Results 1 to 3 of 3

Math Help - Joe Needs A House

  1. #1
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Joe Needs A House

    Joe wants to buy a house but does not want to get a loan. The average price of the dream house is $500,000 and its price is growing at 5 percent per year. How much should Joe invest in a project at the end of each year for the next 5 years in order to accumulate enough money to buy his dream house with cash at the end of the fifth year? Assume the project pays 12 percent rate of return.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, magentarita!

    This is a messy one!


    Joe wants to buy a house but does not want to get a loan.
    The average price of the dream house is $500,000 and its price is growing at 5% per year.
    How much should Joe invest in a project at the end of each year for the next 5 years in order
    to accumulate enough money to buy his dream house with cash at the end of the 5th year?
    Assume the project pays 12% rate of return.
    The price of the house is now $500,000.
    Its price is increasing by 5% per year.
    In five years, its price will be: . 500,\!000(1.05)^5 \:\approx\;\$638,\!140.78
    This is the amount he wants to have at the end of five years.


    Let's walk through his investment plan.

    At the end of year 1, he invests x dollars.

    By the end of year 2, it has grown to 1.12x dollars
    . . and he deposits another x dollars.
    He has 1.12x + x\:=\:(1.12+1)x dollars in his account.

    By the end of year 3, it has grown to 1.12(1.12 + 1)x \:=\:(1.12^2 + 1.12)x dollars
    . . and he deposits another x dollars.
    He has (1.12^2 + 1.12)x + x \:=\:(1.12^2 + 1.12 + 1)x dollars in his account.

    By the end of year 4, it has grown to 1.12(1.12^2+1.12 + 1)x dollars
    . . and he deposits another x dollars.
    He has (1.12^2 + 1.12 + 1)x + x \:=\:(1.12^3 + 1.12^2 + 1.12 + 1)x dollars in his account.

    By the end of year 5, it has grown to 1.12(1.12^3+1.12^2+1.12+1)x dollars
    . . and he deposits another x dollars.
    He has (1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1)x dollars in his account.

    And this should equal his desired goal: $638,140.78.


    \text{We have: }\;\underbrace{(1+1.12+1.12^2+1.12^3+1.12^4)}_{\te  xt{geometric series}}x \;=\;638,\!140.78 .[1]

    The geometric series has first term a = 1, common ratio  r = 1.12, and n = 5 terms.
    . . Its sum is: . \frac{1.12^5-1}{1.12 - 1} \:=\:\frac{1.12^5-1}{0.12}

    Then [1] becomes: . \frac{1.12^5-1}{0.12}\,x \:=\:638,\!140.78 \quad\Rightarrow\quad x \:=\:\frac{0.12}{1.12^5-1}(638,\!140.78)


    Therefore: . x \;=\;100,\!449.5691 \;\approx\;\boxed{\$100,449.57}

    That is the amount he must invest yearly for 5 years to buy his dream house.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    ok....

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    This is a messy one!

    The price of the house is now $500,000.
    Its price is increasing by 5% per year.
    In five years, its price will be: . 500,\!000(1.05)^5 \:\approx\;\$638,\!140.78
    This is the amount he wants to have at the end of five years.


    Let's walk through his investment plan.

    At the end of year 1, he invests x dollars.

    By the end of year 2, it has grown to 1.12x dollars
    . . and he deposits another x dollars.
    He has 1.12+1)x" alt="1.12x + x\:=\1.12+1)x" /> dollars in his account.

    By the end of year 3, it has grown to 1.12^2 + 1.12)x" alt="1.12(1.12 + 1)x \:=\1.12^2 + 1.12)x" /> dollars
    . . and he deposits another x dollars.
    He has 1.12^2 + 1.12 + 1)x" alt="(1.12^2 + 1.12)x + x \:=\1.12^2 + 1.12 + 1)x" /> dollars in his account.

    By the end of year 4, it has grown to 1.12(1.12^2+1.12 + 1)x dollars
    . . and he deposits another x dollars.
    He has 1.12^3 + 1.12^2 + 1.12 + 1)x" alt="(1.12^2 + 1.12 + 1)x + x \:=\1.12^3 + 1.12^2 + 1.12 + 1)x" /> dollars in his account.

    By the end of year 5, it has grown to 1.12(1.12^3+1.12^2+1.12+1)x dollars
    . . and he deposits another x dollars.
    He has (1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1)x dollars in his account.

    And this should equal his desired goal: $638,140.78.


    \text{We have: }\;\underbrace{(1+1.12+1.12^2+1.12^3+1.12^4)}_{\te  xt{geometric series}}x \;=\;638,\!140.78 .[1]

    The geometric series has first term a = 1, common ratio  r = 1.12, and n = 5 terms.
    . . Its sum is: . \frac{1.12^5-1}{1.12 - 1} \:=\:\frac{1.12^5-1}{0.12}

    Then [1] becomes: . \frac{1.12^5-1}{0.12}\,x \:=\:638,\!140.78 \quad\Rightarrow\quad x \:=\:\frac{0.12}{1.12^5-1}(638,\!140.78)


    Therefore: . x \;=\;100,\!449.5691 \;\approx\;\boxed{\$100,449.57}

    That is the amount he must invest yearly for 5 years to buy his dream house.
    Fantastic work as always.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dimensions of a house
    Posted in the Geometry Forum
    Replies: 3
    Last Post: May 12th 2011, 02:47 PM
  2. A farmer and a house
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: August 27th 2010, 06:14 PM
  3. Is there a teacher in the house?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 6th 2009, 04:54 PM
  4. the House Problem
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: March 7th 2009, 05:20 PM
  5. House Points
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 11th 2007, 04:52 PM

Search Tags


/mathhelpforum @mathhelpforum