On the first one, note that it wouldn't make sense for Mary to have to invest 44,298 each year for three years to cover the tuition. The cost will only be 20,000 paid twice; even if she earned zero interest on her investment, a
single savings of 44,298 would be more than enough.
First use the PV function to determine the present value of those two 20K tuition payments, at 7%. This will be the amount Mary needs to have
at the end of the third year, since the PV function delivers the present value as of one period prior to the first payment.
Then use the FV function to figure how much she'll have to put into the investment at the end of the next three years, in order to accumulate to exactly that amount you calculated in the first step. You'll just have to try different inputs into the "payment amount" argument of the FV function, until it outputs the required amount. If you're familiar with Excel's Goal Seek utility, you could use that to speed up the process.
For the second one, first use PV to calc the present value of 60K per year for 20 years. This will give you the amount James must have 14 years from today (one year before the first withdrawal). Then, just as with the first problem, use the FV function to determine what annual payment amount, at 7% and for 10 years, will produce that amount (if his first savings investment is 5 years from today, and the final one will be 14 years from today, that's 10 investment savings in all).
The third one sets up as
Use Excel to solve for
r.
In the fourth one Michael is going to sock away 2 x 5 x 52 = $520 at the end of each year for 40 years. Use the FV function.
You didn't say what the car's price is, in the last one, but let's say it's
F. The problem sets up as
You can solve for
n either by taking logarithms or via a little trial-and-error work in Excel.