Math, asked by Nakshatranigam, 1 year ago

3 year ago the population of a town was 50000 if the annual increase during three successive years 5%, 4% and 3% respectively what is its present population

Answers

Answered by jacobcunningham202
135

population when increase 5 percent

= 50000x5/100

=2500

population when increase 4percent

50000x4/100

=2000

population when increase 3percent

=50000x3/100

=1500

total population =50000+2500+2000+1500

=56238

hope this helped you

#bebrainly

Answered by santy2
15

Answer:

The present population is 56,238

Step-by-step explanation:

The population 3 years ago was 50,000, which in our case is 100%, what we want to know is the number of people living in that area after the first year, when the population increased by 5%. Since the population increased by 5%, we will add the 5% to the 100% ; 100 + 5 = 105%

We will therefore equate as in the following;

50000 = 100%,

105%?

We then cross multiply:

(50000 × 105) ÷ 100

This gives us, 52,500

This was the population in the third year. We are told that the population increased by 4%. In the second year, the 100% population was 52500, what about when it increased by 4%, that is, (100% + 4% )?

So we say, 52500 = 100%,

104%?

We then cross multiply as follows:

(52500 × 104) ÷ 100 = 54600

Thus, the population in the second year is 54,600.

54600 is the 100% population in the second year. We know that the population increased by 3% in the past year, in that ( 100% + 3% )

We therefore work it out as follows

54600 = 100%

103% ?

We then cross multiply

(54600 × 103 ) ÷ 100 = 56,238

56,238 is the present population.

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