SOLUTION: Roberto Invested a total of $10,000 apart at 6% in part and 9% how much did you invest in the lower interest account if the total interest earned in one year was $810?

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Question 1122512: Roberto Invested a total of $10,000 apart at 6% in part and 9% how much did you invest in the lower interest account if the total interest earned in one year was $810?
Found 2 solutions by Shin123, greenestamps:
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
Let a be the amount at 6% and b be the amount at 9%.
system%280.06a%2B0.09b=810%2Ca%2Bb=10000%29
system%280.02a%2B0.03b=270%2Ca%2Bb=10000%29
system%28a%2B1.5b=13500%2Ca%2Bb=10000%29 Subtract the 2 equations to get
0.5b=3500
highlight%28b=7000%29 That means a equals 10000-7000=3000
system%28a=3000%2Cb=7000%29 Check: 3000*0.06=180 7000*0.09=630 180+630=810. Correct!

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a really fast way to solve this kind of problem, without algebra.

(Of course, if a solution using algebra is required, this method won't do....)

(1) $810 return on a $10,000 investment is 8.1%.
(2) 8.1% is 7/10 of the way from 6% to 9%. (Plot the numbers on a number line if you need help seeing this).
(3) Therefore 7/10 of the $10,000 must be invested at the higher rate.

Answer: $7000 at 9%; so $3000 at 6%.