Concept of Percentage
By a certain percent, we mean that many hundredths.
Thus, x percent means x hundredths denoted by x %.
Expressing a percentage as a fraction or decimal number:
i) x % = x/100
ii) 50 % = 50/100 = 1/2 = 0.5
Expressing a fraction or decimal number as percentage:
i) x/y = {(x/y) x 100} %
ii) 1/4 = {(1/4) x 100} % = 25 %
Percentage Increase/Decrease
If the price of a commodity increases R % then the reduction in consumption so as not to increase the expenditure is:
[{R/(100+R)} x 100] %
If the price of a commodity decreases R % then the increase in consumption so as not to decrease the expenditure is:
[{R/(100-R)} x 100] %
Results on Population
Let the population of a town be P now and suppose it increases at the rate of R % per annum, then
1. Population after n years = P(1 + R/100)n
2. Population n years ago = P/{(1 + R/100)n}
Results on Depreciation
Let the present value of a machine P. It depreciates at a rate of R% per annum. Then
1. Value of machine after n years = P(1 - R/100)n
2. Value of the machine n years ago = P/{(1 - R/100)n}
3. If A is R% more than B, then B is less than A by [{R/(100+R)} x 100] %
4. If A is R% less than B, then B is more than A by [{R/(100-R)} x 100] %
Click here for Online Test on Percentage
By a certain percent, we mean that many hundredths.
Thus, x percent means x hundredths denoted by x %.
Expressing a percentage as a fraction or decimal number:
i) x % = x/100
ii) 50 % = 50/100 = 1/2 = 0.5
Expressing a fraction or decimal number as percentage:
i) x/y = {(x/y) x 100} %
ii) 1/4 = {(1/4) x 100} % = 25 %
Percentage Increase/Decrease
If the price of a commodity increases R % then the reduction in consumption so as not to increase the expenditure is:
[{R/(100+R)} x 100] %
If the price of a commodity decreases R % then the increase in consumption so as not to decrease the expenditure is:
[{R/(100-R)} x 100] %
Results on Population
Let the population of a town be P now and suppose it increases at the rate of R % per annum, then
1. Population after n years = P(1 + R/100)n
2. Population n years ago = P/{(1 + R/100)n}
Results on Depreciation
Let the present value of a machine P. It depreciates at a rate of R% per annum. Then
1. Value of machine after n years = P(1 - R/100)n
2. Value of the machine n years ago = P/{(1 - R/100)n}
3. If A is R% more than B, then B is less than A by [{R/(100+R)} x 100] %
4. If A is R% less than B, then B is more than A by [{R/(100-R)} x 100] %
Click here for Online Test on Percentage





