Percentage: Formulas, Problems, Questions & Short Trick

Percentage: Formulas, Problems, Questions & Short Trick: In this page, you will learn about Percentage Math Problems, Short Tricks in English. This topic is very important for 2019 competitive exams preparation. So before proceeding for any exam read these short tricks and explanations carefully about Percentage Questions with the explanation.

Percentage is a way of expressing a number, especially a ratio, as a fraction of 100. The word is derived from the Latin per centum meaning ‘by the hundred’. It is often denoted using the percent sign, ‘%’, or the abbreviation ‘pct.’ For example, 35% (read as ‘thirty-five percent’) is equal to 35/100, or 0.35.

Quicker Methods to Solve the Problems

Example 1: Convert the fraction 3/5 into percent fraction.

Solution: % = 60%

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Example 2: Convert the fraction 3.5/100 into percent fraction

Solution: % = 3.5%

For converting a percentage to a fraction or decimal, divide by hundred.

Example 3: What is the fraction of 60%?

Solution: 

Example 4: If the price of sugar is increased by 25%, find how much percent a family must reduce their consumption of sugar so as not to increase the expenditure of the family?

Solution: Reduction in consumption of sugar

% = 20%

Example 5: If the price of Kerosene oil falls by 10%, find how much perecent can a householder increase its consumption, so as not to decrease expenditure on this item?

Solution: Increase in consumption of Kerosene oil

= % = 11.11%

Example 6: If Ravi’s salary is 50% more than that of Gopal’s, then how much percent is Gopal’s salary less than that of Ravi’s salary?

Solution: Gopal’s salary is less than that of Ravi’s by

= % = %

Example 7: If income of Rekha is 30% less than that of Vina, then how much percent is Vina’s income more than that of Rekha?

Solution: Vina’s income is more than that of Rekha by

= %

% %

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If the original population of a town is P, and the annual increase is r %, then the population after n years is and population before n years = 

If the annual decrease be r %, then the population after n years is  and population before n years = 

Example 8: The population of a certain town increased at a certain rate per cent per annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence?

Solution: Suppose the population increases at r% per annum.

Then, = 456976

∴ 

Population 2 years hence = 456976

= 456976 × = 494265 approximately.

If the value is first increased by x % and then decreased by y% then there is  increase or decrease, according to the +ve or –ve sign respectively.

Example 9: A number is increased by 10%. and then it is decreased by 10%. Find the net increase or decrease per cent.

Solution: % change = 

i.e., 1% decrease.

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Average percentage rate of change over a period.

=  × % [where n = period]

The percentage error = × 100%

If the value is increased successively by x % and y % then the final increase is given by 

If the value is decreased successively by x % and y % then the final decrease is given by 

Example 10: The price of a car is decreased by 10 % and 20 % in two successive years. What percent of price of a car is decreased after two years?

Solution: Put x = – 10 and y = – 20, then

= – 28%

∴ The price of the car decreases by 28%.

The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks

M = 

A candidate scoring x % in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more than the minimum required passing marks. Then the maximum marks

.

Example 11: Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got?

Solution: If Vishal has 15 marks more, he could have scored 40% marks.

Now, 15 marks more than 185 is 185 + 15 = 200

Let the maximum marks be x, then 40% of x = 200

or 

or 

Thus, maximum marks = 500

Quicker method:

Maximum marks

=

= 

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