Results 1 to 7 of 7

Math Help - help needed hope this belongs in this forum

  1. #1
    Newbie
    Joined
    May 2014
    From
    philly
    Posts
    4

    help needed hope this belongs in this forum

    not sure if enough information is given but here goes

    Movie theatre

    2013
    number of people attending in 2013 = x1
    avg ticket price = y1
    2013 revenues x1*y1 = 30.6Million

    2014
    number of people attending in 2014 = x2
    avg ticket price = y2
    2014 revenues x2* y2 = 28Million

    Revenue difference Difference 2014 -2013 =- 2.6 million
    avg ticket price decreased 1.2% in 2014 which contributed 340,000 to the 2.6 M decline
    attendance declined 7.1 in 2014 from 2013

    Solve for x1 and x2

    Thanks all for the help -- my brain is fried trying to solve this so I know 2 equations 2 unknowns but if you could help w/ the calculation I would be eternally grateful. I posted in advance forum also but will delete if someone can help me here
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,347
    Thanks
    900

    Re: help needed hope this belongs in this forum

    Quote Originally Posted by juggler View Post
    not sure if enough information is given but here goes

    Movie theatre

    2013
    number of people attending in 2013 = x1
    avg ticket price = y1
    2013 revenues x1*y1 = 30.6Million

    2014
    number of people attending in 2014 = x2
    avg ticket price = y2
    2014 revenues x2* y2 = 28Million

    Revenue difference Difference 2014 -2013 =- 2.6 million
    avg ticket price decreased 1.2% in 2014 which contributed 340,000 to the 2.6 M decline
    attendance declined 7.1 in 2014 from 2013

    Solve for x1 and x2

    Thanks all for the help -- my brain is fried trying to solve this so I know 2 equations 2 unknowns but if you could help w/ the calculation I would be eternally grateful. I posted in advance forum also but will delete if someone can help me here
    attendance declined 7.1 ... what? from 2013, percent? million?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2014
    From
    philly
    Posts
    4

    Re: help needed hope this belongs in this forum

    attendance declined by 7.1% from 2014 to 2013 -- so Im guessing x1= (1-7.1%)* x2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,347
    Thanks
    900

    Re: help needed hope this belongs in this forum

    Quote Originally Posted by juggler View Post
    attendance declined by 7.1% from 2014 to 2013 -- so Im guessing x1= (1-7.1%)* x2
    I'm not getting any solutions. I also wonder why they include the 340,000.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    539
    Thanks
    243

    Re: help needed hope this belongs in this forum

    I started to help on this same question posted under advanced algebra.

    The student did not say what he was studying so it is a little tricky trying to answer. The main point is that there are four unknowns and so a solution requires four equations, but sufficient information is available. The way the problem is worded it may be designed as an example of something that accountants call analysis of variance.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    539
    Thanks
    243

    Re: help needed hope this belongs in this forum

    There are in fact five equations, but I am having trouble showing that four are independent. I'll look at it more tomorrow.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    539
    Thanks
    243

    Re: help needed hope this belongs in this forum

    I find the subscript notation being used in this problem not as clear as possible.

    REVISED NAMING

    $x = 2013\ attendance.$

    $x + u = 2014\ attendance.$

    $y = 2013\ average\ price.$

    $y + v = 2014\ average\ price.$

    GIVENS

    $xy = 30.60.$

    $(x + u)(y + v) = 28.00 \implies xy + vx + uy + uv = 28 \implies vx + uy + uv = 28- xy = 28 - 30.6 = -\ 2.6.$

    $u = -\ 0.071x.$

    $v = -\ 0.012y.$

    $vx = -\ 0.340.$

    5 equations, 4 unknowns. Question: is system soluble?

    $vx + uy + uv = -\ 2.6 \implies -\ 2.6 = (-\ 0.012y)x + (-\ 0.071x)y + (-\ 0.071x)(-\ 0.012y) = xy(0.000852 -0.071 -0.012) = -\ 0.082148xy.$

    Using the first four equations we can reduce the system to two equations in two unknowns. Unfortunately, those two equations must be either dependent or inconsistent. If they are dependent, all is not lost because we have a fifth equation.

    $-\ 2.6 = -\ 0.082148xy \implies xy = \dfrac{-\ 2.6}{-\ 0.082148} \approx 31.65 \ne 30.6.$ They are in fact inconsistent. The problem is insoluble.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: May 3rd 2012, 06:51 PM
  2. I am hope that my question belongs here.
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: April 15th 2010, 12:06 AM
  3. Math Help Forum: My Last Hope
    Posted in the Advanced Math Topics Forum
    Replies: 5
    Last Post: February 21st 2010, 06:22 AM
  4. im not sure if this belongs here
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 9th 2008, 04:47 AM
  5. I think this belongs here??
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 31st 2007, 06:46 PM

Search Tags


/mathhelpforum @mathhelpforum