If P dollars is invested at r percent compounded annually, at the end of 2 yers it will grow to A = P(1+r)^2. At what interest rate will $1,000 grow to$1,210 in 2 years?

Aug 9, 2018

Interest rate is 10.0%

Explanation:

A=P(1+r)^2 ;A=$1210 , P=$1000

$1210 = 1000 {\left(1 + r\right)}^{2} \therefore {\left(1 + r\right)}^{2} = \frac{1210}{1000}$ or

${\left(1 + r\right)}^{2} = 1.21 \therefore \left(1 + r\right) = \sqrt{1.21} = 1.1$

r=1.1-1=0.1 or 0.1*100=10.0%

Interest rate is 10.0% [Ans]