Glorietta G.
asked • 02/04/23interest and rates
$11,335 is invested, part at 9% and the rest at 6%. If the interest earned from the amount invested at 9% exceeds the interest earned from the amount invested at 6%6% by $865.35, how much is invested at each rate?
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1 Expert Answer
Raymond B. answered • 02/04/23
Tutor
5
(2)
Math, microeconomics or criminal justice
.09N - .07S = 865.35
.09N -.07(11335-N) = 865.35
.09N+.07N = 865.35 + .07(11335)
.16N = 865.35 + 793.45
N = 1658.8/.16= $10,367.50 invested at Nine percent
S = 11335-10367.5= $967.50 invested at Seven percent
check the answer:
.09(10367.5)- .07(967.5)
=933.075- 67.725
= $865.35
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Patrick F.
Here is an answer from chatGPT: "Let x be the amount invested at 9%, then 11,335 - x is invested at 6%. The interest earned at 9% is 0.09x, and the interest earned at 6% is 0.06 * (11,335 - x). The difference between the two interest amounts is 865.35, so we can write the equation: 0.09x - 0.06 * (11,335 - x) = 865.35 Expanding and solving for x, we get: 0.09x - 677.01 = 865.35 0.09x = 1542.36 x = 17,135.11 So, 17,135.11 is invested at 9% and 11,335 - 17,135.11 = -5,799.11 is invested at 6%. Since this result is negative, it's not possible in this context, which means there is a mistake in the problem statement or in the calculation." Now, I can tell you that both answers you received are wrong. Can you find out what is wrong with each of them?02/04/23