Question 993309: Part of $8,000 was invested by an investment club at 10% interest and the rest at 12%. If the annual interest income from these investments is $900, how much was invested at each rate?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Let's call the amount invested at 10% x and the amount at 12% y
x+y= 8000 And we can say that x= 8000-y
The income is:
.10x + .12y = $900
And we said that x= 8000-y, therefore:
.10(8,000 - y) + .12y = 900
800 - .10y+.12y= 900 Subtract 800 on both sides and add the y's
.02y= 100 Divide both sides by .02
y= 5000 This is the amount invested at 12%. And 8,000-5,000= 3,000 at 10%
Proof:
5000×.12= 600
3,000×.10= 300
600+300= 900 We have the right answer.
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