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Question 1154741: In investing &5500 of a couples money, a financial planner put some of it into savings paying 6% annual simple interest. The rest was invested in a riskier minimall development plan paying 12% annual simple interest. The combined interest earned the first year was $522. How much money was invested at each rate
Found 3 solutions by josmiceli, MathTherapy, greenestamps:
Answer by josmiceli(19441)   (Show Source):
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
In investing &5500 of a couples money, a financial planner put some of it into savings paying 6% annual simple interest. The rest was invested in a riskier minimall development plan paying 12% annual simple interest. The combined interest earned the first year was $522. How much money was invested at each rate
Let amount invested at 6% be S
Then amount invested at 12% is: 5,500 - S
We then get: .06S + .12(5,500 - S) = 522
.06S + 660 - .12S = 522
.06S - .12S = 522 - 660
- .06S = - 138
S, or 
You should be able to find the amount invested at 12%
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a solution using the method I like best for "mixture" problems like this.
If you understand the method, and if an algebraic solution is not required, you will get to the answer faster and with much less work than using algebra.
(1) All $5500 invested at 6% would yield $330 interest; all at 12% would yield $660 interest.
(2) The actual interest of $522 is 32/55 of the way from $330 to $660. ($330 to $660 is a difference of $330; $330 to $522 is a difference of $192; 192/330 = 32/55.
(3) That means 32/55 of the total was invested at the higher rate.
ANSWER: 32/55 of $5500, or $3200, at 12%; the other $2300 at 6%.
CHECK: .12(3200)+.06(2300) = 384+138 = 522
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