2001 AMC 12 Problems/Problem 7

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 $140$ benefit tickets for a total of $2001$ . 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?

$\text{(A) }$ <dollar/> $782\qquad \text{(B) }$ <dollar/> $986\qquad \text{(C) }$ <dollar/> $1158\qquad \text{(D) }$ <dollar/> $1219\qquad \text{(E) }$ <dollar/> $1449$

Solution

Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.

Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.

Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single half price ticket costs $\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0\leq h\leq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.

The prime factorization of $4002$ is $4002=2\cdot 3\cdot 23\cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2\cdot 3\cdot 29 = 174$ .

This gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.

In our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $\frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23\cdot 34 = \boxed{(\text{A})782}$ dollars are raised by the full price tickets.

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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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

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2001 AMC 12 Problems/Problem 7

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