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Difference between revisions of "2001 AMC 12 Problems/Problem 7"

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A must be a positive integer. So, we seek a factor of 4002 to set equal to <math>A+140</math> so that we get an integer solution for A that is less than 140. By guess-and-check OR inspection, the appropriate factor is <math>174</math> (<math>2\times3\times29</math>), meaning that A has a value of <math>34</math>. Plug this into the above equation for x to get <math>x = 23</math>.
 
A must be a positive integer. So, we seek a factor of 4002 to set equal to <math>A+140</math> so that we get an integer solution for A that is less than 140. By guess-and-check OR inspection, the appropriate factor is <math>174</math> (<math>2\times3\times29</math>), meaning that A has a value of <math>34</math>. Plug this into the above equation for x to get <math>x = 23</math>.
  
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Therefore, the price of full tickets out of <math>2001</math> is <math>23\times34=\boxed{782}, or answer choice \boxed{A}</math>.
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Therefore, the price of full tickets out of <math>2001</math> is $23\times34=\boxed{(A) 782}.
 +
 
 +
--Edits by Joseph2718 (Reason: Ease of understanding)
  
 
== See Also ==
 
== See Also ==

Revision as of 20:45, 25 April 2020

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

Solution

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 $23\cdot 34 = \boxed{(\text{A})782}$ 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)

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

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