Question 1128120: Steve Devine invested $7200 for one year, part at 10% and the rest at 14% annual interest. His total interest for the year was $960. How much money did he invest at each rate?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
First an outline of the traditional algebraic solution method, which I'll let you finish.
Let x be the amount invested at 10%; then 7200-x is the amount invested at 14%.
The total interest of $960 is 10% of what he invested at 10%, plus 14% of what he invested at 14%:
The equation is easily solved to find the answer to the problem; but with the decimals and the variables the calculations are a bit tedious, and prone to error.
I'll let you finish the problem by that method.
Now here is a method using fractions in which the calculations are much simpler, and faster to perform. If you understand this method, you will get to the answer for this problem and many similar ones with far less work than with the algebraic method.
(1) If all the money were invested at 10%, the interest would be $720; if it were all invested at 14%, the interest would be $1008.
(2) The actual amount of interest, $960, is 5/6 of the way from $720 to $1008.
(1008-720 = 288; 960-720 = 240; 240/288 = 5/6).
(3) Therefore, 5/6 of the money was invested at the higher rate.
ANSWER: 5/6 of the $7200, or $6000, was invested at 14%; 1/6, or $1200, was invested at 10%.
CHECK: .14(6000) +.10(1200) = 840+120 = 960
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