Question

Hello! I’m trying to figure out this problem but have no idea how.

Answer 1

Given: Two investments compounded annually as follows-

$\begin{gathered} P_1=4800 \\ R_1=7\% \end{gathered}$

and,

$\begin{gathered} P_2=7900 \\ R_2=5.7\% \end{gathered}$

Required: To find out the time required by smaller investment to catch up to larger investment.

Explanation: The formula for compound interest is as follows-

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

Putting the values for the first investment (smaller one) we get,

$\begin{gathered} A_1=4800(1+\frac{7}{100})^t \\ =4800(1+0.07)^t \\ =4800(1.07)^t \end{gathered}$

Similarly solving for other investment we get,

$\begin{gathered} A_2=7900(1+0.057)^t \\ =7900(1.057)^t \end{gathered}$

Now lets say in time 't' our investment catch up or becomes equal, i.e.,

$A_1=A_2$$4800(1.07)^t=7900(1.057)^t$$(\frac{7900}{4800})=(\frac{1.057}{1.07})^t$$1.65=(0.98)^t$

Taking logarithm both sides and solving for t gives us,

$t\approx40.7599$$\approx40.8$

Final Answer: The investments would catch up after approximately 40.8 yrs.