SOLUTION: In investing $6,100 of a couple's money, a financial planner put some of it into a savings account paying 4% annual simple interest. The rest was invested in a riskier mini-mall de
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Question 1123940: In investing $6,100 of a couple's money, a financial planner put some of it into a savings account paying 4% annual simple interest. The rest was invested in a riskier mini-mall development plan paying 13% annual simple interest. The combined interest earned for the first year was $532. How much money was invested at each rate? Answer by ikleyn(52803) (Show Source):
Let x = how much money was invested at 13%.
Then the amount invested at 4% is (6100-x) dollars.
Interest from the 13% account = 0.13*x dollars.
Interest from the 4% account = 0.04*(6100-x).
Your equation is
interest + interest = total combined interest, or
0.13x + 0.04*(6100-x) = 532 dollars.
0.13x + 0.04*6100 - 0.04x = 532
(0.13 - 0.04)x = 532 - 0.04*6100
x = = 3200.
Answer. $3200 were invested at 13% and the rest, 6100-3200 = 2900 dollars were invested at 4%.
Check. 0.13*3200 + 0.04*2900 = 532 dollars. ! Correct !
Solved.
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It is a standard and typical problem on investments.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
