Not sure how to go about answering this homework question:
You invested $11,000 in 2 accounts paying 4% and 8% annual interest. If the total interest earned for the year was $760, how much was invested at (a) 4% and (b) 8%?
Thanks for the help!
Not sure how to go about answering this homework question:
You invested $11,000 in 2 accounts paying 4% and 8% annual interest. If the total interest earned for the year was $760, how much was invested at (a) 4% and (b) 8%?
Thanks for the help!
Let $\displaystyle n$ be the amount invested in the account with 4%. Therefore $\displaystyle 11000-n$ is invested in the account with 8%
$\displaystyle I_{4\%}(1) = (0.04n)^1 = 0.04n$
$\displaystyle I_{8\%}(1) = [(11000-n)(0.08)]^1 = 880-0.08n$
These totals are equal to $760. So we add them to give:
$\displaystyle 880-0.08n+0.04n = 760$
Solve for n which will be the answer to part a. To find the answer to part b subtract the value of n from 11,000
$\displaystyle -0.04n = -120$
$\displaystyle n = \$\,3000$
Therefore $3,000 is invested at 4% and $8,000 is invested at 8%
I am not 100% sure about the position of n regarding the power. In this case it's not important for the power 1 doesn't change anything