Percentage

To express x\% as a fraction =\frac{x}{100}
Thus, 3\% = \frac{3}{100} = .03,
48\% = \frac{48}{100} = 0.48 = \frac{12}{25}
To express \frac{a}{b} as a percentage, we have \frac{a}{b} = \frac{a*100}{b}
Thus \frac{1}{4} = \frac{1*4}{100} = \frac{25}{100} = 25%

if the price of commodity increases by R% than the reduction in consumption so as not to increase the expenditure is
\frac{R}{100+R}*100\%

Do – decreases R% ______ do__
\frac{R}{100-R}*100\%

Result on Population– let the population of a town be P now & suppose increases at the rate of (@) R % annum then.
Population after n year = p*(1+\frac{R}{100})*n

Population n year ago = \frac{p}{(1+\frac{R}{100})*n}
Result on depreciation (decrease-) Let the present value of a machine BCp suppose it depreciate @ R% per annum then.
Value of machine after n year = p*(1-\frac{R}{100})*n

Value of machine n year ago = \frac{p}{(1-\frac{R}{100})*n}

if A is R% more than B, B is less than A by

[\frac{R}{(100+R)}*100}]\%
If A is R% less than B, then B is more than A by
[\frac{R}{(100-R)}*100}]\%