Question 1000209
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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


<U>ANSWER</U>.  &nbsp;&nbsp;&nbsp;&nbsp;18,000 at 10%  and  36,000 at 6%


<U>CHECK</U>.  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0.1*18000 + 0.06*(2*18000) = 3960.    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;! correct !


Solved correctly.