Difference between revisions of "2001 AMC 12 Problems/Problem 7"
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== Problem == | == Problem == | ||
− | A charity sells <math>140</math> benefit tickets for a total of < | + | A charity sells <math>140</math> benefit tickets for a total of <cmath>2001</cmath>. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets? |
− | < | + | <math>\text{(A) }</math> <dollar/><math>782\qquad \text{(B) }</math> <dollar/><math>986\qquad \text{(C) }</math> <dollar/><math>1158\qquad \text{(D) }</math> <dollar/><math>1219\qquad \text{(E) }</math> <dollar/><math>1449</math> |
== Solution == | == Solution == |
Revision as of 20:59, 26 April 2015
- The following problem is from both the 2001 AMC 12 #7 and 2001 AMC 10A #14, so both problems redirect to this page.
Problem
A charity sells benefit tickets for a total of . Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?
<dollar/> <dollar/> <dollar/> <dollar/> <dollar/>
Solution
Let's multiply ticket costs by , then the half price becomes an integer, and the charity sold tickets worth a total of dollars.
Let be the number of half price tickets, we then have full price tickets. The cost of full price tickets is equal to the cost of half price tickets.
Hence we know that half price tickets cost dollars. Then a single half price ticket costs dollars, and this must be an integer. Thus must be a divisor of . Keeping in mind that , we are looking for a divisor between and , inclusive.
The prime factorization of is . We can easily find out that the only divisor of within the given range is .
This gives us , hence there were half price tickets and full price tickets.
In our modified setting (with prices multiplied by ) the price of a half price ticket is . In the original setting this is the price of a full price ticket. Hence dollars are raised by the full price tickets.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.