Percentages Questions with Answers | Quantitative Reasoning for GMAT PDF Download

Q.1. If the price of petrol increases by 25% and Raj intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
(a) 10%
(b) 12%
(c) 8%
(d) 6.67%
(e) 12.5%
Ans: (c)
Sol: Let the original price per litre be Rs. x and the original quantity be y litres. His original spending = x × y.
New price = 1.25x. He is willing to spend 1.15 × (x y) = 1.15xy.
So 1.25x × q = 1.15xy ⇒ q = (1.15xy)/(1.25x) = (1.15/1.25) y = 0.92y.
New quantity is 0.92y, a reduction of y - 0.92y = 0.08y which is 8% of the original. Hence the reduction is 8%

Q.2. A shepherd has 1 million sheeps at the beginning of Year 2000. The numbers grow by x% (x > 0) during the year. A famine hits his village in the next year and many of his sheeps die. The sheep population decreases by y% during 2001 and at the beginning of 2002 the shepherd finds that he is left with 1 million sheeps. Which of the following is correct?
(a) x > y
(b) y > x
(c) x = y
(d) Cannot be determined
Ans: (a)
Sol: Start with 1,000,000 sheep. After growth by x% the population becomes 1,000,000 × (1 + x/100). After a decrease by y% it becomes 1,000,000 × (1 + x/100) × (1 - y/100). This final value equals 1,000,000 by the problem statement, so
(1 + x/100)(1 - y/100) = 1.
Expand: 1 + x/100 - y/100 - (xy/10,000) = 1 ⇒ x/100 - y/100 - (xy/10,000) = 0 ⇒ x - y = (xy/100).
Since x > 0 and y ≥ 0, the right-hand side (xy/100) is non-negative, so x - y ≥ 0. Equality (x = y) would force xy = 0, which cannot hold because x > 0. Therefore x > y. (An illustrative numeric case in the original solution also confirms this.)
Q.3. In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes and there are no invalid votes, by how many votes did it lose the election?
(a) 300,000
(b) 168,000
(c) 36,000
(d) 24,000
Ans: (c)
Sol: Let Party D have x% of votes and Party R have (x - 12)%. Since only two parties: x + (x - 12) = 100 ⇒ 2x = 112 ⇒ x = 56%.
So Party R has 44% of total votes. Given 44% = 132,000, total votes T = 132,000 × 100/44 = 300,000.
Margin = 12% of T = 0.12 × 300,000 = 36,000 votes.
Q.4. A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks. Find the maximum marks.
(a) 50
(b) 100
(c) 150
(d) 200
Ans: (b)
Sol: Let maximum marks = x and passing percentage = p%. Given 42% is 12% more than passing, so p = 42% - 12% = 30%.
The difference between pass marks and 20% equals 10 marks: (30% - 20%) of x = 10 ⇒ 10% of x = 10 ⇒ x = 100. Thus maximum marks = 100.
Q.5. When processing flower - nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower - nectar must be processed to yield 1 kg of honey, if nectar contains 50% water, and the honey obtained from this nectar contains 15% water?
(a) 1.5 kgs
(b) 1.7 kgs
(c) 3.33 kgs
(d) None of these
Ans: (b)
Sol: Non-water (solids) fraction in nectar = 50% = 0.50. Solids fraction in honey = 100% - 15% = 85% = 0.85.
To obtain 1 kg honey, solids required = 0.85 × 1 kg = 0.85 kg. Required nectar amount N satisfies 0.50 × N = 0.85 ⇒ N = 0.85/0.50 = 1.7 kg.
Q.6. A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?
(a) 17
(b) 23
(c) 77
(d) None of these
Ans: (b)
Sol: Start with 100 apples. After selling 60%, remaining = 40 apples.
He throws away 15% of this remainder = 0.15 × 40 = 6. Remaining now = 34.
Next day he sells 50% of 34 = 17 and throws away the rest (the thrown away that day) = 17.
Total thrown away = 6 + 17 = 23 apples = 23% of the original 100.
Q.7. If the cost price of 20 articles is equal to the selling price of 16 articles, what is the percentage profit or loss made by the merchant?
(a) 20% Profit
(b) 25% Loss
(c) 25% Profit
(d) 33.33% Loss
Ans: (c)
Sol: Let cost price (CP) of one article = Re.1 ⇒ CP of 20 articles = Rs.20.
Selling price (SP) of 16 articles = Rs.20 ⇒ SP of one article = 20/16 = Rs.1.25.
SP of 20 articles = 20 × 1.25 = Rs.25. Profit = 25 - 20 = Rs.5.
Percentage profit = (5/20) × 100 = 25% profit.
Q.8. 30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
(a) 15%
(b) 20%
(c) 80%
(d) 70%
Ans: (c)
Sol: Men above 50 = 100% - 80% = 20% of men. Football players above 50 = 20% of those = 0.20 × 20% = 4% of all men.
Total football players = 20% of all men. Therefore football players ≤ 50 years = 20% - 4% = 16% of all men.
Proportion of football players who are ≤ 50 = (16% / 20%) × 100 = 80%.
Q.9. If the price of petrol increases by 25%, by how much must a user cut down on his petrol consumption so that his expenditure on petrol remains unchanged?
(a) 25%
(b) 16.67%
(c) 20%
(d) 33.33%
Ans: (c)
Sol: Let original price be Rs.100 per litre and consumption 1 litre ⇒ spend = Rs.100.
New price = Rs.125. To keep spend at Rs.100, new quantity q satisfies 125 × q = 100 ⇒ q = 100/125 = 0.8 litre.
Reduction = 1 - 0.8 = 0.2 = 20%. (Shortcut: For a price increase of 25% = 1/4, the required reduction = 1/(1 + 4) = 1/5 = 20%.)
Q.10. Peter got 30% of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What was the passing marks in the examination?
(a) 35
(b) 250
(c) 75
(d) 85
Ans: (d)
Sol: Let maximum marks = x. Peter's score = 0.30x and he failed by 10, so passing marks = 0.30x + 10.
Paul's score = 0.40x and he scored 15 more than passing: 0.40x = (0.30x + 10) + 15 ⇒ 0.40x - 0.30x = 25 ⇒ 0.10x = 25 ⇒ x = 250.
Peter's marks = 0.30 × 250 = 75. Passing marks = 75 + 10 = 85.


Q.11. A report consists of 20 sheets each of 55 lines and each such line consists of 65 characters. This report is reduced onto sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in number of sheets is closest to:
(a) 20%
(b) 5%
(c) 30%
(d) 35%
Ans: (a)
Sol: Characters per original sheet = 55 × 65. Total characters = 20 × 55 × 65 = 71,500.
New characters per sheet = 65 × 70 = 4,550. Sheets required = 71,500 / 4,550 = 15.714 ⇒ must use 16 sheets.
Reduction = 20 - 16 = 4 sheets ⇒ percentage reduction = (4/20) × 100 = 20%.

Q.12. The number of votes not cast for the PNC Party increased by 25% in the National General Election over those not cast for it in the previous Assembly Polls, and the PNC Party lost by a majority twice as large as that by which it had won the Assembly Polls. If a total 2,60,000 people voted each time, how many voted for the PNC Party in the previous Assembly Polls?
(a) 1,10,000
(b) 1,50,000
(c) 1,40,000
(d) 1,20,000
Ans: (c)
Sol: Total votes = 2,60,000. Let x = votes against PNC in Assembly Polls, so votes for PNC then = 2,60,000 - x. Previous majority = (2,60,000 - x) - x = 2,60,000 - 2x.
In General Election votes against PNC = 1.25x, votes for PNC = 2,60,000 - 1.25x. Lost majority = 1.25x - (2,60,000 - 1.25x) = 2.5x - 2,60,000.
Given lost majority = 2 × previous majority ⇒ 2.5x - 2,60,000 = 2(2,60,000 - 2x) = 5,20,000 - 4x.
So 2.5x - 2,60,000 = 5,20,000 - 4x ⇒ 6.5x = 7,80,000 ⇒ x = 1,20,000.
Votes for PNC in previous Assembly Polls = 2,60,000 - 1,20,000 = 1,40,000.
Q.13. 2/5th of the voters promise to vote for A and the rest promised to vote for B. Of these, on the last day 15% of the voters went back of their promise to vote for A and 25% of voters went back of their promise to vote for B, and A lost by 200 votes. Then, the total number of voters is:
(a) 10000
(b) 11000
(c) 9000
(d) 9500
Ans: (a)
Sol: Let total voters = x. Promised for A = (2/5)x, for B = (3/5)x.
Those who actually vote for A = 85% of A's promises + 25% of B's promises = 0.85 × (2/5)x + 0.25 × (3/5)x = x[(1.7/5) + (0.75/5)] = 0.49x.
Those who actually vote for B = 75% of B's promises + 15% of A's promises = 0.75 × (3/5)x + 0.15 × (2/5)x = x[(2.25/5) + (0.3/5)] = 0.51x.
Difference B - A = 0.02x = 200 ⇒ x = 200 / 0.02 = 10,000 voters.



Q.14. A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance, he purchases oranges. Number of oranges he can purchase is:
(a) 36
(b) 40
(c) 15
(d) 20
Ans: (d)
Sol: Let price of one orange = Rs. x, so total money = 50x. Since 40 mangoes cost 50x, one mango costs 50x/40 = 1.25x.
Taxi fare = 10% of total = 5x. He buys 20 mangoes costing 20 × 1.25x = 25x.
Money left = 50x - 5x - 25x = 20x. At price x per orange he can buy 20 oranges.
Q.15. Forty per cent of the employees of a certain company are men and 75% of the men earn more than Rs. 25,000 per year. If 45% of the company's employees earn more than Rs. 25,000 per year, what fraction of the women employed by the company earn Rs. 25,000 or less per year?
(a) 2/11
(b) 1/4
(c) 1/3
(d) 3/4
Ans: (d)
Sol: Let total employees = x. Men = 0.40x, Women = 0.60x.
Men earning > Rs.25,000 = 75% of men = 0.75 × 0.40x = 0.30x. Total employees earning > Rs.25,000 = 0.45x.
Women earning > Rs.25,000 = 0.45x - 0.30x = 0.15x. Hence women earning ≤ Rs.25,000 = 0.60x - 0.15x = 0.45x.
Required fraction = (0.45x) / (0.60x) = 45/60 = 3/4.
Alternate Sol: Assume 40 men and 60 women. Men earning > Rs.25,000 = 75% of 40 = 30. Total earning > Rs.25,000 = 45% of 100 = 45, so women earning > Rs.25,000 = 45 - 30 = 15. Women earning ≤ Rs.25,000 = 60 - 15 = 45 ⇒ fraction = 45/60 = 3/4.