Math

Tickets to a benefit are priced as follows: children $0.50, men $2.00, and women $3.00.
Find one combination in which 100 people may purchase tickets for $100. Children, men and women have to attend the event.

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  1. first multiply 3 by 30 to get 90. if 30 women go, that is $90, so you have to make $10 on men and children, so have 3 men go to get $6 more, and then you need $4 so divide that by $.50 and you get 8, so 8 children have to go.

    so you now have 30 women, 3 men, and 6 children.
    check your work.
    30 x $3 = $90
    3 x $2 = $6
    8 x $.50= $4

    90+6+4=100
    does it work?
    ~aShLeY

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  2. let the number of children be c
    let the number of men be m
    then the number of women = 100-c-m

    (1/2)c + 2m + 3(100-c-m) = 100
    c + 4m + 6(100-c-m) = 200
    c + 4m + 600 - 6c - 6m = 200
    -5c - 2m = -400
    2m + 5c = 400
    2m = 400 - 5c
    m = 200 - (5/2)c , where c > 40 or else m is > 100

    now pick any even value of c
    e.g. c = 50
    m= 200-125 = 75 , no good since we are already over 100

    let c = 70
    m = 200-(5/2)(70) = 25
    w = 100-25-70 = 5
    so 70 children, 25 men and 5 women

    let c = 72
    m = 20
    w = 8

    See if you can find some others.

    remember, the sum cannot exceed 100 people.

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  3. your answer of 30 women, 3 men and 6 children does not add up to 100 people.

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  4. Another way could be

    78 children
    14 women
    8 men

    Now how can I explain this to an eight year old?

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  5. Sorry Kathy, your last answer does not work
    It satisfies the 100 people condition but does not satisfy the $100 condition.
    78(.50) + 14(3) + 8(2) = $97

    The only way other than just guessing and checking is to set up equations like I did above.

    How to explain this to an eight-year old ????
    Wow, you got me there.

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  6. Thanks for your help!

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  7. Thanks for your help, Reiny!

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