SOLUTION: Bob invested $24000. He invested part of it at 8% interest per year and the rest at 10% interest per year. His total interest income for the year is $2300. How much was invested a
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-> SOLUTION: Bob invested $24000. He invested part of it at 8% interest per year and the rest at 10% interest per year. His total interest income for the year is $2300. How much was invested a
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Question 1145315: Bob invested $24000. He invested part of it at 8% interest per year and the rest at 10% interest per year. His total interest income for the year is $2300. How much was invested at 10%?
Thanks Found 2 solutions by richwmiller, greenestamps:Answer by richwmiller(17219) (Show Source):
For an alternative to the traditional algebraic process for solving "mixture" problems like this, try this...:
(1) $24000 all invested at 8% would return $1920 interest; all at 10% would return $2400 interest.
(2) The actual interest amount, $2300, is 380/480 = 38/48 = 19/24 of the way from $1920 to $2400.
(3) Therefore 19/24 of the money was invested at the higher rate.
ANSWER: 19/24 of $24000, or $19000 at 10%; the rest at 8%.
