Question

14) Mosquitoes are growing at a rate of 10%
a year. If there were 200 mosquitoes in
the begining. Write down the number of
mosquitoes after (1) 3 years (ii) 10 years
(iii)n years.​

Answer 1

200 ×10/100 = 20

220× 10/100= 22

242 ×10/100 = 24.2

266.2

Answer 2

Given:

Initial number of mosquitoes = 200

Growth rate = 10% = 0.1

To find:

The number of  mosquitoes after (1) 3 years (ii) 10 years  (iii)n years.​

Solution:

Exponential growth function.

$P(t)=a(1+r)^t$

where, a is initial value, r is growth rate and t is time period.

Substitute a=200 and r=0.1.

$P(t)=200(1+0.1)^t$

$P(t)=200(1.1)^t$            ...(i)

Put t=3 in (i).

$P(3)=200(1.1)^3$

$P(3)=200(1.331)$

$P(3)=266.2$

$P(3)\approx 266$

Therefore, the number of  mosquitoes after 3 years is 266.

Similarly, put t=10 in (i).

$P(10)=200(1.1)^{10}$

$P(10)=518.748492$

$P(10)\approx 518$

Therefore, the number of  mosquitoes after 10 years is 518.

Similarly, put t=n in (i).

$P(n)=200(1.1)^{n}$

Therefore, the number of  mosquitoes after n years is [tex200(1.1)^{n}[/tex].