### 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?"

## 11 Comments:

Anyone got it yet?

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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.

Marty and Jordan

What we did was enter all possible combinations of 35 and 20 dollar ticket holders into excel. To figure out the revenue for each we put in =(A1*20)+(B1*35),A1 being 20 dollar tickets and B1 being 35 dollar tickets. We dragged it down and got the answer:

There were 22500 $20 ticket holders and 14000 $35 ticket holders.

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to find the awnser first i put in all the ways that people could have the $20 and $35 tickets and scrolled down with excell.....

To find the revenue i just multiplied the amount of people by the price

The awnse ended up being 22500 people had $20 tickets and 14000 people had $35

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

Let X= the nuber of $35 ticket holders.

36500-x= $20 ticket holders

Revenue= Price(number of tickets sold)

$35 TICKET HOLDERS

940000=(35(x))+(20(36500-x))940000=.35x+730000-20x

210000=15x/15

14000=x

$20 TICKET HOLDERS

365000-14000=22500

There will be 14000 $35 ticket holders and 22500 $20 ticket holders.

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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