Question 1120535
<br>
By the traditional algebraic method....<br>
Let x = amount invested at 3%
then 48000-x = amount invested at 7.5%<br>
The total interest is 2880:<br>
{{{.03(x)+.075(48000-x) = 2880}}}<br>
Solving that equation (I leave it to you) gives the amount x invested at 3%.<br>
The fast, easy way to solve "mixture" problems like this that involve two "ingredients"....<br>
(1) The total yield on the investment is {{{2880/48000 = 0.06}}} 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.<br>
Answer: $16,000 at 3%, $32,000 at 7.5%.