A woman invests $31,000, part at 8% and the rest at 9 1/2% annual interest. If the 9 1/2% investment provides $582.50 more income than the 8% investment, how much is invested at each rate?

Question

College Algebra. Need help

Answer 1

Let $x be the amount invested at 8%, the rest being $31000-x.

After a year, the interests are 0.08x and 0.095(31000-x).

0.095(31000-x)-0.08x=582.50, 2945-0.095x-0.08x=582.50, 0.175x=2945-582.50=2362.50.

So x=2362.50/0.175=$13500, and 31000-13500=$17500.

$13500 at 8% and $17500 at 9½%.

Answer 2

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