SOLUTION: The annual interest on an $20,000 investment exceeds the interest earned on a $16,000 investment by $380. The $20,000 is invested at a 0.3% higher rate of interest than the $16,00
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Question 1049110: The annual interest on an $20,000 investment exceeds the interest earned on a $16,000 investment by $380. The $20,000 is invested at a 0.3% higher rate of interest than the $16,000. What is the interest rate of each investment?
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
Let x be the interest rate for the 20,000 and y for the 16,000
The problem says:
x20,000 = y16,000+380 (1)
and
x = y+0.003 Substitute this value for x on equation (1)
>>>>>>>> >>>>>>>>>> >>>>>>>>>
(y+0.003)20,000 = y16,000+380
20,000y+60 = y16,000+380
4,000y = 320
y = 0.08 So the 16,000 are invested at 8%
And since x = y+0.003:
x = .08+0.003 = 0.083 or 8.3%. The 20,000 are invested at 8.3%
:
John
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