SOLUTION: Jolene invests her savings in two bank accounts, one paying 6 percent and the other paying 11 percent simple interest per year. She puts twice as much in the lower-yielding account
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Question 1139184: Jolene invests her savings in two bank accounts, one paying 6 percent and the other paying 11 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 8993 dollars. How much did she invest at each rate?
Let x = amount at 11%, in dollars.
Then the amount at 6% is 2x dollars.
The interest from the 11% amount is 0.11*x dollars.
The interest from the 6% amount is 0.06*(2x) = 0.12x dollars.
Your equation is
interest + interest = total interest, or
0.11*x + 0.12*x = 8993 dollars.
From the equation, express x and calculate the answer
x = = 39100.
Answer. The amount at 11% is $39100; the amount at 6% is twice of it, i.e. 2*39100 = 78200 dollars.
Check. 0.06*78200 + 0.11*39100 = 8993 dollars. ! Correct !
Solved.
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It is a typical and standard problem on investment.
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).
