SOLUTION: Larry Mitchell invested part of his $30,000 advance at 5% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $2,0

Question 1176000: Larry Mitchell invested part of his $30,000 advance at 5% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $2,060, find the amount invested at each rate.
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Using formal algebra....

let x = amount invested at 5%
then 30000-x = amount invested at 9%

The total interest, from x at 5% and (30000-x) at 9%, was 2060:

Solve using basic algebra... but the decimals make it a bit tedious.

For a quick and easy informal solution using logical reasoning and some mental arithmetic, compare the actual interest with the amounts that would have been made if the whole $30,000 had been invested at each rate.

$30,000 at 5% --> $1500
actual interest: $2060
$30,000 at 9% --> $2700

The actual interest of 2060 is 560/1200 = 56/120 = 7/15 of the way from 1500 to 2700 (if it helps, look at the three numbers on a number line....)

That means 7/15 of the total was invested at the higher rate.

ANSWER: 7/15 of $30,000, or $14,000, was invested at 9%; the other $16,000 at 5%.

CHECK: .09(14000)+.05(16000) = 1260+800 = 2060

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