Results 1 to 3 of 3

Math Help - can you help me with this exercise?

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    4

    can you help me with this exercise?

    how much I must deposit now in a 12% account (annual interest) if I want to be able to withdraw $20 000 at the end of each of the next five years.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by gusty View Post
    how much I must deposit now in a 12% account (annual interest) if I want to be able to withdraw $20 000 at the end of each of the next five years.
    are we doing simple interest here?

    If so, use I = PRT

    where I is the simple interest, P is the principal, R is the rate of interest, and T is the time

    at the end of 5 years, I = P(0.12)(5) = 0.6P

    so, at the end, we want to have the amount we invested, P, with the interest, I, to be $20,000

    hence we want, P + I = 20000
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,716
    Thanks
    634
    Hello, gusty!

    How much I must deposit now in a 12% account (annual interest)
    if I want to be able to withdraw $20,000 at the end of each of the next five years?
    There is a formula for this very problem, but we can derive the formula ourselves
    . . if we're willing to invest some time and effort.


    Let P = amount of our initial deposit.

    End of year-1: its value grows to 1.12P dollars,
    . . and $20,000 is withdrawn.
    The balance is: . 1.12P - 20,\!000

    End of year-2: the value grows to (1.12)(1.12P - 20,\!000),
    . . and $20,000 is withdrawn.
    The balance is: . 1.12^2P  - 1.12(20,\!000) -20,\!000

    End of year-3: the value grows to 1.12\left[1.12^2P - 1.12(20,\!000) - 20,\!000\right]
    . . and $20,000 is withdrawn.
    The balance is: . 1.12^3P - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000

    End of year-4: the value grows to 1.12\left[1.12^3P - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000\right]
    . . and $20,000 is withdrawn.
    The balance is: . 1.12^4P - 1.12^3(20,\!000) - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000

    End of year-5: the value grows to 1.12\left[1.12^4 - 1.12^3(20,\!000) - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000\right]
    When the final $20,000 is withdrawn, the balance is zero.

    . . 1.12^5P - 20,\!000(1.12^4) - 20,\!000(1.12^3) - 20,\!000(1.12^2) - 20,\!000(1.12) - 20,\!000 \;=\;0

    . . 1.12^5P \;=\;20,\!000\underbrace{\bigg[1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1\bigg]}_{\text{Geometric series}} .[1]

    On the right, we have a geometric series with: . a = 1,\;r = 1.12,\;n = 5

    . . Its sum is: . S \:=\:\frac{1.12^5-1}{1.12-1}

    Substitute into [1]: . 1.12^5P \;=\;20,\!000\,\frac{1.12^5-1}{0.12} \quad\Rightarrow\quad P \;=\;20,\!000\,\frac{1.12^5-1}{0.12(1.12^5)}

    Therefore: . P \;=\;72,\!095.52405 \;\approx\;\$72,\!095.52

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. An Exercise
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: January 15th 2011, 09:47 AM
  2. Set exercise
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 19th 2010, 09:37 AM
  3. Exercise
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 25th 2009, 03:59 AM
  4. Help with an exercise
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 10th 2009, 11:35 PM
  5. can you help me with this exercise?
    Posted in the Business Math Forum
    Replies: 0
    Last Post: November 8th 2008, 03:18 AM

Search Tags


/mathhelpforum @mathhelpforum