# 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.
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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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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7. Thanks for your help, Reiny!

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