Question

The population of a town was 75000 three years ago. if the population increased by 2%,4% and 5% in the last three years, find its present population.​

Answer 1

Answer:

Step-by-step explanation:

number of population increse=10540-10000

=540

number of females intown=6/100x=540

x= 540*100/6

x= 9000

therefore number of females in town is 9000

Answer 2

$\huge\sf{Answer :- }$

Given :-

• Population of a town 3 years ago = 75000

• Population increased in last 3 years = 2%, 4% and 5%

To Find :-

• Present population of a town

Formula used :-

Present population =

$\small\sf{ P(1+ \frac{ A _{1} }{100} )(1 + \frac{ A _{2}}{100})(1 + \frac{ A_{3} }{100}) }$

Where as,

• P = population of a town 3 years ago

• A1 = 2%, A2 = 4% and A3 = 5%

Solution :-

Let the present population be x

By substituting values according to the formula,

$\small\sf{➨ x= 75000(1 + \frac{2}{100})(1 + \frac{4}{100})(1 + \frac{5}{100}) }$

$\large\sf{➙ x = 75000 \times \frac{102}{100} \times \frac{104}{100} \times \frac{105}{100} }$

$\large\sf{➙ x = 75000 \times \frac{1113840}{1000000} }$

$\large\sf{➙ x = 0.75 \times 111384}$

$\large\sf{➠ x = 83538}$

Hence , The present population of a town is 83538