Question 1108863: A total of $3600 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $385, how much was invested at each rate?
6 %+
11%=
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Let "part of it" be x. So, x was invested at 6% or 0.06x
The remainder is invested at 11%, and that amount = 3600 - x
Put it all together:
0.06x + 0.11(3600 - x) = 385
0.06x + 396 - 0.11x = 385
-0.05x = -11 divide, and remember that -/- = + so when you divide you can ignore the - sign, just divide 11/0.05
x = 220 this is how much is invested at 6%
3600 - 220 = 3380 this is the amount invested at 11%
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Check:
220 * 0.06 = 13.2
3380 * 0.11 = 371.80
13.2 + 371.80 = 385 Correct.
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