Creative solutions
As I just said in class, we don't use creativity often enough to solve 'typical' unit problems. Take a few minutes with the people in your group, and discuss electronically (either as comments or separate posts), the solution to a problem like this.....use any resources that you can find.
Record, in detail, your 'thought' process in solving this problem...Have fun!
"A playoff football game draws 36 500 fans. Depending on where their seats were, the ticket prices for the game were $35 or $20. The home
team realized ticket revenue of $940 000 for the game. How many of
each type of ticket were sold?"
posted by Ryan Maksymchuk @ 1:08 PM
5 Comments:
Anyone got it yet?
Laura and Caity:
First we identified the independent variable (x) as the # of $35 ticket holders.
If there were x # of $35 ticket holders, there had to have been 36 500 - x $20 ticket holders.
So... revenue = sales(price)
and in this case;
940 000 =(35x)+[(20)(36500-x]
Here is what our solution looks like:
1.Let x= # of $35 ticket holders
2.Let 36000-x= # of $20 ticket holders
3.R=S(P)
4.940 000=(35x)+[(20)(36500-x)]
5.940 000=35x + 730 000-20x
6.940 000=15x + 730 000
7.210 000 = 15x
------------ <--(divide);)
15
8.X = 14 000
. IF X= 14 000, THERE
. . ARE 14 000 $35 TICKET HOLDERS.
AND THERE ARE 36 500 - 14 000 WHICH EQUALS 22 500 $20 TICKET HOLDERS.
THAT GIVES YOU A TOTAL REVENUE OF $940 000.
I Let a represent= $35 dollar tickets
b represent= $20 dollar tickets
L1= a+b=36500
L2= 35a+20b=940000
so..
a=36500-b
35(36500-b)+20b=940000
1277500-35b+20b=940000
1277500-15b=940000
-15b=-337500
b=22500
so then if..
a=36500-b
=36500-22500
=14000
At the playoff football game $14,000 of $35 dollar tickets were sold and $22,500 of $20 dollar tickets were sold.
x represents= $35 tickets
y represents= $20 tickets
x+y=36500
x=36500-y
35(36500-y)+20y=940000
1277500-35y+20y=940000
1277500-15y=-337500
-15y=-337500
y=22500
x=36500-y
x=36500-22500
x=14000
Tyler & Mark
We made a simple formula on a piece of paper.
940,000=(a*20)+(b*35)
Then using the variables, we "guessed and checked" using the calculator.
First we cut the number of tickets in half and let one half of the tickets be $20 and the others $35.
This didn't give us the answer of couse but gave us a better idea of how many of each ticket there was.
Then we did the $20 tickets 1/3 of the total number of tickets and the $35 tickets at 2/3. We plugged in our new values into the vaiables.
This time is gave us even a better idea of how many tickets there were for each.
We contiued this until we got the right variables.
# of $35 tickets=14,000
# of $20 tickets=22,500
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