# You invest $500 in an account that earns interest compounded monthly. Use a table or graph to find the least annual interest rate (to the nearest tenth of a percent) that the account would have to...

You invest $500 in an account that earns interest compounded monthly. Use a table or graph to find the least annual interest rate (to the nearest tenth of a percent) that the account would have to earn if you want to have a balance of $600 in 4 years.

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### 2 Answers

It sounds like they want you to take a stab at what you would get in 4 years from $500 at different interest rates compounded monthly. You'll need to use the discrete compounding equation here:

`A = P(1+r/n)^(nt)`

Here, A is the future value of the money, P is the principle (original amount you put in, $500), r is the annual interest rate, n is the number of intervals in the year that you are compounding (here, 12 because monthly), and t is the number of years (4).

Plugging in all of the numbers gets you the following relation with respect to rate:

` `

`A = 500(1+r/12)^(12*4)`

`A = 500(1+r/12)^48`

We can't very easily simplify this further, but we can use it to find the graph:

The red line is y = 600. The black line is the graph of our equation. The x-axis (rate) starts at 0.00 and goes to 0.05 (0% - 5%). The y-scale starts at 0 and goes to 700. Where the two lines intersect is where we would expect the correct rate. Based on the graph, it looks to be around 0.045 or 4.5%. Let's make a table of values around 4.5% to see if we are correct. We will plug these values in to our relationship between A and the rate and examine the result.

Rate Result in 4 years

4.3% $593.66

4.4% $596.03

4.5% $598.41

4.6% $600.80

4.7% $603.20

It looks like our graph estimate was a bit off! Clearly, if you wanted $600, you would want at least **4.6% interest** on your money.

As a side note, you can solve exactly for the correct rate. Follow the steps below to solve algebraically with only the estimate at the end limited by your calculator's precision.

`600 = 500 (1+r/12)^48`

`6/5 = (1+r/12)^48`

`root(48)(6/5) = 1+r/12`

`root(48)(6/5) - 1 = r/12`

`r = 12(root(48)(6/5) - 1) ~~ 0.045667 = 4.5667%`

` `Looks like we weren't too far off with **4.6%** from the table!

thank you very much