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

Do – decreases R% ______ do__

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

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