: Let $C = 100$. $B = 100 - 25\% = 75$. $A = 75 + 20\%$ of $75 = 75 + 15 = 90$. Difference between $C$ and $A = 100 - 90 = 10$. $A$ is $10\%$ less than $C$.

Question 9

9. In a survey, it was found that $60\%$ of people like Tea, $50\%$ like Coffee, and $10\%$ like neither. What percentage of people like both?

A) $10\%$

B) $30\%$

C) $15\%$

D) $20\%$

Answer

Correct Answer: D

Solution: Total people = $100\%$. People who like at least one = $100\% - 10\% = 90\%$. Using $n(A \cup B) = n(A) + n(B) - n(A \cap B)$: $90\% = 60\% + 50\% - n(\text{Both})$ $n(\text{Both}) = 110\% - 90\% = 20\%$.

Question 10

10. If the length of a rectangle is increased by $40\%$ and the breadth is decreased by $20\%$, find the percentage change in its area.

A) $10\%$ increase

B) $12\%$ increase

C) $20\%$ increase

D) $8\%$ decrease

Answer

Correct Answer: B

Solution: Area = $L \times B$. Using successive percentage formula: $x + y + \frac{xy}{100}$ $x = +40$, $y = -20$ Change = $40 - 20 + \frac{(40)(-20)}{100}$ Change = $20 - 8 = +12\%$. Area increases by $12\%$.

Accepted Answer

Correct Answer: A

Solution: Let total voters = $x$. Votes cast = $0.9x$. Valid votes = $0.9 \times 0.9x = 0.81x$. Winner got $54\%$ of valid votes, so Loser got $46\%$. Difference = $(54 - 46)\%$ of $0.81x = 8\%$ of $0.81x$. $0.08 \times 0.81x = 1,620$ $0.0648x = 1,620 \Rightarrow x = \frac{1,620}{0.0648} = 25,000$.

Competitive Exams : Quantitative Aptitude - Percentage (Set 2) | Concept of Maths