Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - math problem

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    51

    math problem

    in the start of the year you put 20000 in your bankaccount that has a intrest of 4%. You lift every year 2500euros How much do you have left after 8 lifts?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    From
    Berkeley, Illinois
    Posts
    364
    Quote Originally Posted by Rambo View Post
    in the start of the year you put 20000 in your bankaccount that has a intrest of 4%. You lift every year 2500euros How much do you have left after 8 lifts?
    Let's do this in 2 parts. Since you requested a series, and not the annuity formulas, I'll do it that way for you.

    First, take the accumulated value of 20,000 each year at 4% for 8 years.

    Do your series with 20000*1.04^1 + 2500*1.04^2 + ......

    You should get 9.582795311

    Multiply the series by your 20000 payment, and you have accumulated 191655.9062 at time 8.

    That's part 1. For part 2, do the same, but with your 2500 payment instead.
    Try that and let me know.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    no i do not get 9.????

    can you write the whole thing again please
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    Quote Originally Posted by mathceleb View Post
    Let's do this in 2 parts. Since you requested a series, and not the annuity formulas, I'll do it that way for you.

    First, take the accumulated value of 20,000 each year at 4% for 8 years.

    Do your series with 20000*1.04^1 + 2500*1.04^2 + ......

    You should get 9.582795311

    Multiply the series by your 20000 payment, and you have accumulated 191655.9062 at time 8.

    That's part 1. For part 2, do the same, but with your 2500 payment instead.
    Try that and let me know.
    with lift i mean withdraw
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2008
    Posts
    51

    is this correct?

    200000*1.04^8=27371.38

    27371.38/(1/(1.04)^1+1/(1.04)^2+1/(1.04)^3+1/(1.04)^4+1/(1.04)^5+1/(1.04)^6+1/(1.04)^7+1/(1.04)^8))=28472.00641

    27371.38-28472.00641=1100.626

    Answer is 1100.626

    IS THIS CORRECT????
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2008
    From
    Berkeley, Illinois
    Posts
    364
    Quote Originally Posted by Rambo View Post
    200000*1.04^8=27371.38

    27371.38/(1/(1.04)^1+1/(1.04)^2+1/(1.04)^3+1/(1.04)^4+1/(1.04)^5+1/(1.04)^6+1/(1.04)^7+1/(1.04)^8))=28472.00641

    27371.38-28472.00641=1100.626

    Answer is 1100.626

    IS THIS CORRECT????
    I misunderstood your original question text.

    Because you take out payments at the end of the year, and you invested the 20000 at the beginning of the year, your investment is 20000*1.04^9

    The beginning of the 9th year is the end of the 8th year.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    Quote Originally Posted by mathceleb View Post
    I misunderstood your original question text.

    Because you take out payments at the end of the year, and you invested the 20000 at the beginning of the year, your investment is 20000*1.04^9

    The beginning of the 9th year is the end of the 8th year.
    you withdraw the money 8 times not 9 times!!!????

    Is my math correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Feb 2008
    From
    Berkeley, Illinois
    Posts
    364
    Quote Originally Posted by Rambo View Post
    you withdraw the money 8 times not 9 times!!!????

    Is my math correct?
    Rambo,

    Your initial deposit is at time 0. Beginning of Year. Your first deposit is at the end of year 1. Therefore, your initial deposit has already had 1 year of interest. So, when your last withdrawal at the end of year 8 occurs, your initial deposit at time 0 has already had 9 years to accumulate interest.

    Also, you are accumulating in this problem, not discounting. Therefore, your series;

    (1/(1.04)^1+1/(1.04)^2+1/(1.04)^3+1/(1.04)^4+1/(1.04)^5+1/(1.04)^6+1/(1.04)^7+1/(1.04)^8))= should not have 1/'s in it.

    it should be:

    (1.04)^1+(1.04)^2+(1.04)^3+(1.04)^4+(1.04)^5+(1.04 )^6+(1.04)^7

    Since your withdrawals are at the end of the month, your time is 1 less.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    I put the money to the account in THE BEGINNING of the year and 8 years forward i withdraw 2500euros.
    And if you take 2500euro every year 8 years forward form the account it is discounting!!!!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Feb 2008
    From
    Berkeley, Illinois
    Posts
    364
    Quote Originally Posted by Rambo View Post
    I put the money to the account in THE BEGINNING of the year and 8 years forward i withdraw 2500euros.
    And if you take 2500euro every year 8 years forward form the account it is discounting!!!!
    No, you are projecting both balances to end of year 8 with interest.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    Quote Originally Posted by mathceleb View Post
    No, you are projecting both balances to end of year 8 with interest.
    Yes and then counting them - eachother.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    please solve the whole thing using series so i see what you mean write out every calculation please.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    Quote Originally Posted by mathceleb View Post
    Rambo,

    Your initial deposit is at time 0. Beginning of Year. Your first deposit is at the end of year 1. Therefore, your initial deposit has already had 1 year of interest. So, when your last withdrawal at the end of year 8 occurs, your initial deposit at time 0 has already had 9 years to accumulate interest.

    Also, you are accumulating in this problem, not discounting. Therefore, your series;

    (1/(1.04)^1+1/(1.04)^2+1/(1.04)^3+1/(1.04)^4+1/(1.04)^5+1/(1.04)^6+1/(1.04)^7+1/(1.04)^8))= should not have 1/'s in it.

    it should be:

    (1.04)^1+(1.04)^2+(1.04)^3+(1.04)^4+(1.04)^5+(1.04 )^6+(1.04)^7

    Since your withdrawals are at the end of the month, your time is 1 less.

    how do you assume the withdrawals are at the end of month?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    Quote Originally Posted by mathceleb View Post
    No, you are projecting both balances to end of year 8 with interest.
    My opinion is that the withdrawals are 8 when it says so in the text!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Mar 2008
    Posts
    51
    so the answer is 20000*1.04^9=28466.236

    28466.236(1.04)^1+...+(1.04)^8=32028.10557

    28466.236-32028.10557=3561.90 is the answer????
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Math problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 7th 2010, 06:00 PM
  2. Math problem
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: March 2nd 2010, 07:36 AM
  3. Math Problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 19th 2009, 09:45 AM
  4. math problem
    Posted in the Algebra Forum
    Replies: 5
    Last Post: March 3rd 2009, 11:54 PM
  5. Math B problem (please help)
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 3rd 2005, 12:04 AM

Search Tags


/mathhelpforum @mathhelpforum