Difference between revisions of "2001 AMC 12 Problems/Problem 7"
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In our modified setting (with prices multiplied by <math>2</math>) the price of a half price ticket is <math>\frac{4002}{174} = 23</math>. In the original setting this is the price of a full price ticket. Hence <math>23\cdot 34 = \boxed{(\text{A})782}</math> dollars are raised by the full price tickets. | In our modified setting (with prices multiplied by <math>2</math>) the price of a half price ticket is <math>\frac{4002}{174} = 23</math>. In the original setting this is the price of a full price ticket. Hence <math>23\cdot 34 = \boxed{(\text{A})782}</math> dollars are raised by the full price tickets. | ||
+ | ==Solution 2== | ||
+ | |||
+ | Let the cost of the full price ticket be <math>x</math>, let the number of full price tickets be <math>A</math> And half price tickets be <math>B</math> | ||
+ | |||
+ | Multiplying everything by two first to make cancel out fractions. | ||
+ | |||
+ | |||
+ | We have | ||
+ | |||
+ | <math>2Ax+Bx=4002</math> | ||
+ | |||
+ | And we have <math>A+B=140\implies B=140-A</math> | ||
+ | |||
+ | Plugging in, we get <math>\implies 2Ax+(140-A)(x)=4002</math> | ||
+ | |||
+ | Simplifying, we get <math>Ax+140x=4002</math> | ||
+ | |||
+ | Factoring out the <math>x</math>, we get <math>x(A+140)=4002\implies x=\frac{4002}{A+140}</math> | ||
+ | |||
+ | Obviously, we see that the fraction has to simplify to an integer. | ||
+ | |||
+ | Hence, <math>A+140</math> has to be a factor of 4002. | ||
+ | |||
+ | By inspection, we see that the prime factorization of <math>4002=2\times3\times23\times29 | ||
+ | |||
+ | We see that </math>A=34<math> through inspection. We also find that </math>H=23<math> | ||
+ | |||
+ | Hence, the price of full tickets out of </math>2001<math> is </math>23\times34=782$. | ||
== See Also == | == See Also == | ||
Revision as of 22:21, 19 May 2017
- The following problem is from both the 2001 AMC 12 #7 and 2001 AMC 10A #14, so both problems redirect to this page.
Contents
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?
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.
Solution 2
Let the cost of the full price ticket be , let the number of full price tickets be And half price tickets be
Multiplying everything by two first to make cancel out fractions.
We have
And we have
Plugging in, we get
Simplifying, we get
Factoring out the , we get
Obviously, we see that the fraction has to simplify to an integer.
Hence, has to be a factor of 4002.
By inspection, we see that the prime factorization of $4002=2\times3\times23\times29
We see that$ (Error compiling LaTeX. ! Missing $ inserted.)A=34H=23200123\times34=782$.
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.