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Math Help - Annual Compounding Interest

  1. #1
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    Annual Compounding Interest

    Jane has $6 and Sarah has $8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.

    a-10.50
    b-They will never have the same amount of money
    c-17.23
    d-17.95

    I just want the answer.

    Thanks
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    Jane has $6 and Sarah has $8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.

    a-10.50
    b-They will never have the same amount of money
    c-17.23
    d-17.95

    I just want the answer.

    Thanks
    It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.

    6(1+.11)^t=8(1+.08)^t

    \log 6(1.11)^t=\log 8(1.08)^t

    Etc. and so forth to find that t \approx 10.4997388
    Last edited by masters; December 25th 2008 at 08:34 AM. Reason: I only solved for time (t) instead of amount (a). Thanks soroban.
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  3. #3
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    Hello, magentarita!

    I got a different result . . .


    Jane has $6 and Sarah has $8. Over the next few years,
    Jane invests her money at 11%. Sarah invests her money at 8%.
    When they have the same amount of money, how much will they have?
    Assume annual compounding interest, and round to the nearest cent.

    a)\;\$10.50 \qquad b)\;\text{never equal}\qquad c)\;\$17.23 \qquad d)\;\$17.95

    At the end of n years, Jane will have: . 6\left(1.11^n\right) dollars.
    At the end of n years, Sarah will have: . 8\left(1.08^n\right) dollars.

    So we have: . 6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}

    Take logs: . \ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)

    . . Hence: . n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\  frac{1.11}{1.08}\right)} \;=\;10.4997388


    They will have the same amount of money in about {\color{blue}10\tfrac{1}{2}} years.

    At that time, they will have: . 6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d  )}

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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    I got a different result . . .



    At the end of n years, Jane will have: . 6\left(1.11^n\right) dollars.
    At the end of n years, Sarah will have: . 8\left(1.08^n\right) dollars.

    So we have: . 6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}

    Take logs: . \ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)

    . . Hence: . n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\  frac{1.11}{1.08}\right)} \;=\;10.4997388


    They will have the same amount of money in about {\color{blue}10\tfrac{1}{2}} years.

    At that time, they will have: . 6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d  )}

    You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
    Last edited by masters; December 25th 2008 at 03:21 PM.
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  5. #5
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    yes...

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    I got a different result . . .


    At the end of n years, Jane will have: . 6\left(1.11^n\right) dollars.
    At the end of n years, Sarah will have: . 8\left(1.08^n\right) dollars.

    So we have: . 6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}

    Take logs: . \ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)

    . . Hence: . n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\  frac{1.11}{1.08}\right)} \;=\;10.4997388


    They will have the same amount of money in about {\color{blue}10\tfrac{1}{2}} years.

    At that time, they will have: . 6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\$17.95}\;\;{\color{red}(d  )}
    Yes, the answer is (d) and I want to thank you.
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  6. #6
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    ok..

    Quote Originally Posted by masters View Post
    It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.

    6(1+.11)^t=8(1+.08)^t

    \log 6(1.11)^t=\log 8(1.08)^t

    Etc. and so forth to find that t \approx 10.4997388
    I got the answer from Soroban but I thank you for your effort.
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  7. #7
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    Smile yes...

    Quote Originally Posted by masters View Post
    You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
    Don't feel bad. No one is perfect except God and He is not taking any math courses lately.
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