Question

Three years ago, the population of a city was 50000. If the annual increase during three successive years be 5%, 8% and 10% respectively, find the present population of the city.

Answer 1

Answer: The present population of the city is 61,500 persons.

Step-by-step explanation:

Increase in population is given for 3 successive years.

When population increased 5%, the number of persons added newly will be,

= 50000 x $\frac{5}{100}$

=2500

When population increased 8%, the number of persons added newly will be,

50000 x $\frac{8}{100}$

=4000

When population increased 10%, the number of persons added newly will be,

=50000 x $\frac{10}{100}$

=5000

Total population = 50000+2500+4000+5000

=61500 persons.

Answer 2

Ans

P=50,000

R1=5%

R2=8%

R3=10%

THEN,

Pt=P(1+R1/100)(1+R2/100)(1+R3/100)

Pt=50000(1+5/100)(1+8/100)(1+10/100)

Pt=62370

therefore, population after three yrs is 62370

Step-by-step explanation: