☛ It is an extension of unitary method only difference being, it is expressed in terms of 100 and not 1.
☛ Per cent means ‘per 100’ and is denoted as %.
Eg. Ramesh scored 88% in his exam and the total marks of the exam was 1200 marks. How much did he score?
Given that Ramesh scored 88% marks it means that if the total marks were 100, he scored 88.
If the total marks were 1200, he would have scored:
88/100 x 1200 = 976 marks
☛ For ease of calculations, percentage may have to be converted into decimal fractions and vice versa.
☛ x % is expressed as x/100 and on simplifying we get its corresponding fraction.
Eg. 25% is 25/100 which can be simplified as 1/4
☛ Similarly, to express a/b as a percent,we multiply by 100.
Eg. 1/5 in percentage terms is:
1/5 x 100 = 20%
Suppose a boy scored 450 marks out of 600 marks in his exam. Express his marks in percentage.
Soln:. We have to find the marks, person would have obtained out of 100.
450/600 x 100 = 75%
☛ If two quantities a and b are multiplied and the product has to remain constant,
and one of the quantity is increased by y%, then the other quantity has to be decreased by 100y/(100+y) %.
Eg. Ram buys 10 articles worth Rs. 80 each. How many articles will he buy with the same amount if the price of each item is hiked to Rs. 100 per article.
Soln:. Amount with Ram = No. of articles x Cost per article.
Amount with Ram is a constant. ie. 10 x 80 = Rs. 800
% Increase in price = 20/80 x 100 = 25%
Hence % decrease in no. of articles = (25 x 100/125)% = 20%.
Initially, number of articles = 10.
Decrease of 20 % = decrease of 20/100 x 10 = 2 articles. Therefore, he will be able to buy 8 articles.
Time speed Distance problems are example for this concept.
☛ Similarly, if a is decreased by x% b will be increased by 100 x/(100-x) %.
Appreciation / Depreciation Problems:
☛If a person buys a property for amount A and it appreciates x% per year, the worth after n years,
= A [1+ x/100]n
The value compounds. It is the same method used for finding Compound Interest
Similarly, machines depreciate over time
☛ A machine bought for amount A depreciates at x% per year, the worth of machine after n years is :