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Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest.
They reported that the annual earnings from both investments were the same amount that would have been earned
by the total loan if it had been invested at 8%. Find the amount loaned at each rate.
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The solution in the post by @mananth is incorrect,
since the governing equation was setup incorrectly in his post.
I came to bring a correct and accurate solution.
Let x be the amount invested at 9% interest.
Then (36000-x) dollars invested at 6%.
Write the total interest equation
0.09x + 0.06*(36000-x) = 0.08*36000.
Simplify and find x
0.09x + 0.06*36000 - 0.06x = 0.08*36000,
0.09x - 0.06x = 0.08*36000 - 0.06*36000,
0.03x = 0.02*36000
x = = 24000.
ANSWER. $24000 was invested at 9% and the rest, 36000-24000 = 12000 dollars, was invested at 6%.
CHECK. The total interest is 0.09*24000 + 0.06*12000 = 2880 dollars.
Calculated by another way, it is 0.08*36000 = 2880 follars, the same amount.
The solution is confirmed to be correct.
Solved correctly.