2001 AMC 12 Problems/Problem 7

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 $140$ 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?

$\textbf{(A) } \textdollar 782 \qquad \textbf{(B) } \textdollar 986 \qquad \textbf{(C) } \textdollar 1158 \qquad \textbf{(D) } \textdollar 1219 \qquad \textbf{(E) }\ \textdollar 1449$

Solutions

Solution 1

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 $23\cdot 34 = \boxed{\textbf{(A) }782}$ dollars are raised by the full price tickets.

Solution 2

Let the cost of the full price ticket be $x$ , the number of full-price tickets be $A$ , and the number of half-price tickets be $B$

Let's multiply both sides of the equation that naturally follows by 2. We have

$2Ax+Bx=4002$

And we have $A+B=140\implies B=140-A$

Plugging in, we get $\implies 2Ax+(140-A)(x)=4002$

Simplifying, we get $Ax+140x=4002$

Factoring out the $x$ , we get $x(A+140)=4002\implies x=\frac{4002}{A+140}$

We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)

Thus, $A+140$ must be a factor of 4002.

Consider the prime factorization of $4002$ : $2\times3\times23\times29$

A must be a positive integer. So, we seek a factor of 4002 to set equal to $A+140$ so that we get an integer solution for A that is less than 140. By guess-and-check OR inspection, the appropriate factor is $174$ ( $2\times3\times29$ ), meaning that A has a value of $34$ . Plug this into the above equation for x to get $x = 23$ .

Therefore, the price of full tickets out of $2001$ is $23\times34=\boxed{(A) 782}$ .

--Edits by Joseph2718 (Reason: Ease of understanding)

Solution 3

Let $f$ equal the number of full price tickets, and let $h$ equal the number of half price tickets. Additionally, suppose that the price of $f$ is $p$ . We are trying to solve for $fx.$

Since the total number of tickets sold is $140$ , we know that $f+p=140.$ The sales from full price tickets ( $fx$ ) plus the sales from half price tickets $\Big(\frac{hx}{2}$ , because each half price ticket costs $\frac{x}{2}$ dollars $\Big)$ equals $2001.$ Then we can write $fx + \frac{hx}{2}=2001.$

Substituting $h=140-f$ into the second equation, we get $fx+\frac{(140-f)x}{2}=fx+\frac{140x-fx}{2}=fx+\frac{140x}{2}=2001.$

Multiplying by 2 and subtracting $140x$ gives us $fx=4002-140x.$

Since the problem states that $x$ is a whole number, $140x$ will be some integer multiple of $140$ that ends in a $0$ . Thus, $4002-140x$ will end in a $2$ . Looking at the answer choices, only $\boxed{(\text{A})782}$ satisfies that condition.

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

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