Question 753066: I am not sure where to start with this problem.
John invested $5200 in three interest-bearing accounts, earning 5%, 8% and 9% annually. If John invested twice as much money at 5% as he did at 9% and earned $356 last year, how much did he invest at 8%?
Answer by MathTherapy(10555) (Show Source):
You can put this solution on YOUR website! I am not sure where to start with this problem.
John invested $5200 in three interest-bearing accounts, earning 5%, 8% and 9% annually. If John invested twice as much money at 5% as he did at 9% and earned $356 last year, how much did he invest at 8%?
Let the amount he invested at 8% be E
Let the amount he invested at 9% be N
Then the amount he invested at 5% = 2N
Therefore, E + N + 2N = 5,200 ----- E + 3N = 5,200 ----- eq (i)
Also, .08E + .09N + .05(2N) = 356 ---- .08E + .09N + .1N = 356 ---- .08E + .19N = 356 ---- eq (ii)
Solve this system of equations to determine E: the amount invested at 8%.
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