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Question 1000209: 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 3960 dollars.
How much did he invest at each rate?
Found 3 solutions by mananth, ikleyn, josgarithmetic:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! investment in 10% --------x
Investment in 6% ----------2x
10%x+6%(2x) = 3960
multiply by 100
10x+12x=396000
22x=396000
x= 396000/22
x=18000
10000 @ 10%
20,000 @6%
Answer by ikleyn(53742)  (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 3960 dollars.
How much did he invest at each rate?
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The answer in the post by @mananth, 10,000 at 10% and 20,000 at 6%, is incorrect
I came to bring a correct solution.
investment in 10% --------x
Investment in 6% ----------2x
10%x+6%(2x) = 3960
multiply by 100
10x + 12x = 396000
22x = 396000
x = 396000/22
x=18000
ANSWER. 18,000 at 10% and 36,000 at 6%
CHECK. 0.1*18000 + 0.06*(2*18000) = 3960. ! correct !
Solved correctly.
Answer by josgarithmetic(39790)  (Show Source):
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