Question 818417: A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 9900 dollars. How much did he invest at each rate?
Your answer is
Found 2 solutions by mananth, MathTherapy:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 10% ------------------$x
6%--------------------$2x
Total interest -----------9900
10%x+6%*2x=9900
0.1x+0.06x=9900
0.16x=9900
x=61875
Amount invested at 10%= 61875
amount invested at 6% = 123,750
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 9900 dollars. How much did he invest at each rate?
Your answer is
Amount invested at 10%: $
Amount invested at 6%: 2($45,000), or $
You can do the check!!
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