SOLUTION: A combined total of $48,000 is invested in two bonds that pay 3% and 7.5% simple interest. The annual interest is $2,880.00. How much is invested in each bond?

Question 1120535: A combined total of $48,000 is invested in two bonds that pay 3% and 7.5% simple interest. The annual interest is $2,880.00. How much is invested in each bond?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

By the traditional algebraic method....

Let x = amount invested at 3%
then 48000-x = amount invested at 7.5%

The total interest is 2880:

Solving that equation (I leave it to you) gives the amount x invested at 3%.

The fast, easy way to solve "mixture" problems like this that involve two "ingredients"....

(1) The total yield on the investment is or 6%.
(2) The average return of 6% is 2/3 of the way from the lower rate of 3% to the higher rate of 7.5%: (6-3)/(7.5-3) = 3/4.5 = 2/3.
(3) Therefore 2/3 of the $48,000 (= $32,000) must have been invested at the higher rate.

Answer: $16,000 at 3%, $32,000 at 7.5%.
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