algebra people!!, need help pleez |
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 Sep 6 2005, 08:09 PM
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#1
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 x___aka K0K0  Group: Member Posts: 466 Joined: Jan 2005 Member No: 89,266  |
Can someone please explain how to do this? I need a memory refresh because it slipped my mind over the summer.
A man invested $6000, part of it at 5% simple interest and the rest at 7% simple interest. If his annual interest income is $372, how much did he invest at each rate? |
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| *mipadi* |
Sep 6 2005, 08:22 PM
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#2
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Well, you know he invested $6000 at two rates, and he got $372 in interest. Let the amount at one rate be equal to x; the amount at the second rate, then, is 6000-x, because it's the total ($6000) minus the amount at the other rate (x). Then just solve for x. Your equation should look something like this:
$372 = .07x + .05(6000-x) |
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Sep 6 2005, 09:25 PM
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#3
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x___aka K0K0 Group: Member Posts: 466 Joined: Jan 2005 Member No: 89,266 |
Hmm.. alright. So is it $36.00 ?
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Sep 6 2005, 09:32 PM
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#4
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x___aka K0K0 Group: Member Posts: 466 Joined: Jan 2005 Member No: 89,266 |
Oops hold on, I`m not done...
that`s for 5% right? how do i figure out the 7%? |
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| *mipadi* |
Sep 7 2005, 02:03 PM
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#5
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Well, if you solve the expression in the parentheses, you will find that:
372 = .07x + 300 - .05x So now you just solve for x. Once you get x, you're not done; you know that x amount was invested at 7%, and 6000 (the total) was invested at 5%; so x was invested at 5%, and 6000 - x was invested at 7%. |
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