Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
m (→‎Solution 3)
 
(37 intermediate revisions by 16 users not shown)
Line 1: Line 1:
−
{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #7]] and [[2001 AMC 10 Problems|2001 AMC 10A #14]]}}
+
{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #7]] and [[2001 AMC 10 Problems|2001 AMC 10 #14]]}}
  
 
== Problem ==
 
== Problem ==
 +
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 <cmath>2001</cmath>. 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>\textbf{(A) } \textdollar 782 \qquad \textbf{(B) } \textdollar 986 \qquad \textbf{(C) } \textdollar 1158 \qquad \textbf{(D) } \textdollar 1219 \qquad \textbf{(E) }\ \textdollar 1449</math>
−
 
 
−
<math>\text{(A) }</math> <dollar/><math>782\qquad \text{(B) }</math> <dollar/><math>986\qquad \text{(C) }</math> <dollar/><math>1158\qquad \text{(D) }</math> <dollar/><math>1219\qquad \text{(E) }</math> <dollar/><math>1449</math>
 
−
 
 
−
== Solution ==
 
  
 +
=Solutions=
 +
=== 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.
  
 
Let <math>h</math> be the number of half price tickets, we then have <math>140-h</math> full price tickets. The cost of <math>140-h</math> full price tickets is equal to the cost of <math>280-2h</math> half price tickets.  
 
Let <math>h</math> be the number of half price tickets, we then have <math>140-h</math> full price tickets. The cost of <math>140-h</math> full price tickets is equal to the cost of <math>280-2h</math> half price tickets.  
  
−
Hence we know that <math>h+(280-2h) = 280-h</math> half price tickets cost <math>4002</math> dollars. Then a single half price ticket costs <math>\frac{4002}{280-h}</math> dollars, and this must be an integer. Thus <math>280-h</math> must be a divisor of <math>4002</math>. Keeping in mind that <math>0\leq h\leq 140</math>, we are looking for a divisor between <math>140</math> and <math>280</math>, inclusive.
+
Hence we know that <math>h+(280-2h) = 280-h</math> half price tickets cost <math>4002</math> dollars. Then a single full price ticket costs <math>\frac{4002}{280-h}</math> dollars, and this must be an integer. Thus <math>280-h</math> must be a divisor of <math>4002</math>. Keeping in mind that <math>0\leq h\leq 140</math>, we are looking for a divisor between <math>140</math> and <math>280</math>, inclusive.
  
 
The prime factorization of <math>4002</math> is <math>4002=2\cdot 3\cdot 23\cdot 29</math>. We can easily find out that the only divisor of <math>4002</math> within the given range is <math>2\cdot 3\cdot 29 = 174</math>.  
 
The prime factorization of <math>4002</math> is <math>4002=2\cdot 3\cdot 23\cdot 29</math>. We can easily find out that the only divisor of <math>4002</math> within the given range is <math>2\cdot 3\cdot 29 = 174</math>.  
Line 19: Line 18:
 
This gives us <math>280-h=174</math>, hence there were <math>h=106</math> half price tickets and <math>140-h = 34</math> full price tickets.
 
This gives us <math>280-h=174</math>, hence there were <math>h=106</math> half price tickets and <math>140-h = 34</math> 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.
+
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{\textbf{(A) }782}</math> dollars are raised by the full price tickets.
 +
 
 +
=== Solution 2 ===
 +
 
 +
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>
 +
 
 +
Let's multiply both sides of the equation that naturally follows by 2. We have
 +
 
 +
<cmath>2Ax+Bx=4002</cmath>
 +
 
 +
And we have <math>A+B=140\implies B=140-A</math>
 +
 
 +
Plugging in, we get <math>\implies 2Ax+(140-A)(x)=4002</math>
 +
 
 +
Simplifying, we get <math>Ax+140x=4002</math>
 +
 
 +
Factoring out the <math>x</math>, we get <math>x(A+140)=4002\implies x=\frac{4002}{A+140}</math>
 +
 
 +
We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)
 +
 
 +
Thus, <math>A+140</math> must be a factor of 4002.
 +
 
 +
Consider the prime factorization of <math>4002</math>: <math>2\times3\times23\times29</math>
 +
 
 +
<math>A</math> must be a positive integer. So, we seek a factor of <math>4002</math> to set equal to <math>A+140</math> so that we get an integer solution for <math>A</math> that is less than <math>140</math>. By guess-and-check OR inspection, the appropriate factor is <math>174</math> (<math>2\times3\times29</math>), meaning that <math>A</math> has a value of <math>34</math>. Plug this into the above equation for <math>x</math> to get <math>x = 23</math>.
 +
 
 +
Therefore, the price of full tickets out of <math>2001</math> is <math>23\times34=\boxed{\textbf{(A) }782}</math>.
 +
 
 +
--Edits by Joseph2718 (Reason: Ease of understanding)
 +
 
 +
=== Solution 3 ===
 +
 
 +
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>f \cdot p</math>.
 +
 
 +
Since the total number of tickets sold is <math>140</math>, we know that <cmath>f+h=140.</cmath> The sales from full-price tickets (<math>f \cdot p</math>) plus the sales from half-price tickets <math>\Big(\frac{h \cdot p}{2}</math>, because each hall-price ticket costs <math>\frac{p}{2}</math> dollars<math>\Big)</math> equals <math>2001.</math> Then we can write <cmath>fx + \frac{hx}{2}=2001.</cmath>
 +
 
 +
Substituting <math>h=140-f</math> into the second equation, we get <cmath>f \cdot p +\frac{(140-f)p}{2}=f \cdot p+\frac{140p-f\cdot p}{2}=\frac{f\cdot p+140p}{2}=2001.</cmath>
 +
 
 +
Multiplying by <math>2</math> and subtracting <math>140p</math> gives us <cmath>f\cdot p=4002-140p.</cmath>
 +
 
 +
Since the problem states that <math>p</math> is a whole number, <math>140p</math> will be some integer multiple of <math>140</math> that ends in a <math>0</math>. Thus, <math>4002-140p</math> will end in a <math>2</math>. Looking at the answer choices, only <math>\boxed{\textbf{(A) }782}</math> satisfies that condition.
 +
 
 +
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/x7SYSp2lYrE?si=If1SLv7j8iNdM10l
 +
 
 +
~Thesmartgreekmathdude
  
 
== 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 17:32, 25 August 2024

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

--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 $p$ 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.

Video Solution by Daily Dose of Math

https://youtu.be/x7SYSp2lYrE?si=If1SLv7j8iNdM10l

~Thesmartgreekmathdude

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_7&oldid=225776"