Bob wants $50,000 at the end of 7 years in order to buy a car. if his bank pays 4.2% interest, compounded annually, how much must he deposit each year in order to reach his goal?
Accepted Solution
A:
The correct answer is $1452.50 per year.
Explanation: We use the formula [tex]P(1+\frac{r}{n})^{nt}+PMT(\frac{[(1+\frac{r}{n})^{nt}-1]}{\frac{r}{n}})\times (1+\frac{r}{n})[/tex],
where P is the amount of principal invested, r is the interest rate as a decimal number, n is the number of times per year the interest is compounded, PMT is the monthly deposit added, and t is the number of years.
Since the amount of principal is not stated, we will assume that Bob is depositing the same amount every following year as he does the first year, so we will let PMT=P.
Our interest rate, r, is 4.2%; 4.2%=4.2/100=0.042.
The number of times the interest is compounded annually, n, is 1. The amount of time, t, is 7.
We know he wants $50,000. This gives us the equation 50000=P(1+0.042/1)⁽¹ˣ⁷⁾+P{[(1+0.042/1)⁽¹ˣ⁷⁾]/(0.042/1)}*(1+0.042/1).
Simplifying this a bit, we have 50000=P(1.042)⁷+P((1.042⁷)/0.042)*(1.042).
We can factor out P, giving us 50000=P[1.042⁷+((1.042⁷)/0.042)*1.042].
This then gives us 50000=P(34.4234).
Divide both sides: 50000/34.4234 = (P(34.4234))/34.4234,