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Question 91817: Hello,
I need help with this problem:
Rhonda invested $4725 for one year, part at 11% annual interest, and the balance at 8% annual interest. She earned twice as much interest from the 11% investment as from the 8% investment. How much did she invest at each rate?
Thank you
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Rhonda invested $4725 for one year, part at 11% annual interest, and the balance at 8% annual interest. She earned twice as much interest from the 11% investment as from the 8% investment. How much did she invest at each rate?
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Let amt invested at 11% be x ; interest on this is 0.11x dollars
Amt invested at 8% is "4725-x"; interest is 0.08(4725-x) = 378-0.08x dollars
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EQUATION:
11% int. = 2*8% int
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0.11x = 2*(378 - 0.08x)
0.11x = 756 - 0.16x
0.27x = 756
x = $2800 (amount invested at 11%)
4725-2800 = $1925 (amount invested at 8%)
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Cheers,
Stan H.
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