Math Homework Problem

Question

An executive invests $28,000, some at 8% and the rest at 5% annual interest. If he receives an annual return of $1,970, how much is invested at each rate?

Answer 1

Let a = the fraction invested at 8% and let b = the fraction invested at 5%

Then Sum1 @ 8% = 28000a(1.08)

Sum2 @ 5% = 28000b(1.05)

Then S1 + S2 = 29,970 (original sum + interest from both accounts)

So, a + b = 1

and 30,240a + 29,400b = 29,970

Solving for a and b we get a = .6786, b = .3214 meaning

Money invested at 8% = 28000a = 19,000

and money invested at 5% = 28000b = 9,000

Answer 2

Hi Ty,

This is one of my favorite questions to ask my algebra students!

So we have two accounts that I can invest a total of $28,000. Let x be the amount of money you invest in the account at 8% and y be the amount of money you invest in the account at 5%. This means that x + y = 28,000.

As far as the account goes, the formula for simple interest is I = Prt, where P is the amount invested into the account, r is the interest in decimal form, and t is the time in years.

For the 8% account, P = x, r = 0.08, and t = 1 since it is an annual interest. The interest for this account is

I = Prt

I = x(0.08)(1)

I = 0.08x

For the 5% account, P = y, r = 0.05, and t = 1 since it is an annual interest. The interest for this account is

I = Prt

I = y(0.05)(1)

I = 0.05y

The two interests from the account added together gives me a return of $1,970. So,

0.08x + 0.05y = 1970

Well we need a single variable to be able to solve this equation and x + y = 28,000. We can solve this equation for either x or y. Let's say we solve for y: y = 28,000 - x. Plugging this in,

0.08x + 0.05(28,000 - x) = 1970

0.08x + 1400 - 0.05x = 1970

0.03x + 1400 = 1970

0.03x = 570

x = 19,000 (the amount invested at 8%)

To find the amount invested at 5%, we need to find y:

y = 28,000 - 19,000

y = 9,000 (the amount invested at 5%)

Hope this helps! :)