Question

Compound Interest Application

Compound interest is given by the formula A = P ( 1 + r ) t . Where A is the balance of the account after t years, and P is the starting principal invested at an annual percentage rate of r , expressed as a decimal.

Wyatt is investing money into a savings account that pays 2% interest compounded annually, and plans to leave it there for 15 years. Determine what Wyatt needs to deposit now in order to have a balance of $40,000 in his savings account after 15 years.

Wyatt will have to invest $___________ now in order to have a balance of $40,000 in his savings account after 15 years. Round your answer UP to the nearest dollar.

Answer 1

Answer:

$29,721

Step-by-step explanation:

We have been given that Wyatt is investing money into a savings account that pays 2% interest compounded annually, and plans to leave it there for 15 years. We are asked to find the amount deposited by Wyatt in order to have a balance of $40,000 in his savings account after 15 years.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nT}$, where,

A = Final amount after T years,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

T = Time in years.

Let us convert our given interest rate in decimal form.

$2\%=\frac{2}{100}=0.02$

Upon substituting our given values in compound interest formula, we will get:

$\ 40,000=P(1+\frac{0.02}{1})^{1*15}$

$\ 40,000=P(1+0.02)^{15}$

$\ 40,000=P(1.02)^{15}$

$\ 40,000=P\times 1.3458683383241296$

Switch sides:

$P\times 1.3458683383241296=\ 40,000$

$\frac{P\times 1.3458683383241296}{ 1.3458683383241296}=\frac{\ 40,000}{1.3458683383241296}$

$P=\ 29,720.5891995$

Upon rounding our answer to nearest dollar, we will get:

$P\approx \ 29,721$

Therefore, Wyatt will have to invest $29,721 now in order to have a balance of $40,000 in his savings account after 15 years.