SOLUTION: Bob invested some money at 5% simple interest and some at 9% simple interest. The amount invested at the higher rate was twice the amount invested at the lower rate. If the total

Algebra ->  Finance -> SOLUTION: Bob invested some money at 5% simple interest and some at 9% simple interest. The amount invested at the higher rate was twice the amount invested at the lower rate. If the total      Log On


   

Question 190386This question is from textbook algebra for college students
: Bob invested some money at 5% simple interest and some at 9% simple interest. The amount invested at the higher rate was twice the amount invested at the lower rate. If the total interst on the investments for 1 year was $920, then how much did he invest at each rate?
I know what the answer is but I have tried every forumla and can't figure out where the numbers come from. The anser is $4000 @ 5% and $8000 @ 9%. Your help is greatly appreciated! This question is from textbook algebra for college students

Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!


Let x = amount inv. @ 5%
And, 2x = amount for 9% (twice that of lower rate 5%)

Then it follows,
Interest on 5% + Interest on 9% = $920
0.05x%2B0.09%282x%29=920
0.05x%2B0.18x=920
0.23x=920 ----> cross%280.23%29x%2Fcross%280.23%29=cross%28920%294000%2Fcross%280.23%29
x = $4,000.00 , amount for 5%

Also, 2*($4,000) = $8,000.00 , amount for 9%


Let's check,
0.05%284000%29%2B0.09%288000%29=920
200%2B720=920
920=920

Thank you,
Jojo