Results 1 to 4 of 4

Math Help - Help with a algebra word problem!!!!

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    babbitt
    Posts
    3

    Help with a algebra word problem!!!!

    the problem is:
    Cycle Center has bicycles and tricycles in the storeroom.There are at least two of each,and there are more bicycles then tricycles.There are 23 wheels altogether.How many bicycles and tricycles are in the storeroom?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Oct 2012
    From
    Maryland
    Posts
    22
    Thanks
    4

    Re: Help with a algebra word problem!!!!

    You know you have the equation: 23 = 3*Tricycles + 2*Bicycles
    Since you know each has at least two, take these out.
    This leaves you with:
    13 = 3*(Remaining Tricycles) + 2*(Remaining Bicycles)
    Since the number is small, trial and error works best. You have 2 possible combinations to make 13.
    1. 1 tricycle, 5 bicycles
    2. 3 tricycles, 2 bicycles

    Since we know there are more bicycles, option number 1 is correct. Now add back in the first 2, leaving you with:
    3 tricycles and 7 bicycles.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,864
    Thanks
    744

    Re: Help with a algebra word problem!!!!

    Hello, kellym!

    This problem is so small, we can solve it without Algebra.


    Cycle Center has bicycles and tricycles in the storeroom.
    There are at least two of each, and there are more bicycles then tricycles.
    There are 23 wheels altogether.
    How many bicycles and tricycles are in the storeroom?

    Let B = number of bicycles: B \ge 2.
    Let T = number of tricycles: T \ge 2.
    . . And: B > T.


    B bicycles have 2B wheels.
    T tricycles have 3T wheels.
    . . There are 23 wheels: . 2B + 3T \:=\:23

    Solve for B\!:\;B \:=\:\frac{23-3T}{2} \;=\;11 - T - \frac{T-1}{2}

    From the fraction, we see that T must be odd.

    There are only four possible cases:

    . . \begin{array}{ccc} \text{Case} & T & B \\ \hline (1) & 1 & 10 \\ (2) & 3 & 7 \\ (3) & 5 & 4 \\ (4) & 7 & 1\end{array}


    Since T \ge 2, we can eliminate case (1).
    Since  B > T, we can eliminate cases (3) and (4).

    Therefore: . T = 3,\:B = 7
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2012
    From
    babbitt
    Posts
    3

    Re: Help with a algebra word problem!!!!

    Thank you soo much for your help !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Algebra word problem help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 14th 2012, 03:21 PM
  2. Help on algebra word problem.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 13th 2011, 04:29 PM
  3. Algebra Word Problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 25th 2010, 05:50 PM
  4. Algebra word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 22nd 2008, 04:57 PM
  5. algebra word problem
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 15th 2008, 08:11 AM

Search Tags


/mathhelpforum @mathhelpforum