Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Percentage Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- The present population of a village is 4550. If the male and female population increases by 10% and 15% successively and the population will become 5650 then find the present population of males in that village?
- 5229
- 5231
- 5246
- 5253

Answer: (a) 5229

Solution: Let the number of male present population in that village is x.

Then the number of female present population in the village = \((4550 - x)\)

\(110 \ \% \ of \ x + 15 \ \% \ of \ (4550 - x)\) = \(5650\)

\(\frac{110x}{100} + \frac{15}{100} \times (4550 - x)\) = \(5650\)

\(110x + 15 \times (4550 - x)\) = \(5650 \times 100\)

\(110x + 68250 - 15x = 565,000\)

\(95x = 496,750\)

\(x = 5228.9\)

\(\approx 5229\)

Hence the male present population in the village is 5229.

Solution: Let the number of male present population in that village is x.

Then the number of female present population in the village = \((4550 - x)\)

\(110 \ \% \ of \ x + 15 \ \% \ of \ (4550 - x)\) = \(5650\)

\(\frac{110x}{100} + \frac{15}{100} \times (4550 - x)\) = \(5650\)

\(110x + 15 \times (4550 - x)\) = \(5650 \times 100\)

\(110x + 68250 - 15x = 565,000\)

\(95x = 496,750\)

\(x = 5228.9\)

\(\approx 5229\)

Hence the male present population in the village is 5229.

- A woman spent $2500 on festive shopping, $2000 on buying a refrigerator, and the remaining 40% of the total amount saved for other expenditures. What was the total amount she had?
- $7200
- $7500
- $7600
- $7700

Answer: (b) $7500

Solution: Let the total amount was $x then.$$ (100 - 40) \ \% \ of \ x = 2500 + 2000 $$ $$ 60 \ \% \ of \ x = 4500 $$ $$ \frac{60x}{100} = 4500 $$ $$ x = \frac{4500 \times 100}{60} $$ $$ x = 7500 $$ Hence the total amount was $7500.

Solution: Let the total amount was $x then.$$ (100 - 40) \ \% \ of \ x = 2500 + 2000 $$ $$ 60 \ \% \ of \ x = 4500 $$ $$ \frac{60x}{100} = 4500 $$ $$ x = \frac{4500 \times 100}{60} $$ $$ x = 7500 $$ Hence the total amount was $7500.

- The monthly income of a person was $25,000 and his monthly expenditure was $18,000. If his income increased by 20% and his expenditure increased by 15% after one year then find the percentage increase in his savings after one year?
- 28.21 %
- 29.33 %
- 30.56 %
- 32.85 %

Answer: (d) 32.85 %

Solution: The amount saved by the person in the first year = \((25000 - 18000)\) = \(\$ \ 7000\)

Increased income after one year = \(120 \ \% \ of \ 25000\)

= \(\left[\frac{120}{100} \times 25000\right]\)

= \(30,000\)

Increased expenditure after one year = \(115 \ \% \ of \ 18000\)

= \(\left[\frac{115}{100} \times 18000\right]\)

= \(\$ \ 20,700\)

Saved amount after one year = \((30,000 - 20,700)\) = \(\$ \ 9300\)

Increased saving amount after one year = \((9300 - 7000)\) = \(\$ \ 2300\)

Increased savings after one year in percentage = \(\left[\frac{2300}{7000} \times 100\right]\) = \(32.85 \ \%\)

Solution: The amount saved by the person in the first year = \((25000 - 18000)\) = \(\$ \ 7000\)

Increased income after one year = \(120 \ \% \ of \ 25000\)

= \(\left[\frac{120}{100} \times 25000\right]\)

= \(30,000\)

Increased expenditure after one year = \(115 \ \% \ of \ 18000\)

= \(\left[\frac{115}{100} \times 18000\right]\)

= \(\$ \ 20,700\)

Saved amount after one year = \((30,000 - 20,700)\) = \(\$ \ 9300\)

Increased saving amount after one year = \((9300 - 7000)\) = \(\$ \ 2300\)

Increased savings after one year in percentage = \(\left[\frac{2300}{7000} \times 100\right]\) = \(32.85 \ \%\)

- Ram got 25% of the maximum marks in an exam and failed by 15 marks. Krishna who got 35% in the same exam, got 10 marks more than passing marks. What were the passing marks?
- 74.5 Marks
- 77.5 Marks
- 79.5 Marks
- 80.5 Marks

Answer: (b) 77.5 Marks

Solution: Let the maximum marks was \(x\) then.$$ (25 \ \% \ of \ x) + 15 = (35 \ \% \ of \ x) - 10 $$ $$ \frac{25x}{100} + 15 = \frac{35x}{100} - 10 $$ $$ 15 + 10 = \frac{35x}{100} - \frac{25x}{100} $$ $$ 25 = \frac{10x}{100} $$ $$ x = 250 $$ Hence the minimum passing marks. $$ = (25 \ \% \ of \ 250) + 15 $$ $$ = \frac{25}{100} \times 250 + 15 $$ $$ = 77.5 \ Marks $$

Solution: Let the maximum marks was \(x\) then.$$ (25 \ \% \ of \ x) + 15 = (35 \ \% \ of \ x) - 10 $$ $$ \frac{25x}{100} + 15 = \frac{35x}{100} - 10 $$ $$ 15 + 10 = \frac{35x}{100} - \frac{25x}{100} $$ $$ 25 = \frac{10x}{100} $$ $$ x = 250 $$ Hence the minimum passing marks. $$ = (25 \ \% \ of \ 250) + 15 $$ $$ = \frac{25}{100} \times 250 + 15 $$ $$ = 77.5 \ Marks $$

- In the U.S. Presidential election between two candidates, 80% of voters cast their votes, out of which 5% of the votes were declared invalid. A candidate got 12,550 votes which were 70% of the total valid votes. Find the total number of votes enrolled for the election?
- 23,410
- 23,430
- 23,590
- 23,660

Answer: (c) 23,590

Solution: Let the total number of votes enrolled for the election was \(x\), then

Number of votes cast = \(80 \ \% \ x\) = \(\frac{80x}{100}\)

Valid electoral votes = \(95 \ \% \ of \ \frac{80x}{100}\)

A candidate got 70% of the total valid votes.

\(70 \ \% \ of \ \left[95 \ \% \ of \ \frac{80x}{100}\right]\) = 12550

\(\frac{70}{100} \times \frac{95}{100} \times \frac{80x}{100}\) = 12550

\(\frac{70 \times 95 \times 80x}{100 \times 100 \times 100}\) = 12550

\(x = 23590.23\)

\(\approx 23590\)

Hence 23,590 votes enrolled for the election.

Solution: Let the total number of votes enrolled for the election was \(x\), then

Number of votes cast = \(80 \ \% \ x\) = \(\frac{80x}{100}\)

Valid electoral votes = \(95 \ \% \ of \ \frac{80x}{100}\)

A candidate got 70% of the total valid votes.

\(70 \ \% \ of \ \left[95 \ \% \ of \ \frac{80x}{100}\right]\) = 12550

\(\frac{70}{100} \times \frac{95}{100} \times \frac{80x}{100}\) = 12550

\(\frac{70 \times 95 \times 80x}{100 \times 100 \times 100}\) = 12550

\(x = 23590.23\)

\(\approx 23590\)

Hence 23,590 votes enrolled for the election.

Lec 1: Introduction to Percentage
Exercise-1
Lec 2: Percentage Case-1
Exercise-2
Lec 2: Percentage Case-2
Exercise-3
Lec 2: Percentage Case-3 & Case-4
Exercise-4
Exercise-5
Exercise-6
Exercise-7
Exercise-8
Exercise-9
Exercise-10