A group of people planned to rent?

rent large house

A group of people planned to rent a large beach house for a weekend trip. Each person was to share the 800 dollar cost equally. However, two people were unable to go and this increased the cost for each person by 20 dollars. how many persons were in the original group?

I need help with how to set this type of problem up. any feedback would be greatly appreciated! ty.

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January 30th, 2010


3 Responses to “A group of people planned to rent?”

  1. manjyomesando1 Says:
    February 1st, 2010 at 4:44 pm

    800 = xc = (x-2)(c+20)
    xc = xc – 2c + 20x – 40
    2c = 20(x – 2)
    c = 10(x – 2) = 800/x
    x(x – 2) = 80
    x^2 – 2x – 80 = 0
    (x – 10)(x + 8) = 0
    x > 0 no de, x = 10

    answer = 10, and costed each $80, but with only 8 of them going, it raises to $100.

  2. kevin_c_k Says:
    February 2nd, 2010 at 8:30 am

    10 were in the original group… 8 (of course) in the new group.

  3. Lisa T Says:
    February 4th, 2010 at 6:38 am

    Let x = the original number of people.

    Then the cost c = 800/x

    When the two people are unable to go then c + 20 = 800/(x -2)

    substitute c= 800/x in for c so that your equation only has one variable

    800/x + 20 = 800/(x-2) multiply by the common denominator
    x(x-2)

    800(x-2) +20x(x-2) = 800x distribute

    800x – 1600 + 20x^2 -40x = 800x subtract 800x from both sides

    20x^2 -40x -1600 = 0 divide by 20

    x^2 -2x -80 =0 factor
    (x – 10)(x + 8)= 0 solve
    x=10 or x=-8 you can’t have a negative amount of people

    so 10 people were orginally going.

    Check your work 800/10 = 80 dollars each

    800/8 = 100 dollars each which is 20 dollars more