# Percentage Questions

**FACTS AND FORMULAE FOR PERCENTAGE QUESTIONS**

**I.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% = x/100.

Thus, 20% = 20/100 = 1/5;

48% = 48/100 = 12/25, etc.

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

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

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

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is $\left[\frac{R}{\left(100-R\right)}\times 100\right]\%$

**III. Results on Population : **Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :

1. Population after n years = $P{\left(1+\frac{R}{100}\right)}^{n}$

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

**IV. Results on Depreciation :** Let the present value of a machine be P. Suppose it depreciates at the rate 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 = $\frac{P}{{\left(1-{\displaystyle \frac{R}{100}}\right)}^{n}}$

**V.** If A is R% more than B, then B is less than A by

$\left[\frac{R}{\left(100+R\right)}\times 100\right]\%$

If A is R% less than B , then B is more than A by

$\left[\frac{R}{\left(100-R\right)}\times 100\right]\%$

A) ₹ 75 lakh | B) ₹ 12.5 lakh |

C) ₹ 25 lakh | D) ₹ 125 lakh |

A) 36% decrease | B) 36% increase |

C) 40% increase | D) 40% decrease |