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- anonymous

Can anyone do this??? The annual interest on a 19000 investment exceeds the interest earned on a 18000 investment by 204. the 19000 is invested at a .6% higher rater than the 18000. What is the interest rate of each investment?

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- anonymous

- katieb

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- anonymous

i messed this up somehow let me start again

- anonymous

ok. i'm quite confused

- anonymous

yeah you should be because i was being stupid. let me start over

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- anonymous

ok

- anonymous

put the rate of the 18000 investment at r
the rate of the 19000 investment is r+.006
we have to work with numbers, not percents

- anonymous

then you know that the return on 18000 is
\[18000\times r\] and the return on 19000 is
\[19000\times (r+.006)\] and you know that the second one is $204 more than the first, so
\[18000\times r +204=19000(r+.006)\] now we can solve for r

- anonymous

\[18000r+204=19000r + .006\times 19000\]
\[18000r + 204=19000 r + 114\]
\[204-114=19000r -18000r\]
\[90=1000r\]
\[r=\frac{90}{1000}=\frac{9}{100}=9\%\]

- anonymous

love to know where they got 9% interest , i would go invest there

- anonymous

i have to have the interest for both. do i add or subtract .6%

- anonymous

?

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