Question 867641: Cyril invested part of 185000 at 7% and the rest at 9%. If the interest from the 9% investment is 1450 more than the interest earned at the 7% investment, how much did he invest at the 9% rate?
Answer by checkley79(3341) (Show Source):
You can put this solution on YOUR website! .07x=.09(185,00-x)-1,450
.07x=16,650-.09x-1,450
.07x+.09x=16,650-1,450
.16x=15,200
x=15,200/.16
x=95,000 invested @ 7%.
185,000-95,000=90,000 invested @ 9%
Prof:
.07*95,000=.09*90,000-1,450
6,650=8,100-1,450
6,650=6,650
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