Question

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

Answer 1

★ Given :

Number of mosquitoes = 200

Growth rate = 10%

★ To find :

I) number of mosquitoes in 3 years.

ii) Number of mosquitoes in 10 years

iii) Number of mosquitoes in n years..

★ Solution :

We know that,

$A = P\left(1 + \frac{R}{100}\right)^t$

Here,

P = 200 , R = 10 and t = 3 years

by putting values

$= 200 \times \left(1 + \frac{10}{100}\right)^{3}$

$= 200 \times \left(1 + \frac{1}{10}\right)^{3}$

$= 200 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$

$= \dfrac{266200}{1000}$

$\purple{\leadsto 266.2}$

ii) Here,

P = 200 , R = 10 , t = 10

Applying the same formula

$= 200 \times \left(1 + \frac{10}{100}\right)^{10}$

$= 200 \times \left(1 + \frac{1}{10}\right)^{10}$

$= 200 \times \left(\frac{11}{10}\right)^{10}$

$= \frac{200 \times 1331 \times 1331 \times 1331 \times 11}{10^{10}}$

$\purple{\leadsto 518. 748 = 519(approx)}$

iii) for n years

= $\purple{P \left(1 + \frac{R}{100}\right)^{n}}$

Where,

P = original population

R = rate of growth

n = number of years..

Answer 2

$\rule{200}{2}$

ANSWER:

The number of mosquitoes after

• (i) 3 years ≈ 262
• (ii) 10 years ≈ 519
• (iii) n years = 200(1.1)ⁿ

$\rule{200}{2}$

GIVEN:

• Mosquitoes are growing at a rate of 10% a year

• 200 mosquitoes are in the beginning.

$\rule{200}{2}$

TO FIND:

The number of mosquitoes after

• (i) 3 years
• (ii) 10 years
• (iii) n years.

$\rule{200}{2}$

EXPLANATION:

Let x be the number of mosquitoes after one year.

x/200 × 100 = 10%

x/2 = 10

x = 20 mosquitoes

Then mosquitoes increased 20 for the first year.

$\rule{200}{1}$

Here we can use the formula:

$\large{\bold{A=P\left(1+ \frac{r}{100}\right)^t}}$

(i) The number of mosquitoes after 3 years

P = 200

r = 10

t = 3

$A=200\left(1+ \dfrac{10}{100}\right)^3$

$A=200\left(1+ \dfrac{1}{10}\right)^3$

$A=200\left( \dfrac{10+1}{10}\right)^3$

$A=200\left(\dfrac{11}{10}\right)^3$

$A=200\left(\dfrac{1331}{1000}\right)$

$A=\left(\dfrac{1331}{5}\right)$

$A\approx262$

$\rule{200}{1}$

(ii) The number of mosquitoes after 10 years:

P = 200

r = 10

t = 10

$A=200\left(1+ \dfrac{10}{100}\right)^{10}$

$A=200\left(1+ \dfrac{1}{10}\right)^{10}$

$A=200\left(\dfrac{10+1}{10}\right)^{10}$

$A=200\left(\dfrac{11}{10}\right)^{10}$

$A=200\left(1.1\right)^{10}$

$A=200\left(2.59\right)$

$A\approx518$

$\rule{200}{1}$

(iii) The number of mosquitoes after n years:

P = 200

r = 10

t = n

$A=200\left(1+ \dfrac{10}{100}\right)^n$

$A=200\left(\dfrac{10+1}{10}\right)^{n}$

$A=200\left(\dfrac{11}{10}\right)^{n}$

$A=200\left(1+ \dfrac{1}{10}\right)^n$

$A=200\left(1.1\right)^n$

$\rule{200}{2}$