Question 211194: A man invested $6,000, part of it at 5% simple intrest and the rest at 7% simple intrest. If his annual intrest income is $372, how much did he invest in each rate?
Answer by math problem solving(30)   (Show Source):
You can put this solution on YOUR website! Let x be the amount of money invested at 5%.
Then the amount invested at 7% is 6000 - x.
The interest earned from the money invested at 5% is x*5%
The interest earned from the money invested at 7% is (6000 - x)*7%
As the total interest earned from the two investments is given as $372, we have:
x*5% + (6000 - x)*7% = 372
Solving for x, we get
5x + 7(6000 - x) = 37200 (multiply both sides by 100)
5x + 42000 - 7x = 37200
-2x + 42000 = 37200
-2x = -4800
x = 2400
So $2,400 is invested at 5%, and $6,000 - $2,400 = $3,600 is invested at 7%.
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