2001 AMC 12 Problems/Problem 7

Revision as of 16:36, 9 August 2022 by Rithvikv (talk | contribs) (Solution 3: Fixed a mistake where an extra variable was used)
The following problem is from both the 2001 AMC 12 #7 and 2001 AMC 10 #14, so both problems redirect to this page.


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?

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


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


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 $23\times34=\boxed{\textbf{(A) }782}$.

--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 $f \cdot p$.

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

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

Multiplying by $2$ and subtracting $140p$ gives us \[f\cdot p=4002-140p.\]

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

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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