Results 1 to 4 of 4

Math Help - i need help with a word problem

  1. #1
    Newbie
    Joined
    Jun 2007
    From
    texas
    Posts
    6

    Unhappy i need help with a word problem

    a video store charges non-member $3 to rent a video. a membership cost $21 and then videos cost only $1.50 to rent. how many videos would you need to rent in order to jusify a membership?

    i know that you could rent 7 videos to equal to $21 which is enough for a membership. but i dont know if thats what they meant by justifying the membership. can someone help me?
    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 lilmama-14 View Post
    a video store charges non-member $3 to rent a video. a membership cost $21 and then videos cost only $1.50 to rent. how many videos would you need to rent in order to jusify a membership?

    i know that you could rent 7 videos to equal to $21 which is enough for a membership. but i dont know if thats what they meant by justifying the membership. can someone help me?
    here's how i interpret the problem. how many videos should a nonmember buy to make having a membership worthwhile in the first place. if say we were just going to rent one video and never rent from that store again, then it doesn't make sense to have a membership, since we would end up spending $22.50 to get the video with membership instead of just $3 without the membership. the same holds for 2, videos, and 3 and so on. we want to find how many videos we must rent so that becoming a member would end up cheaper.

    first let's find where the member and the nonmember spends the same amount of money.

    Let v be the number of videos rented.

    a nonmember will spend 3v dollars, since he pays $3 for each video.
    a member would pay 1.5v dollars, since he would pay $1.50 for each video, PLUS the membership fee of $21. so equating these two we get:

    3v = 1.5v + 21

    => 3v - 1.5v = 21 ..........get the v's on one side by subtracting 1.5v from both sides

    => 1.5v = 21 .................calculate the left hand side

    => v = 21/1.5 ...............divide both sides by 1.5

    => v = 14

    So at the 14th video rented, the member and non member pays the same amount of money in total. if we rent any more videos than this, it would be cheaper to be a member, since we would pay $1.50 for each additional video above 14, as opposed to $3 for each video above 14 if we are nonmembers.

    So renting more than 14 videos justifies the membership
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2007
    From
    Cambridge, UK
    Posts
    41
    Hi there, lilmama.

    Firstly, I think "justifying the membership" means:

    "How many videos must you rent, before it becomes cheaper to be a member?"

    So, thinking about this, let's do a little bit of maths. Say you rent N videos. How much did it cost?

    If you are not a member, you just pay $3 for every video.
    So, the total cost is $ 3 N.

    If you become a member first, you have to pay $21 first, then you pay
    So, the total cost is $ (1.5N + 21).

    We want to find a number of videos N such that the cost of becoming a member and renting the videos is less than the cost of renting them without membership.

    We can write this with an inequality:

    1.5N + 21 < 3N

    Let's change this equation, by doing the same thing to both sides. First subtract 1.5N:

    21 < 1.5N

    Now divide by 1.5:

    14 < N
    or N > 14.

    The answer is, you must rent 14 videos to justify membership.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2007
    From
    texas
    Posts
    6

    Smile

    Quote Originally Posted by Jhevon View Post
    here's how i interpret the problem. how many videos should a nonmember buy to make having a membership worthwhile in the first place. if say we were just going to rent one video and never rent from that store again, then it doesn't make sense to have a membership, since we would end up spending $22.50 to get the video with membership instead of just $3 without the membership. the same holds for 2, videos, and 3 and so on. we want to find how many videos we must rent so that becoming a member would end up cheaper.

    first let's find where the member and the nonmember spends the same amount of money.

    Let v be the number of videos rented.

    a nonmember will spend 3v dollars, since he pays $3 for each video.
    a member would pay 1.5v dollars, since he would pay $1.50 for each video, PLUS the membership fee of $21. so equating these two we get:

    3v = 1.5v + 21

    => 3v - 1.5v = 21 ..........get the v's on one side by subtracting 1.5v from both sides

    => 1.5v = 21 .................calculate the left hand side

    => v = 21/1.5 ...............divide both sides by 1.5

    => v = 14

    So at the 14th video rented, the member and non member pays the same amount of money in total. if we rent any more videos than this, it would be cheaper to be a member, since we would pay $1.50 for each additional video above 14, as opposed to $3 for each video above 14 if we are nonmembers.

    So renting more than 14 videos justifies the membership
    thank you so much for hlping me with that problem cause i didnt get it at all
    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