Percentage Examples

8/8/2019 Percentage Examples

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ICSE SOLVED EXAMPLES FOR CLASS 7

Expressing a Quantity as Percentage of another Quantity

To express x as the percentage of y, we use the following formula:

Percentage = (x/y) x 100%

Example 1: Find the number whose 35% is 105.

Solution:

35% of the number = 105

1% of the number = 105/35

Therefore, 100% of the number = 3 x 100 = 300

Increase and Decrease Percentage

Increase% = (Increase/Original) x 100%

Decrease% = (Decrease/Original) x 100%

Example 1: The price of rice is decreased from Rs. 16 per kg to Rs.14 per kg. Find the

percentage decrease in price.

Solution:

Original price = Rs.16/- Current price = Rs.14/-

Decrease = Rs.16 Rs. 14 = Rs. 2.

Decrease in % = (2/16) x 100% = 25/2 = 12 = 0.5

Example 2: The population of Nanded, last year was 2500. This year it is 3000. Find the

percentage increase.

Solution:

Increase of population: 3000 2500 = 500

Increase in % = (Increase/Original) x 100% = (500/2500) x 100% = 20%

Example 3: In a mixture of milk and water, are mixed in the ratio 4:1. Find the percentage of

milk in the mixture.

Solution:

Milk : Water = 4:1Therefore, % of milk = 4/4+1 = 4/5 part = (4/5) x 100% = 80%

Example 4: In a month Omkar has decrease his weight by 16%. In the previous month his

weight was 75 kg. Find his present weight.

Solution:

Initial Weight = 75kg

Decrease in weight = (16% of 75kg) = (16100) x 75kg = 12kg.

So, his present weight = 75 12 = 63kg


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Otherwise, theres one more way to solve this problem.

Weight is decreased by 16%. (Take this as a point)

Therefore, his present weight = (100 16) % of original weight = 84% of original weight

Present Weight = (84/100) x 75kg = 63kg.

Example 5: The salary of Mr. Modi is raised by 8%. If his present salary is Rs.10368. Find his

original salary.

Solution:

Original salary = 100%

% Increase in salary = 8%

Present salary = (100 - 8)% = 108%

According to the problem 108% = Rs. 10368. Therefore, 1% = Rs.(10368/108) x 100 = Rs.9500

Important Note: The quantity, on which the percentage is calculated is taken as 100%. In the

previous problem, the original salary is taken as 100%.

Example 6:A number when reduced by 10% becomes 360. Find the number.Solution:

Let the number be 100%, then the amount to be decreased is 10%. Then, the amount becomes

90%. According to the question, 90% is equal to 360. That is 90% = 360.

Therefore, 1% = 360/90 = 4.

Therefore, 100% = 100x4 = 400.

Thus, the number is 400.

Example 7: Ruchita saves 25% of her monthly income. If her monthly income expenditure is

Rs.8,400. Find her monthly savings.Solution:

Let the monthly income be 100%, and the income that she saves for her monthly income is 25%.

Then her monthly expenditure is (100 25)% = 75%. According to the problem, 75% = Rs.8500.

Therefore, 1% = Rs.8500/75.

Therefore, 25% = Rs. (8400/75) x 25% = Rs.2,800.

Example 8: A student has to score 40% of the maximum marks to pass. Sonia scored 165 marks

and failed by 35 marks. Calculate the maximum marks.

Solution:

Marks scored by Sonia = 165.

Storage of marks from pass marks = 35.

Thus, minimum pass marks = 165 + 35 = 200.

100% maximum marks = (200/40) x 100 = 500.

Thus, maximum marks he has to score is 500.


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Example 9: If As income is 25% more than Bs, how much percent is Bs income less than As?

Solution:

Let Bs income be 100. Therefore, As income is Rs(100+25)=125.

Therefore, when As income is Rs.125. Bs income is less by 25%.

Therefore the less % = (25/125) x 100% = 20% Thus, Bs income is 20% less than As income.

Example 10: Shashank and Sharang contested for a part in students union. In all, there were

3600 votes of which 75% were polled. If Shashank got 60% of the polled vote, how many didSharang get?

Solution:

Total Number of Votes = 3600.

Number of votes were polled = 75% of 3600 = (75/100) x 3600 = 2700.

Number of votes obtained by Shashank = 60% of 2700 = (60/100) x 2700 = 1620.

Number of votes obtained by Sharang = 2700 1620 = 1080.

Therefore, Shashank obtained more votes than Sharang.

Example 11: 5% out of the total number of 2000 examiners did not appear in the examination.60% of the examiners who sat in the examination passed in the first and second divisions. If the

ratio of the examiners who passed in first and second division is 2:3, find the number of

examiners who passed in first division.

Solution:

Total number of examiners = 2000.

Number of examiners who did not appear = 5% of 2000. (5/100) x 2000 = 100.

Therefore, Number of examiners who sat in the examination = 2000 100 = 1900.

Number of examiners who passed in first and second divisions = 60% of 1900

( = (60/100) x 1900 = 1140.Number of examiners who passed in first division = (2/2+3) x 1140 = (2/3) x 1140 = 456.

Thus, 456 examiners passed in first division.

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