SOLUTION: How can $60,000 be invested, part at 11% annual simple interest and the remainder at 9% annual simple interest, so that the interest earned by the two accounts will be equal?
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-> SOLUTION: How can $60,000 be invested, part at 11% annual simple interest and the remainder at 9% annual simple interest, so that the interest earned by the two accounts will be equal?
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Question 906683: How can $60,000 be invested, part at 11% annual simple interest and the remainder at 9% annual simple interest, so that the interest earned by the two accounts will be equal?
The amount to invest at 11% is? $ Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! We know the total amount of money invested. $60000
x+y=60000,
We know that the difference in interest earned by the two accounts is $0
0.11*x-0.09*y=0
x=60000-y
We substitute for x
0.11*(60000-y)-0.09*y=0
We multiply out
6600-0.11y-0.09*y=0
We combine like terms.
6600=0.2*y
Isolate y
y=6600/0.2
y=33000 at 9%
Calculate x
x=60000-33000
x=27000 at 11%
Interest earned at 11% is 2970
Interest earned at 9% is 2970
We check
0.11*27000-0.09*33000=0
2970-2970=0
0=0
Since this statement is TRUE and neither amount is negative then all is well
