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Question 279618: 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.
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! 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
let x be invested at 6%
36000-x will be invested at 9%
0.06x=0.09(36000-x) ( the interests on both are the same.)
0.06x= 3240. 00- 0.09x
0.15x= 3240
x=$21600 at 6 %
$36000-$21600= $24400 at 9%
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
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.
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