Question

Samantha invests `$1000` at `6%` per annum compounded quarterly. Mark invests `$1200` at `5.5%` per annum compounded quarterly. When will the balances in their accounts be equal?

Answer 1

The balances in the accounts will be equal in `37` years.

Explanation:

For problems such as these, we use the formula `A = P(1 + r/n)^(nt)`, where P is the amount of money you begin with, `r` is the rate of interest, `n` is the frequency the interest is compounded and A is the final amount. `t` is the time in years.

We have to write a system of equations and solve for `A` and for `t`.

Equation 1:

`A = 1000(1 + (0.06)/4)^(4t)`

Equation 2:

`A = 1200(1 + (0.055)/4)^(4t)`

Substitute:

`1200(1 + (0.055)/4)^(4t) = 1000(1 + (0.06)/4)^(4t)`

`1200(1.01375)^(4t) = 1000(1.015)^(4t)`

`1200(1.01375)^(4t) - 1000(1.015)^(4t) = 0`

Solve using a graphing calculator. If you're using a TI-83 or a TI-84, change the t's to x's. Enter `y_1 = 1200(1.01375)^(4x) - 1000(1.015)^(4x)` and `y_2 = 0`. Then press CALC`->` Intersect.

You will of course want a positive intersection. Once you have moved your cursor sufficiently, press ENTER twice before the calculator will say guess? followed by INTERSECTION.

The result it gives you should say `36.99`, or `37` years.

Hopefully this helps!