Results 1 to 5 of 5

Math Help - Word Problem

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    27

    Word Problem

    Is it ok if I ask another one? Its a different type of question.

    In a family there are two cars. In a given week, the first car gets an average of 20 MPG, and the second car gets 15 MPG. The two cars combined drive a total of 1000 miles in that week, for a total gas consumption of 60 gallons. How many gallons were consumed by each of the cars that week.

    and this is what I did:

    x= First car @ 20MPG
    y= Second car @ 15MPG

    x+y=60
    20x+15y=60
    -20x -20x

    15y=-20x+60
    /15 /15 /15

    y=-20/14x+4
    y=-4/3x+4

    20x+15(4/3x+4)=60
    20x+20x+60=60
    -60 -60
    20x+20x=0
    40x=0
    /40 /40
    x=40
    y=20
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by GeinoD View Post
    Is it ok if I ask another one?
    of course it's ok. it would be a forum if it wasn't

    Its a different type of question.

    In a family there are two cars. In a given week, the first car gets an average of 20 MPG, and the second car gets 15 MPG. The two cars combined drive a total of 1000 miles in that week, for a total gas consumption of 60 gallons. How many gallons were consumed by each of the cars that week.

    and this is what I did:

    x= First car @ 20MPG
    y= Second car @ 15MPG

    x+y=60
    20x+15y=60
    -20x -20x

    15y=-20x+60
    /15 /15 /15

    y=-20/14x+4
    y=-4/3x+4

    20x+15(4/3x+4)=60
    20x+20x+60=60
    -60 -60
    20x+20x=0
    40x=0
    /40 /40
    x=40
    y=20
    check your answers to see if they make sense.

    car x gets 20 MPG, so if it uses 40 gallons, it means it drove 20*40 = 800 miles

    car y gets 15 MPG, so if it uses 20 gallons, it means it drove 15*20 = 300 miles

    300 + 800 = 1100 miles. which is not what we are told they drove. so your solution can't be right.

    let's start over...

    Let x be the number of gallons used by the 20 MPG car
    Let y be the number of gallons used by the 15 MPG car

    clearly we have x + y = 60 .................(1)

    (ah, you got that equation! nice job!)

    also, for each gallon, car-x drives 20 miles, so the number of miles car-x drives is 20x, similarly, the number of miles car-y drives is 15y, since the cars drove a total of 1000 miles, we have

    20x + 15y = 1000 ....................(2)

    (aha! that's where you messed up)

    now continue
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2008
    Posts
    27
    ok, I got it


    x+y=60

    20x+15y=1000

    x+y=60
    -x -x
    y=-x+60

    20x+15(-x+60)=1000

    20x-15x+900=1000
    -900 -900
    5x=100
    /5 /5
    x=20

    20+y=60
    -20 -20
    y=40

    x=20
    y=40
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by GeinoD View Post
    ok, I got it


    x+y=60

    20x+15y=1000

    x+y=60
    -x -x
    y=-x+60

    20x+15(-x+60)=1000

    20x-15x+900=1000
    -900 -900
    5x=100
    /5 /5
    x=20

    20+y=60
    -20 -20
    y=40

    x=20
    y=40
    you got it, you're a genius!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2008
    Posts
    27
    Thank You very much for the help.
    I am so thankful I found this site.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. word problem - the snow plough problem?
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 13th 2011, 01:02 PM
  2. Replies: 3
    Last Post: January 2nd 2011, 08:20 PM
  3. 1 Word problem and 1 function problem
    Posted in the Algebra Forum
    Replies: 8
    Last Post: April 21st 2010, 08:01 AM
  4. Replies: 2
    Last Post: January 10th 2009, 05:49 AM

Search Tags


/mathhelpforum @mathhelpforum