PERCENT AND PERCENTAGE - Q&A
EXERCISE 7(A)
1. Evaluate:
(i) 55% of 160 + 24% of 50 - 36% of 150
Solution:
= (55/100) × 160 + (24/100) × 50 - (36/100) × 150
= (0.55 × 160) + (0.24 × 50) - (0.36 × 150)
= 88 + 12 - 54
= 100 - 54
= 46
(ii) 9.3% of 500 - 4.8% of 250 - 2.5% of 240
Solution:
= (9.3/100) × 500 - (4.8/100) × 250 - (2.5/100) × 240
= (9.3 × 5) - (4.8 × 2.5) - (2.5 × 2.4)
= 46.5 - 12 - 6
= 46.5 - 18
= 28.5
2. (i) A bag contains 8 white and 12 black balls, the percentage of black balls in the bag is:
Solution:
Total balls = 8 (white) + 12 (black) = 20 balls.
Percentage of black balls = (Number of black balls / Total balls) × 100%
= (12 / 20) × 100%
= 12 × 5%
= 60%
3. (i) A number is increased from 125 to 150; find the percentage increase.
Solution:
Original number = 125
New number = 150
Increase = 150 - 125 = 25
Percentage Increase = (Increase / Original number) × 100%
= (25 / 125) × 100%
= (1 / 5) × 100%
= 20%
(ii) A number is decreased from 125 to 100; find the percentage decrease.
Solution:
Original number = 125
New number = 100
Decrease = 125 - 100 = 25
Percentage Decrease = (Decrease / Original number) × 100%
= (25 / 125) × 100%
= 20%
(iii) A number is first increased by 20%, then the resulting number is decreased by 20%. On the whole the original number is increased/decreased by:
Solution:
Let the original number be 100.
Increase by 20%: 100 + 20% of 100 = 100 + 20 = 120.
Now, decrease 120 by 20%: 120 - 20% of 120
= 120 - (20/100 × 120)
= 120 - 24 = 96.
Net change from 100 to 96 is a decrease of 4.
Percentage decrease = 4%.
Answer: (b) 4% decreased
4. Find:
(i) 45 is what percent of 54?
Solution:
Let 45 be x% of 54.
(x/100) × 54 = 45
x = (45 × 100) / 54
x = (5 × 100) / 6 (Dividing 45 and 54 by 9)
x = 500 / 6 = 250 / 3
x = 83 1/3%
(ii) 2.7 is what percent of 18?
Solution:
Required Percentage = (2.7 / 18) × 100%
= (27 / 180) × 100%
= (3 / 20) × 100%
= 3 × 5%
= 15%
5. Answer the following:
(i) 252 is 35% of a certain number, find the number.
Solution:
Let the number be x.
35% of x = 252
(35/100) × x = 252
x = (252 × 100) / 35
x = 7.2 × 100
x = 720
(ii) If 14% of a number is 315; find the number.
Solution:
Let the number be y.
(14/100) × y = 315
y = (315 × 100) / 14
y = 22.5 × 100
y = 2250
(iii) After paying 20% of the income, a man is left with 160; then the income of the man is:
Solution:
If he pays 20%, he is left with (100 - 20) = 80% of his income.
80% of Income = 160
(80/100) × Income = 160
Income = (160 × 100) / 80
Income = 2 × 100 = 200.
Answer: (c) 200
(iv) If A + B + C = 400 in which A is 40% and B is 45%, then the exact quantity of C in the whole is:
Solution:
Percentage of A + Percentage of B = 40% + 45% = 85%.
Therefore, Percentage of C = 100% - 85% = 15%.
Quantity of C = 15% of 400
= (15/100) × 400
= 15 × 4 = 60.
Answer: (b) 60
(v) A number is decreased by 20%. If the resulting number is 800, the original number is:
Solution:
Let the original number be x.
Decreased number = (100 - 20)% of x = 80% of x.
(80/100) × x = 800
x = (800 × 100) / 80
x = 1000.
Answer: (c) 1000
6. Find the percentage change, when a number is changed from:
(i) 80 to 100
Solution:
Initial value = 80, Final value = 100.
Increase = 100 - 80 = 20.
Percentage Increase = (Increase / Initial value) × 100%
= (20 / 80) × 100% = (1/4) × 100% = 25%.
(ii) 100 to 80
Solution:
Initial value = 100, Final value = 80.
Decrease = 100 - 80 = 20.
Percentage Decrease = (Decrease / Initial value) × 100%
= (20 / 100) × 100% = 20%.
(iii) 6.25 to 7.50
Solution:
Increase = 7.50 - 6.25 = 1.25.
Percentage Increase = (1.25 / 6.25) × 100%
= (125 / 625) × 100%
= (1/5) × 100% = 20%.
7. An auctioneer charges 8% for selling a house. If the house is sold for 2,30,500. Find the charges of the auctioneer.
Solution:
Selling price = 2,30,500.
Charges = 8% of 2,30,500
= (8/100) × 2,30,500
= 8 × 2,305
= 18,440.
8. Out of 800 oranges, 50 are found rotten. Find the percentage of good oranges.
Solution:
Total oranges = 800.
Rotten oranges = 50.
Good oranges = 800 - 50 = 750.
Percentage of good oranges = (750 / 800) × 100%
= 750 / 8
= 93.75% (or 93 3/4%)
9. A cistern contains 5 thousand litres of water. If 6% water is leaked, find how many litres of water would be left in the cistern.
Solution:
Total water = 5000 litres.
Water leaked = 6% of 5000 = (6/100) × 5000 = 300 litres.
Water left = 5000 - 300 = 4700 litres.
10. A man spends 87% of his salary. If he saves 325; find his salary.
Solution:
Percentage spent = 87%.
Percentage saved = 100% - 87% = 13%.
Given savings = 325.
Let Salary be x.
13% of x = 325
(13/100) × x = 325
x = (325 × 100) / 13
x = 25 × 100 = 2,500.
11. (i) A number 3.625 is wrongly read as 3.265; find the percentage error.
Solution:
Original number = 3.625
Wrong number = 3.265
Error = 3.625 - 3.265 = 0.360
Percentage Error = (Error / Original number) × 100%
= (0.360 / 3.625) × 100%
= (360 / 3625) × 100%
= 36000 / 3625
= 9.93% (approx)
(ii) A number 5.78 × 103 is wrongly written as 5.87 × 103 find the percentage error.
Solution:
Original number = 5.78 × 1000 = 5780
Wrong number = 5.87 × 1000 = 5870
Error = 5870 - 5780 = 90
Percentage Error = (90 / 5780) × 100%
= 900 / 578
= 1.56% (approx)
12. In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18,336 votes. Find the total number of votes polled and the votes secured by each candidate.
Solution:
Winner secured = 58%.
Loser secured = 100% - 58% = 42%.
Difference in vote percentage = 58% - 42% = 16%.
Given difference = 18,336 votes.
Let total votes be x.
16% of x = 18,336
x = (18,336 × 100) / 16 = 1,14,600.
Votes secured by Winner = 58% of 1,14,600
= 0.58 × 1,14,600 = 66,468.
Votes secured by Loser = 42% of 1,14,600
= 0.42 × 1,14,600 = 48,132.
13. In an election between two candidates, one candidate secured 47% of votes polled and lost the election by 12,366 votes. Find the total votes polled and the votes secured by the winning candidate.
Solution:
Losing candidate secured = 47%.
Winning candidate secured = 100% - 47% = 53%.
Margin of loss (Difference) = 53% - 47% = 6%.
Given margin = 12,366 votes.
6% of Total Votes = 12,366
Total Votes = (12,366 × 100) / 6 = 2,06,100.
Votes for Winning Candidate = 53% of 2,06,100
= (53/100) × 2,06,100
= 53 × 2,061 = 1,09,233.
14. The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is 8,000, find its cost:
(i) after one year
Solution:
Depreciation = 15% of 8,000 = (15/100) × 8000 = 1,200.
Cost after 1 year = 8,000 - 1,200 = 6,800.
(ii) after 2 years.
Solution:
Value at start of 2nd year = 6,800.
Depreciation in 2nd year = 15% of 6,800 = (15/100) × 6,800 = 1,020.
Cost after 2 years = 6,800 - 1,020 = 5,780.
15. In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks; find the maximum marks.
Solution:
Marks obtained = 65.
Marks needed to pass = 65 + 3 = 68.
Pass percentage = 40%.
Let Maximum marks be M.
40% of M = 68
(40/100) × M = 68
M = (68 × 100) / 40 = 680 / 4 = 170.
16. In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35%, find the maximum marks.
Solution:
Marks needed to pass = 125 + 15 = 140.
Pass percentage = 35%.
Let Maximum marks be x.
35% of x = 140
x = (140 × 100) / 35
x = 4 × 100 = 400.
17. In an objective type paper of 150 questions, John got 80% correct answers and Mohan got 64% correct answers.
(i) How many correct answers did each get?
Solution:
Total questions = 150.
John's correct answers = 80% of 150 = (80/100) × 150 = 120.
Mohan's correct answers = 64% of 150 = (64/100) × 150 = 96.
(ii) What percent is Mohan's correct answers to John's correct answers?
Solution:
Percentage = (Mohan's score / John's score) × 100%
= (96 / 120) × 100%
= (4 / 5) × 100%
= 80%.
18. The number 8,000 is first increased by 20% and then decreased by 20%. Find the resulting number.
Solution:
First Increase: 20% of 8,000 = 1,600.
Number becomes 8,000 + 1,600 = 9,600.
Then Decrease: 20% of 9,600 = (20/100) × 9,600 = 1,920.
Resulting number = 9,600 - 1,920 = 7,680.
19. The number 12,000 is first decreased by 25% and then increased by 25%. Find the resulting number.
Solution:
First Decrease: 25% of 12,000 = 3,000.
Number becomes 12,000 - 3,000 = 9,000.
Then Increase: 25% of 9,000 = (25/100) × 9,000 = 2,250.
Resulting number = 9,000 + 2,250 = 11,250.
20. The cost of an article is first increased by 20% and then decreased by 30%, find the percentage change in the cost of the article.
Solution:
Let original cost = 100.
After 20% increase = 120.
After 30% decrease on 120 = 30% of 120 = 36.
New Cost = 120 - 36 = 84.
Change = 100 - 84 = 16 (Decrease).
Percentage Change = 16% Decrease.
21. The cost of an article is first decreased by 25% and then further decreased by 40%. Find the percentage change in the cost of the article.
Solution:
Let original cost = 100.
After 25% decrease = 75.
After further 40% decrease on 75 = 40% of 75 = (40/100) × 75 = 30.
New Cost = 75 - 30 = 45.
Total decrease = 100 - 45 = 55.
Percentage Change = 55% Decrease.
EXERCISE 7(B)
1. Multiple Choice Type: Choose the correct answer from the options given below.
(i) Out of two students A and B, A does 10 questions and B does 30 questions in the same time. The percentage of number of questions done by B to the number of questions done by A is:
Solution:
(Questions by B / Questions by A) × 100%
= (30 / 10) × 100% = 300%.
Answer: (b) 300%
(ii) In an election, there are only two candidates A and B. A gets 60% of the total votes polled and wins the election by 960 votes. What is the number of total votes polled?
Solution:
A gets 60%, so B gets 40%.
Difference = 20%.
20% of Total = 960.
Total = (960 × 100) / 20 = 4800.
Answer: (d) 4800
(iii) If A is 20% less than B, then B is:
Solution:
Let B = 100. Then A = 80.
B is 20 more than A.
Percentage = (20 / A) × 100% = (20/80) × 100% = 25%.
So B is 25% more than A.
Answer: (d) 25% more than A
(iv) A student has to obtain 35% of the total marks to pass. He got 25% of the total marks and failed by 80 marks. The total of marks is:
Solution:
Difference in % = 35% - 25% = 10%.
10% of Total = 80 marks.
Total = 800.
Answer: (b) 800
(v) A mixture of milk and water contains 4 parts of milk and 1 part of water. The percentage of milk in the mixture is:
Solution:
Total parts = 4 + 1 = 5.
Milk % = (4/5) × 100% = 80%.
Answer: (d) 80%
2. A man bought a certain number of oranges; out of which 13 percent were found rotten. He gave 75% of the remaining in charity and still had 522 oranges left. Find how many oranges had he bought?
Solution:
Let total oranges = x.
Rotten = 13%. Good = 87%.
Remaining after rotten = 0.87x.
Given 75% of remaining to charity, so 25% of remaining is left.
25% of (87% of x) = 522.
(25/100) × (87/100) × x = 522
(1/4) × (87/100) × x = 522
x = (522 × 400) / 87
x = 6 × 400 = 2400.
He bought 2400 oranges.
3. 5% pupil in a town died due to some disease and 3% of the remaining left the town. If 2,76,450 pupil are still in the town, find the original number of pupil in the town.
Solution:
Let original number = x.
After 5% died, remaining = 95% of x.
After 3% of remaining left, final remaining = 97% of (95% of x).
(97/100) × (95/100) × x = 2,76,450
x = (2,76,450 × 100 × 100) / (97 × 95)
2,76,450 / 97 = 2850.
2850 / 95 = 30.
x = 30 × 100 × 100 = 3,00,000.
4. In a combined test in English and Physics; 36% candidates failed in English; 28% failed in Physics and 12% in both; find:
(i) the percentage of passed candidates.
Solution:
Total Failed % = n(E) + n(P) - n(E ∩ P)
= 36% + 28% - 12% = 52%.
Passed % = 100% - 52% = 48%.
(ii) the total number of candidates appeared, if 208 candidates have failed.
Solution:
52% of Total = 208.
Total = (208 × 100) / 52 = 400 candidates.
5. In a combined test in Maths and Chemistry, 84% candidates passed in Maths, 76% in Chemistry and 8% failed in both. Find:
(i) the percentage of failed candidates.
Solution:
Failed in Maths = 100 - 84 = 16%.
Failed in Chemistry = 100 - 76 = 24%.
Failed in Both = 8%.
Total Failed = 16% + 24% - 8% = 32%.
(ii) if 340 candidates passed in the test, then, how many candidates had appeared in the test?
Solution:
Percentage Passed = 100% - 32% (Total Failed) = 68%.
68% of Total = 340.
Total = (340 × 100) / 68 = 500 candidates.
6. A's income is 25% more than B's. Find out by how much percent is B's income less than A's?
Solution:
Let B's income = 100. A's income = 125.
B is less than A by 25.
Percentage less = (25 / A's income) × 100%
= (25 / 125) × 100% = 20%.
7. Mona is 20% younger than Neetu. By how much percent is Neetu older than Mona?
Solution:
Let Neetu's age = 100. Mona's age = 80.
Neetu is older by 20.
Percentage older = (20 / Mona's age) × 100%
= (20 / 80) × 100% = 25%.
8. If the price of sugar is increased by 25% today, by what percent should it be decreased tomorrow to bring the price back to the original?
Solution:
Let original price = 100. New price = 125.
To bring back to 100, decrease by 25.
Percentage decrease on New Price = (25 / 125) × 100% = 20%.
9. A number increased by 15% becomes 391. Find the number.
Solution:
Let number be x.
115% of x = 391.
x = (391 × 100) / 115 = 340.
10. A number decreased by 23% becomes 539. Find the number.
Solution:
Let number be x.
(100 - 23)% of x = 539 => 77% of x = 539.
x = (539 × 100) / 77 = 700.
11. Two numbers are respectively 20 percent and 50 percent more than a third number. What percent of the first number is the second number?
Solution:
Let 3rd number = 100.
1st number = 120.
2nd number = 150.
Percentage = (2nd / 1st) × 100% = (150 / 120) × 100% = 125%.
12. Two numbers are respectively 20 percent and 50 percent of a third number. What percent of the first number is the second number?
Solution:
Let 3rd number = 100.
1st number = 20.
2nd number = 50.
Percentage = (50 / 20) × 100% = 250%.
13. Two numbers are respectively 30 percent and 40 percent less than a third number. What percent of the first number is the second number?
Solution:
Let 3rd number = 100.
1st number = 70.
2nd number = 60.
Percentage = (60 / 70) × 100% = 600/7 % = 85 5/7%.
14. Mohan gets 1,350 from Geeta and 650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit, what percent does he get from Rohit?
Solution:
Total money = 1350 + 650 = 2000.
Percentage from Rohit = (650 / 2000) × 100%
= 65 / 2 = 32.5%.
15. The monthly income of a man is 16,000. 15 percent of it is paid as income-tax and 75% of the remainder is spent on rent, food, clothing, etc. How much money is still left with the man?
Solution:
Income = 16,000.
Tax = 15% of 16,000 = 2,400.
Remainder = 16,000 - 2,400 = 13,600.
Spent 75% of Remainder, so Left = 25% of Remainder.
Money Left = 25% of 13,600 = (1/4) × 13,600 = 3,400.
16. During 2003, the production of a factory decreased by 25%. But during 2004, it (production) increased by 40% of what it was at the beginning of 2004. Calculate the resulting change (increase or decrease) in production during these two years.
Solution:
Let initial production (start of 2003) = 100.
End of 2003 = 100 - 25 = 75.
Start of 2004 = 75.
During 2004, Increase = 40% of 75 = (40/100) × 75 = 30.
End of 2004 = 75 + 30 = 105.
Net change = 105 - 100 = 5 Increase.
Resulting change = 5% Increase.
17. Last year, oranges were available at 24 per dozen; but this year, they are available at 50 per score. Find the percentage change in the price of oranges. (1 score = 20)
Solution:
Last year: Price of 12 oranges = 24 => Price of 1 orange = 2.
This year: Price of 20 oranges (1 score) = 50 => Price of 1 orange = 50/20 = 2.5.
Increase = 2.5 - 2 = 0.5.
Percentage Increase = (0.5 / 2) × 100% = 25%.
18. (i) Increase 180 by 25%.
Solution:
New value = 180 × (1 + 25/100) = 180 × 1.25 = 225.
(ii) Decrease 140 by 18%.
Solution:
New value = 140 × (1 - 18/100) = 140 × 0.82 = 114.8.
19.(i) A number when increased by 23% becomes 861; find the number.
Solution:
Let number be x. 123% of x = 861.
x = (861 × 100) / 123 = 700.
(ii) A number when decreased by 16% becomes 798; find the number.
Solution:
Let number be x. 84% of x = 798.
x = (798 × 100) / 84 = 950.
20. The price of sugar is increased by 20%. By what percent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same?
Solution:
Let original price = 100. New price = 120.
To keep expenditure 100, consumption must reduce by 20 (from 120 down to 100 cost worth).
Percentage Decrease = (20 / 120) × 100% = 100/6 % = 16 2/3%.
Test yourself
1. Multiple Choice Type: Choose the correct answer.
(i) A number, whose 4% is 6, is:
4% of x = 6 => x = 600/4 = 150.
Answer: (c) 150
(ii) What percent of 50 is 10?
(10/50) × 100% = 20%.
Answer: (a) 20%
(iii) 18 hours as a percentage of 3 days is:
3 days = 3 × 24 hours.
Percentage = (18 / (3 × 24)) × 100%.
Answer: (c) (18 / 3×24) × 100%
(iv) An alloy contains 30% of copper, 30% of zinc and rest nickel. The amount of nickel in 400 gm of alloy is:
Nickel % = 100 - (30+30) = 40%.
Amount = 40% of 400gm.
Answer: (a) 40% of 400 gm
(v) A number is first decreased by 40% and then increased by 40%. The equivalent change is:
Use formula x + y + (xy/100). -40 + 40 + (-1600/100) = -16%.
Answer: (c) 16% decrease
(vi) Out of 700 eggs, 20% are rotten. The number of good eggs is:
Good = 80%. 80% of 700 = 560.
Answer: (b) 560
(vii) 80% of 200 - 50 is equal to:
160 - 50 = 110.
Answer: (b) 110
(viii) A number 80 is wrongly taken as 100. The percentage error is:
Error = 20. Base (Original) = 80.
(20/80) × 100% = 25%.
Answer: (b) 25%
(ix) The price of an article was 680 last year. This year its price is 816. The percentage change in the price is:
Change = 816 - 680. Base = 680.
Answer: (c) (816-680)/680 × 100% increases
(x) Statement 1: To change a number in percentage to a ratio, write it as fraction with denominator 100 and then reduce it to the lowest terms if possible. Statement 2: A percentage can be converted to a fraction by removing sign of % dividing by 100.
Both are standard definitions.
Answer: (a) Both the statements are true.
(xi) Assertion (A): 9% of √0.0169 is 0.0117. Reason (R): To find a percentage of a quantity, we change the percentage to a fraction or a decimal and multiply it by the quantity.
√0.0169 = 0.13. 9% of 0.13 = 0.09 × 0.13 = 0.0117. True.
Answer: (a) (1) Both true and R explains A
(xii) Assertion (A): If we decrease 120 by 12 1/2% then decreased amount = 105. Reason (R): Percentage change = (Actual change / Original quantity) × 100%.
12.5% of 120 = (1/8) × 120 = 15. 120 - 15 = 105. True.
Answer: (b) (2) Both true but R is just formula, doesn't explain calculation step specifically? Or (a)? Usually R provides the formula used. Let's select (a).
(xiii) Assertion (A): By increasing 320 by 20%, we obtain the increased amount = 348. Reason (R): To increase a quantity by a percentage...
20% of 320 = 64. 320 + 64 = 384. Assertion is False.
Answer: (d) (4) A is false, R is true
(xiv) Assertion (A): The sum of two numbers is 28/25 of the first number. Then the second number is 12% of the first number. Reason (R): To express one quantity as a percentage of the other...
x + y = (28/25)x = 1.12x. So y = 0.12x = 12% of x. True.
Answer: (a) (1)
2. A family spends 30% of its income on house rent and 60% of the rest on house hold expenses. If the total savings of the family is 12,600 per month, find the total monthly income of the family.
Solution:
Let income = 100.
Rent = 30. Remainder = 70.
Expenses = 60% of 70 = 42.
Savings = 70 - 42 = 28.
28% of Income = 12,600.
Income = (12,600 × 100) / 28 = 45,000.
3. Geeta saves 20% of her monthly salary and saves 43,500 per month. Find her monthly expenditure.
Solution:
Savings = 20% of Salary = 43,500.
Expenditure = 80% of Salary.
Expenditure = 4 × Savings = 4 × 43,500 = 1,74,000.
4. In an examination, 92% of the candidates passed and 96 failed. Find the number of canditates who appeared for this exam.
Solution:
Fail percentage = 100% - 92% = 8%.
8% of Total = 96.
Total = (96 × 100) / 8 = 1,200.
5. A number is increased by 30% and then this increased number is decreased by 30%. Find the net change.
Solution:
Let number = 100.
Inc 30% -> 130.
Dec 30% of 130 = 39.
New number = 130 - 39 = 91.
Net change = 100 - 91 = 9 Decrease. (9%)
6. A number is decreased by 30% and then this decreased number is increased by 30%. Find the net change as percent.
Solution:
Let number = 100.
Dec 30% -> 70.
Inc 30% of 70 = 21.
New number = 70 + 21 = 91.
Net change = 9% Decrease.
7. The population of a village increases by 10% per year. If the present population of the village is 24,000; find it at the end of 2 years.
Solution:
Population after 2 years = 24,000 × (1.1) × (1.1)
= 24,000 × 1.21
= 29,040.
8. The cost of a machine decreases by 10% per year. If its present cost is 24,000; find its value at the beginning of 3rd year.
Solution:
Value at beginning of 3rd year means after 2 years.
Value = 24,000 × (0.9) × (0.9)
= 24,000 × 0.81
= 19,440.
9. The price of sugar has been increased by 50%. By how much percent can the consumption of the sugar be decreased in order to keep the expenditure on sugar the same.
Solution:
Let original price = 100. New price = 150.
To keep expense same, reduce consumption by 50.
% Reduction = (50 / 150) × 100%
= 1/3 × 100%
= 33 1/3%.