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 <math>2001</math>. 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? | + | A charity sells <math>140</math> benefit tickets for a total of <math>\$2001</math>. 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) } \textdollar 782 \qquad \text{(B) } \textdollar 986 \qquad \text{(C) } \textdollar 1158 \qquad \text{(D) } \textdollar 1219 \qquad \text{(E) }\ \textdollar 1449</math> | <math>\text{(A) } \textdollar 782 \qquad \text{(B) } \textdollar 986 \qquad \text{(C) } \textdollar 1158 \qquad \text{(D) } \textdollar 1219 \qquad \text{(E) }\ \textdollar 1449</math> |
Latest revision as of 20:07, 25 January 2021
- The following problem is from both the 2001 AMC 12 #7 and 2001 AMC 10 #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?
Solutions
Solution 1
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.
Solution 2
Let the cost of the full price ticket be , the number of full-price tickets be , and the number of half-price tickets be
Let's multiply both sides of the equation that naturally follows by 2. We have
And we have
Plugging in, we get
Simplifying, we get
Factoring out the , we get
We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)
Thus, must be a factor of 4002.
Consider the prime factorization of :
A must be a positive integer. So, we seek a factor of 4002 to set equal to so that we get an integer solution for A that is less than 140. By guess-and-check OR inspection, the appropriate factor is (), meaning that A has a value of . Plug this into the above equation for x to get .
Therefore, the price of full tickets out of is .
--Edits by Joseph2718 (Reason: Ease of understanding)
Solution 3
Let equal the number of full price tickets, and let equal the number of half price tickets. Additionally, suppose that the price of is . We are trying to solve for
Since the total number of tickets sold is , we know that The sales from full price tickets () plus the sales from half price tickets , because each half price ticket costs dollars equals Then we can write
Substituting into the second equation, we get
Multiplying by 2 and subtracting gives us
Since the problem states that is a whole number, will be some integer multiple of that ends in a . Thus, will end in a . Looking at the answer choices, only satisfies that condition.
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.