Math, asked by jyothsna22, 10 months ago

three years ago the population of a town was 50000 .if the annual increase during three successive years was at the rate 4%,5% and 4% per annum respectively,find the present population.​

Answers

Answered by venupillai
5

Answer:

The present population is 56,784

Step-by-step explanation:

Original population of the town = 50,000

At the end of the first year,

Population = Original population + 4% of Original Population

                   = (1 + 4/100) * Original Population

                    = 1.04 * 50000

                     = 52000

Similarly, at the end of the second year,

Population = 52000 * 1.05                     (As growth is 5%)

                   = 54600

Likewise, at the end of the third year,

Population = 54600 * 1.04

                  = 56784

In general,

If P is the original population growing at rates R1, R2, R3, ......Rn in each of the n years respectively. Then, population at the end of n years (say Pn) will be:

Pn = P*(1 + R1/100)*(1 + R2/100)*(1 + R3/100)* ........(1 + Rn/100)

(Note that R1, R2, etc are expressed in percentage)

In our case, P = 50,000, R1 = 4, R2 = 5 and R3 = 4

Final Population = 50000(1.05)(1.04)(1.05)

                            = 50000(1.13568)

                            = 56784

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