Results 1 to 6 of 6

Math Help - compound interest

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    18

    compound interest

    Mary wants to take out a loan. Suppose she can afford to make monthly payments of 200 dollars and the bank charges interest at an annual rate of 5 percent, compounded monthly. What is the maximum amount that Mary could afford to borrow if the loan is to be paid off eventually?

    I am not sure how to do a problem with "eventually."
    I have tried to approach this like a regular compound interest problem, but it is not working.

    Yn = Yo (1 + i )^n

    b / (1-a) = 200 / (1 - 1.041667) = -4799.96

    i = 0.5 / 12 = 0.041667
    n = ????

    -4799.96 + (Yo - (-4799.96)(1.041667^n) = ????

    How can I do this without knowing how much she took out, or how long it will take to pay off?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by Babs0201 View Post
    Mary wants to take out a loan. Suppose she can afford to make monthly payments of 200 dollars and the bank charges interest at an annual rate of 5 percent, compounded monthly. What is the maximum amount that Mary could afford to borrow if the loan is to be paid off eventually?

    I am not sure how to do a problem with "eventually."
    I have tried to approach this like a regular compound interest problem, but it is not working.

    Yn = Yo (1 + i )^n

    b / (1-a) = 200 / (1 - 1.041667) = -4799.96

    i = 0.5 / 12 = 0.041667
    n = ????

    -4799.96 + (Yo - (-4799.96)(1.041667^n) = ????

    How can I do this without knowing how much she took out, or how long it will take to pay off?
    i = \frac{0.05}{12}

    What you want is to figure out what loan generates $200 of interest over the course of a month. So L + 200 = L(1 + \frac{0.05}{12}).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    18

    Smile

    Thank you for your time!

    so I am getting 47961.6 for the amount she can borrow.
    I solved for L.
    L + 200 = L + 0.00417L

    200 / 0.00417 = 47961.6
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by Babs0201 View Post
    Thank you for your time!

    so I am getting 47961.6 for the amount she can borrow.
    I solved for L.
    L + 200 = L + 0.00417L

    200 / 0.00417 = 47961.6
    Now, the minor technicality is that if she borrows exactly that much money, she'll never pay it off. Anything less will suffice (assuming she lives long enough)!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,814
    Thanks
    703
    Hello, Babs!

    Icemanfan has an elegant solution . . . impressive!

    I deliberately did this "the long way" . . .


    Mary wants to take out a loan.
    Suppose she can afford to make monthly payments of $200 dollars
    and the bank charges 5% annually, compounded monthly.
    What is the maximum amount that Mary could afford to borrow
    if the loan is to be paid off eventually?

    This is an Amortization problem.

    Formula: . A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n-1}\quad \text{where: }\;\begin{Bmatrix}A & = & \text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}


    We have: . A = \$200,\;i \,=\,\frac{5\%}{12} \,=\,\frac{1}{240}

    Then:. . 200 \;=\;P\,\frac{\frac{1}{240}\left(1 + \frac{1}{240}\right)^n}{\left(1 + \frac{1}{240}\right)^n - 1} \quad\Rightarrow\quad 200\left[\left(1 + \frac{1}{240}\right)^n - 1\right] \;=\;\frac{P}{240}\left(1 + \frac{1}{240}\right)^n

    . . . . 48,\!000\left(1 + \frac{1}{240}\right)^n - 48,000 \;=\;P\left(1 + \frac{1}{240}\right)^n

    . . . . 48,\!000\left(1 + \frac{1}{240}\right)^n - P\left(1 + \frac{1}{240}\right)^n \;=\;48,000

    Factor: . \left(1 + \frac{1}{240}\right)^n(48,\!000 - P) \;=\;48,\!000 \quad\Rightarrow\quad \left(1 + \frac{1}{240}\right)^n \;=\;\frac{48,\!000}{48,\!000-P}


    We have the form: . X^n \;=\;\frac{48,\!000}{48,\!000-P}

    Since the left side is positive, we see that P < 48,\!000


    Therefore, she can borrow any amount less than $48,000.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Of course, as P approaches $48,000, the problem gets silly.

    If she borrowed $47,990, we have: . \left(\frac{241}{240}\right)^n \:=\:\frac{48,\!000}{10} \:=\:4,\!800

    Take logs: . \ln\left(\frac{241}{240}\right)^n \:=\:\ln(4,\!800) \quad\Rightarrow\quad n\!\cdot\!\ln\left(\frac{241}{240}\right) \;=\;\ln(4,\!800)

    Hence: . n \;=\;\frac{\ln(4,\!800)}{\ln\left(\frac{241}{240}\  right)} \;=\;2038.564336\text{ months}\;\approx\;170\text{ years}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2008
    Posts
    18
    Thank you so much. It really helped.

    By looking at your solutions I figured out a shorter way to do it.

    simply take the interest 0.05 and divid it by the months (12) and multiply it by the unknown amout she is borrowing K. Set it equal to 200. so...

    (0.05/12)*k = 200

    k = 48000, which is the amount she will never be able to pay off with the amount she is paying now!!!!

    Thanks again for helping to explain this!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compound interest at changing interest rates
    Posted in the Business Math Forum
    Replies: 2
    Last Post: October 21st 2010, 04:55 AM
  2. compound interest
    Posted in the Business Math Forum
    Replies: 11
    Last Post: March 12th 2010, 08:05 AM
  3. Compound Interest
    Posted in the Business Math Forum
    Replies: 4
    Last Post: November 24th 2009, 06:18 PM
  4. Compound interest
    Posted in the Business Math Forum
    Replies: 1
    Last Post: October 31st 2009, 08:14 PM
  5. Compound Interest
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 6th 2009, 04:24 AM

Search Tags


/mathhelpforum @mathhelpforum