Concept of Percentage:- By a certain percent, we mean that many hundredths. Thus, x percent means x hundredths, written as x%.

To express x% as a fraction:- We have, x% = \(\frac{x}{100}\)

Thus, \(20\%\) = \(\frac{20}{100}\) = \(\frac{1}{5}\);

\(48 \%\) = \(\frac{48}{100}\) = \(\frac{12}{25}\) etc.,

To express \(\frac{a}{b}\) as a percent:- We have, \(\frac{a}{b}\) = \(\left(\frac{a}{b} \times 100\right)\%\)

Thus, \(\frac{1}{4}\) = \(\left(\frac{1}{4} \times 100\right)\%\) = 25%;

0.6 = \(\frac{6}{10}\) = \(\frac{3}{5}\) = \(\left(\frac{3}{5}\times100\right)\%\) = 60% 

Important Formulas

If the price of a commodity increase by R%, then the reduction in consumption so as not to increase the expenditure is \(\left[\frac{R}{(100+R)} \times 100\right]\%\)

If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is \(\left[\frac{R}{(100-R)} \times 100\right]\%\)

Results on Population and Results on Depreciation

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 = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{n}\)

2. Population n years ago = \(\frac{\mathrm{P}}{\left(1+\frac{\mathrm{R}}{100}\right)^{n}}\)

Results on Depreciation:- Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum, then:-

1. Value of the machine after n years = \(P\left(1-\frac{R}{100}\right)^{n}\)

2. Value of the machine n years ago = \(\frac{\mathbf{P}}{\left(1-\frac{\mathbf{R}}{100}\right)^{n}}\)