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 "2010 AMC 12B Problems/Problem 8"

(→‎Solution 2)
(→‎Solution 2)
Line 8: Line 8:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Let <math>a</math> be Andrea's score. We know that she was the highest on our team, so <math>a < 37</math>.  
+
Let <math>a</math> be Andrea's place. We know that she was the highest on our team, so <math>a < 37</math>.  
  
 
Since <math>a</math> is the median, there are <math>a-1</math> to the left and right of the median, so the total number of people is <math>2a-1</math> and the number of schools is <math>(2a-1)/3</math>. This implies that <math>2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}</math>.  
 
Since <math>a</math> is the median, there are <math>a-1</math> to the left and right of the median, so the total number of people is <math>2a-1</math> and the number of schools is <math>(2a-1)/3</math>. This implies that <math>2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}</math>.  

Revision as of 09:38, 14 July 2018

Problem 8

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$th and $64$th, respectively. How many schools are in the city?

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$

Solution

There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$, because there wouldn't be a $64$th place if there were. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \Rightarrow \boxed{B}$.

Solution 2

Let $a$ be Andrea's place. We know that she was the highest on our team, so $a < 37$.

Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$. This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$.

Also, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \implies a \ge 33$.

Putting it all together: $33 \le a < 37$ and $a \equiv 2 \pmod{3}$, so clearly $a = 35$, and the number of schools as we got before is $(2a-1)/3 = 69/3 = \boxed{23}$.

~adihaya

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

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=2010_AMC_12B_Problems/Problem_8&oldid=96192"