## Percentage

### Introduction

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 = $$\frac{\mathbf{P}}{\left(1-\frac{\mathbf{R}}{100}\right)^{n}}$$