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Math Help - Another Word Problem

  1. #1
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    Another Word Problem

    Hi again once again I need help setting up the equation for this, it is similar in spirit to my last question but still I can't figure out how to set up the equation , please help.

    Bob has made $1980 annual interest from his two bank accounts. One of his bank accounts yields 6% interest and the other yields 9% interest , how much money would he have to have in each account?
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  2. #2
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    I tried to work this out all I can think of is  0.06x + 0.09y = 1980 but than that just seems useless, cause there’s nothing you can do any further.

    Anyone else?
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  3. #3
    Super Member Matt Westwood's Avatar
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    Unless there's something subtle that we've missed, this question can't be solved.

    Unless it means he made $1980 from each of his accounts, in which case the equations are:

    0.06x = 1980
    0.09y = 1980

    A solution to either of these would be a valid solution for the question where $1980 is the total, you'd assume that either x or y would be zero. Or the amount could be anywhere in between.
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  4. #4
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    Quote Originally Posted by Matt Westwood View Post
    Unless there's something subtle that we've missed, this question can't be solved.

    Unless it means he made $1980 from each of his accounts, in which case the equations are:

    0.06x = 1980
    0.09y = 1980

    A solution to either of these would be a valid solution for the question where $1980 is the total, you'd assume that either x or y would be zero. Or the amount could be anywhere in between.
    yeah I was thinking the same thing, maybe the op mis read the question
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  5. #5
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    Quote Originally Posted by Matt Westwood View Post
    Unless there's something subtle that we've missed, this question can't be solved.

    Unless it means he made $1980 from each of his accounts, in which case the equations are:

    0.06x = 1980
    0.09y = 1980

    A solution to either of these would be a valid solution for the question where $1980 is the total, you'd assume that either x or y would be zero. Or the amount could be anywhere in between.
    Yes I did not include all of the info as I don't have my book with me (at an internet cafe). I'm pretty sure that this was the original question , if Im wrong I apologize in advance for wasting your time and thank u for your help--would this be solvable??

    [Paraphrase]Bob invested $25,000 into two bank accounts which gave him a combined annual interest of $1980. One account gave him a 6% annual return and the other gave him a 9%. How much would he have to have put in each of his accounts in order to get this amount of interest?
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  6. #6
    Member courteous's Avatar
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    C_1*1.06 + (25000-C_1)*1.09=1980+25000
    C_1*1.06 + 25000*1.09-C_1*1.09=26980
    C_1*0.03=25000*1.09-26980
    C_1*0.03=270
    C_1=9000
    C_2=16000

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  7. #7
    A riddle wrapped in an enigma
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    Quote Originally Posted by courteous View Post
    C_1*1.06 + (25000-C_1)*1.09=1980+25000
    C_1*1.06 + 25000*1.09-C_1*1.09=26980
    C_1*0.03=25000*1.09-26980
    C_1*0.03=270
    C_1=9000
    C_2=16000

    Here's another look at the "setup"

    Let x = amount invested at 6%
    Let 25000 - x = amount invested at 9%

    .06x + .09(25000 - x) = 1980

    6x + 225000 - 9x = 198000

    -3x = -27000

    x = $9,000 (amount invested at 6%0

    $25,000 - $9,000 = $16,000 (amount invested at 9%)
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