RESULTS ON POPULATION :-
1. Let the present population of a city is P. Suppose it increases at the rate of R% per annum. Then,
Its population after n years = $P\times\left ( 1+\frac{R}{100} \right )^{n} $
Its population n years ago = $\frac{P}{\left ( 1+\frac{R}{100} \right )^{n}}$
NOTE :- If the annual decrease be R % , then,
Its population after n years = $P\times\left ( 1-\frac{R}{100} \right )^{n}$
For eg. 1. If the present population of a city is 15625 and increases at the rate of 4 % per annum. Find the population after 3 yrars?
sol. population after 3 years = $15625\, \left ( 1+\frac{4}{100} \right )^{3}$
= $15625\, \left ( 1+\frac{1}{25} \right )^{3}$
= $15625\,\times\,\frac{26}{25}\,\times\,\frac{26}{25}\,\times\,\frac{26}{25}$
= 17576.
Example 2. During one year, the population of a town increased by 5 % and during the next year, the population decreased by 5 %. If the total population is 9975 at the end of the second year, then what was the population size in the begining of the first year?
sol. population in the begining of the first year
= $\frac{9975}{\left(1+\frac{5}{100}\right)\left( 1-\frac{5}{100}\right)}$
= $\frac{9975}{\left(1+\frac{1}{20}\right)\left( 1-\frac{1}{20}\right)}$
= $\frac{9975}{\left(\frac{21}{20}\right)\left(\frac{19}{20}\right)}$
= $\left(9975\times\frac{20}{21}\times\frac{20}{19}\right)$
= 10000.
1. Let the present population of a city is P. Suppose it increases at the rate of R% per annum. Then,
Its population after n years = $P\times\left ( 1+\frac{R}{100} \right )^{n} $
Its population n years ago = $\frac{P}{\left ( 1+\frac{R}{100} \right )^{n}}$
NOTE :- If the annual decrease be R % , then,
Its population after n years = $P\times\left ( 1-\frac{R}{100} \right )^{n}$
For eg. 1. If the present population of a city is 15625 and increases at the rate of 4 % per annum. Find the population after 3 yrars?
sol. population after 3 years = $15625\, \left ( 1+\frac{4}{100} \right )^{3}$
= $15625\, \left ( 1+\frac{1}{25} \right )^{3}$
= $15625\,\times\,\frac{26}{25}\,\times\,\frac{26}{25}\,\times\,\frac{26}{25}$
= 17576.
Example 2. During one year, the population of a town increased by 5 % and during the next year, the population decreased by 5 %. If the total population is 9975 at the end of the second year, then what was the population size in the begining of the first year?
sol. population in the begining of the first year
= $\frac{9975}{\left(1+\frac{5}{100}\right)\left( 1-\frac{5}{100}\right)}$
= $\frac{9975}{\left(1+\frac{1}{20}\right)\left( 1-\frac{1}{20}\right)}$
= $\frac{9975}{\left(\frac{21}{20}\right)\left(\frac{19}{20}\right)}$
= $\left(9975\times\frac{20}{21}\times\frac{20}{19}\right)$
= 10000.
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