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.
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.
your answer of 30 women, 3 men and 6 children does not add up to 100 people.
Another way could be
78 children
14 women
8 men
Now how can I explain this to an eight year old?
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.
Thanks for your help!
Thanks for your help, Reiny!
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