Mosquitoes are growing at a rate of 10% a year. If there were 200 mosq

1.	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» ★ 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..