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

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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #7]] and [[2001 AMC 10 Problems|2001 AMC 10A #14]]}}
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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #7]] and [[2001 AMC 10 Problems|2001 AMC 10 #14]]}}
  
 
== Problem ==
 
== Problem ==
 
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A charity sells <math>140</math> benefit tickets for a total of <math>\$2001</math>. 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?
A charity sells <math>140</math> benefit tickets for a total of <math>2001</math>. 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?
 
  
 
<math>\text{(A) } \textdollar 782 \qquad \text{(B) } \textdollar 986 \qquad \text{(C) } \textdollar 1158 \qquad \text{(D) } \textdollar 1219 \qquad \text{(E) }\ \textdollar 1449</math>
 
<math>\text{(A) } \textdollar 782 \qquad \text{(B) } \textdollar 986 \qquad \text{(C) } \textdollar 1158 \qquad \text{(D) } \textdollar 1219 \qquad \text{(E) }\ \textdollar 1449</math>
  
== Solution ==
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== Solutions ==
 
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=== Solution 1 ===
 
Let's multiply ticket costs by <math>2</math>, then the half price becomes an integer, and the charity sold <math>140</math> tickets worth a total of <math>4002</math> dollars.
 
Let's multiply ticket costs by <math>2</math>, then the half price becomes an integer, and the charity sold <math>140</math> tickets worth a total of <math>4002</math> dollars.
  
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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.
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=== Solution 2 ===
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 +
Let the cost of the full price ticket be <math>x</math>, the number of full-price tickets be <math>A</math>, and the number of half-price tickets be <math>B</math>
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 +
Let's multiply both sides of the equation that naturally follows by 2. We have
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<cmath>2Ax+Bx=4002</cmath>
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And we have <math>A+B=140\implies B=140-A</math>
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Plugging in, we get <math>\implies 2Ax+(140-A)(x)=4002</math>
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Simplifying, we get <math>Ax+140x=4002</math>
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Factoring out the <math>x</math>, we get <math>x(A+140)=4002\implies x=\frac{4002}{A+140}</math>
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We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)
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Thus, <math>A+140</math> must be a factor of 4002.
 +
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Consider the prime factorization of <math>4002</math>: <math>2\times3\times23\times29</math>
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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>.
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Therefore, the price of full tickets out of <math>2001</math> is <math>23\times34=\boxed{(A) 782}</math>.
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--Edits by Joseph2718 (Reason: Ease of understanding)
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=== Solution 3 ===
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Let <math>f</math> equal the number of full price tickets, and let <math>h</math> equal the number of half price tickets. Additionally, suppose that the price of <math>f</math> is <math>p</math>. We are trying to solve for <math>fx.</math>
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Since the total number of tickets sold is <math>140</math>, we know that <cmath>f+p=140.</cmath> The sales from full price tickets (<math>fx</math>) plus the sales from half price tickets <math>\Big(\frac{hx}{2}</math>, because each half price ticket costs <math>\frac{x}{2}</math> dollars<math>\Big)</math> equals <math>2001.</math> Then we can write <cmath>fx + \frac{hx}{2}=2001.</cmath>
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Substituting <math>h=140-f</math> into the second equation, we get <cmath>fx+\frac{(140-f)x}{2}=fx+\frac{140x-fx}{2}=fx+\frac{140x}{2}=2001.</cmath>
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Multiplying by 2 and subtracting <math>140x</math> gives us <cmath>fx=4002-140x.</cmath>
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Since the problem states that <math>x</math> is a whole number, <math>140x</math> will be some integer multiple of <math>140</math> that ends in a <math>0</math>. Thus, <math>4002-140x</math> will end in a <math>2</math>. Looking at the answer choices, only <math>\boxed{(\text{A})782}</math> satisfies that condition.
  
 
== See Also ==
 
== See Also ==
 
 
{{AMC12 box|year=2001|num-b=6|num-a=8}}
 
{{AMC12 box|year=2001|num-b=6|num-a=8}}
 
{{AMC10 box|year=2001|num-b=13|num-a=15}}
 
{{AMC10 box|year=2001|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:07, 25 January 2021

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?

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

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

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.

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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