Question

how to find principal and annual interest rate if time to double is 11 years and future value after 10 years is 2100?

Answer 1

Let's begin by listing out the given information:

Amount after 10 years = $2,100

Time to double = 11 years; A = 2P

The formula for compound interest is given by:

`\begin{gathered} A=P(1+(r)/(n))^(nt) \\ After\text{ 10 year}s\text{, we have:} \\ A=2,100,n=1(annually),t=10 \\ 2100=P\mleft(1+(r)/(1)\mright)^(1\cdot10) \\ 2100=P(1+r)^(10)-----1 \\ \\ After\text{ 11 year}s\text{, we have:} \\ A=2P,n=1,t=11 \\ 2P=P(1+(r)/(n))^(1\cdot11) \\ 2P=P(1+(r)/(1))^(11) \\ 2P=P(1+r)^(11) \\ \text{Divide both sides by }P\text{:} \\ (2P)/(P)=(P(1+r)^(11))/(P) \\ 2=(1+r)^(11)----2 \\ (1+r)=\sqrt[11]{2} \\ 1+r=1.065 \\ r=1.065-1=0.065 \\ r=0.065=6.5\text{ \%} \\ r=6.5\text{ \%} \\ \\ Substitute\text{ the value of r into equation 1, we have:} \\ 2100=P(1+r)^(10) \\ 2100=P(1+0.065)^(10) \\ P(1+0.065)=\sqrt[10]{2100}\Rightarrow P(1.065)=\sqrt[10]{2100}\Rightarrow P=\frac{\sqrt[10]{2100}}{1.065} \\ P=2.018 \end{gathered}`