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(13209)   (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
|
|
|
