SOLUTION: Please only answer if you'll give final answer
Last year, Bill had
$20,000
to invest. He invested some of it in an account that paid
6%
simple interest per year, and he
Algebra.Com
Question 1104748: Please only answer if you'll give final answer
Last year, Bill had
$20,000
to invest. He invested some of it in an account that paid
6%
simple interest per year, and he invested the rest in an account that paid
5%
simple interest per year. After one year, he received a total of
$1170
in interest. How much did he invest in each account?
1. First account?:
2. Second account?:
Found 3 solutions by Alan3354, josgarithmetic, greenestamps:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Tutoring is not doing it for you.
It's showing you how to do it.
--------------
The "final answer" is your problem.
If you don't like that, request a refund.
Answer by josgarithmetic(39623) (Show Source): You can put this solution on YOUR website!
This works like a mixture problem.
x, how much in 6% rate account
20000-x, how much in 5% rate account
Amount of interest earned was $1170.
, equation to account for the two interest amounts.
-
-----------to invest in the 6% rate account.
------------to invest at the 5% rate.
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
As the other tutor said, this is essentially a mixture problem. You are mixing a 5% interest investment with a 6% interest investment and getting an overall interest rate of something between 5% and 6%.
Like any other mixture problem involving two parts, this can be solved by the method of alligation. The version of that method that I like to use goes like this:
(1) The overall interest rate is or 5.85%.
(2) 5.85% is 85% of the way from 5% to 6%.
(3) Therefore 85% of the total $20000 was invested at the higher rate.
So the amount invested at 6% is 85% of $20,000, or $17,000; the amount invested at 5% was $3000.
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