SOLUTION: suppose that 18,000 is invested in two types of bonds for one year. Part of the money is invested in a high-risk, high growth fund that earned 24% annual interest. The rest is inve

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: suppose that 18,000 is invested in two types of bonds for one year. Part of the money is invested in a high-risk, high growth fund that earned 24% annual interest. The rest is inve      Log On

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Question 236329: suppose that 18,000 is invested in two types of bonds for one year. Part of the money is invested in a high-risk, high growth fund that earned 24% annual interest. The rest is invested in a safe fund that earned 5% annual interest. If $2320 in interest is earned, how much was invested at each rate?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
18000 was invested.
interest earned was 2320

x = amount invested at 24%
y = amount invested at 5%

two equations you have to work with are:

x+y = 18000
.24x + .05y = 2320

solve for y in the first equation to get:

y = 18000 - x

replace y with 18000 - x in the second equation to get:

.24*x + .05*(18000-x) = 2320

remove parentheses to get:

.24*x + .05*18000 - .05*x = 2320

simplify and combine like terms to get

.19*x + 900 = 2320

subtract 900 from both sides to get:

.19*x = 2320 - 900 = 1420

divide both sides by .19 to get:

x = 7473.684211

replace x with 7473.684211 in x + y = 18000 to get:

7473.684211 + y = 18000

subtract 7473.684211 from both sides to get:

y = 18000 - 7473.684211 = 10526.31579

you have

x = 7473.684211
y = 10526.31579

x + y = 7473.684211 + 10526.31579 = 18000

.24x + .05y = .24*7473.684211 + .05*10526.31579 = 2320

Answer is confirmed.

Answer is:

You invest $7473.68 at 24% interest and you invest $10526.32 at 5% interest for one year to get $2320 interest by the end of the year.