Results 1 to 3 of 3

Math Help - Setting up problem.

  1. #1
    Ash
    Ash is offline
    Junior Member
    Joined
    Nov 2006
    Posts
    56

    Setting up problem.

    I think I did problem #2 right, but I'm confused about problem #1.
    I find word problems confusing!!!

    #1 The sponsors of a Boys and Girls Club took some members to a play and bought seven more tickets for members than for adults. The members' tickets cost $1.25 and the adults' tickets cost $2.75. If a total of $28.75 was paid for all the tickets, how many members went to the play?


    #2 The total receipts for a college basketball game were $674, with a student ticket selling for $1.25 and a non-students ticket selling for 2.75. If there were 136 more students then non-students in attendance, how many people were at the game?


    2.75 + 1.25(n+136)=674

    2.75 + 1.25+170=674

    172.75 + 1.25n=674

    1.25n=501.25
    n=401

    There were 401 people at the game???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Ash View Post
    I think I did problem #2 right, but I'm confused about problem #1.
    I find word problems confusing!!!

    #1 The sponsors of a Boys and Girls Club took some members to a play and bought seven more tickets for members than for adults. The members' tickets cost $1.25 and the adults' tickets cost $2.75. If a total of $28.75 was paid for all the tickets, how many members went to the play?


    #2 The total receipts for a college basketball game were $674, with a student ticket selling for $1.25 and a non-students ticket selling for 2.75. If there were 136 more students then non-students in attendance, how many people were at the game?


    2.75 + 1.25(n+136)=674

    2.75 + 1.25+170=674

    172.75 + 1.25n=674

    1.25n=501.25
    n=401

    There were 401 people at the game???
    #2 The total receipts for a college basketball game were $674, with a student ticket selling for $1.25 and a non-students ticket selling for 2.75. If there were 136 more students then non-students in attendance, how many people were at the game?

    Umm, let us see.

    2.75 + 1.25(n+136)=674
    On that first term alone, the "275", you mean there is only 1 non-student in the game? Wrong, isn't it?
    That's okay, you're not good at it yet. You'd learn.

    Here is one way to do #2.

    What does the Problem asks us to find? The number of people at the game.
    Who are these people? Students and non-students.

    We don't know how many are each, so
    let x = number of students
    and y = number of non-students.

    "If there were 136 more students then non-students in attendance"
    So, x = y +136.
    Call that Eq. (1).

    "... with a student ticket selling for $1.25 and a non-students ticket selling for 2.75"
    So, if a student bought a ticket, that student paid $1.25. Hence, since there are x number of students at the game, then all these students paid x*1.25 = 1.25x dollars.
    Likewise, the y number of non-students paid y*2.75 = 2.75y dollars.

    "The total receipts for a college basketball game were $674,..."
    That means
    1.25x +2.75y = 674
    Call that Eq.(2)

    So we have 2 equations with two unknowns:
    x = y +136 --------------(1)
    1.25x +2.75y = 674 ------(2)

    It's easy now, eh?
    Let me continue, anyway.

    Substitute the value of x in (1) into (2),
    1.25(y +136) +2.75y = 674
    1.25y +170 +2.75y = 674
    1.25y +2.75y = 674 -170
    4y = 504
    y = 504 /4 = 126 non-students -----**
    So, in Eq.(1),
    x = y +136
    x = 126 +136 = 262 students --------**

    Therefore, there are 262 +126 = 388 people at the game. ------answer.

    ---------
    Check,
    262(1.25) +126(2.75) =? 674
    327.50 +346.50 =? 674
    674 =? 674
    Yes, so, OK.

    =============================================
    #1 The sponsors of a Boys and Girls Club took some members to a play and bought seven more tickets for members than for adults. The members' tickets cost $1.25 and the adults' tickets cost $2.75. If a total of $28.75 was paid for all the tickets, how many members went to the play?

    Find: number of members who went to play.

    Who played? Members and adults ( adults are non-members, I understand).
    So, let x = number of members who played.

    Umm, this Problem #1 is similar to the Problem #2 above. So just follow the procedure there.

    Let y = number of adults who played.

    As for the total of number of tickets bought,
    x = y +7 ---------------(3)

    As for the total cost of all the tickets bought,
    1.25x +2.75y = 28.75 --------(4)

    Since we want to find only the x actually, then we we find the y from (3), or we express the the y in terms of x, in Eq.(3),
    y = x -7
    Substitute that into 4),
    1.25x +2.75(x -7) = 28.75
    1.25x +2.75x -19.25 = 28.75
    1.25x +2.75x = 28.75 +19.25
    4x = 48
    x = 48/4 = 12 members

    Therefore, 12 members went to play. ------answer.

    ---------------------
    Check,
    y = x -7 = 12 -7 = 5 non-members.

    12(1.25) +5(2.75) =? 28.75
    28.75 =? 28.75
    Yes, so, OK.

    ================================
    Word problems are relatively easy to solve if you can interpret them correctly. Be patient and be "exact" with them. Then, check your answers against the meaning of the Problem, to be sure your answers are correct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Ash
    Ash is offline
    Junior Member
    Joined
    Nov 2006
    Posts
    56
    Thank you, for taking your time and explaining it very clearly!!!
    I understand a lot better now in how to set them up....They're still not my favorite thing to do, but hopefully with more practice they'll become a breeze.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Setting up problem.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: October 1st 2009, 03:58 PM
  2. Need help setting up this problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 20th 2009, 03:37 AM
  3. Need help setting up this problem....
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 4th 2008, 08:10 PM
  4. Help setting up problem.
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: January 25th 2007, 10:01 PM
  5. Help with setting up a problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 8th 2006, 11:00 AM

Search Tags


/mathhelpforum @mathhelpforum