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## #1 2006-01-26 18:54:36

AA
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### AMC 12 2001 Problem

"A charity sells 140 benefit tickets for a total of \$2001. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?"

## #2 2006-01-26 21:31:50

MathsIsFun
Registered: 2005-01-21
Posts: 7,664

### Re: AMC 12 2001 Problem

P = Full Price
A = Tickets sold at full price
B = Tickets sold at half price

AP + B(P/2) = 2001
A+B=140, so B = 140-A

Substitute B = 140-A: AP + (140-A)(P/2) = 2001
Rearrange: AP + 140P/2 -AP/2 = 2001
Simplify: AP/2 + 140P/2 = 2001
Then: A + 140 = 4002/P
Then: A = 4002/P - 140

Now, we know that "P" must be a whole dollar amount, and also "A" must obviously be a whole number.

So, we can try different values of "P" and reject all values that don't give a whole number for "A"

Also we can narrow our search for "P" knowing that 140 tickets were sold:
- if 140 tickets were sold at full price, then the full price would be: 2001/140 = \$14.3
- if 140 tickets were sold at half price, then the full price would be: \$28.6

So, just test out the values of "P" between \$15 and \$28, and find any value(s) where A is a whole number.

I will have a go at this later

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #3 2006-01-26 22:41:16

MathsIsFun
Registered: 2005-01-21
Posts: 7,664

### Re: AMC 12 2001 Problem

This is how it works out:

Try: P=15, A = 4002/15 - 140 = 126.8    ...NO
Try: P=16, A = 4002/16 - 140 = 110.125    ...NO
...
Try: P=23, A = 4002/23 - 140 = 34   ...YES

And that was the only one that gave me a whole number for "A"

TEST:
\$23 × 34 + \$11.50 × 106 = \$782 + \$1219 = \$2001    ...OK

And the full-priced tickets raised \$782

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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