Difference between revisions of "2010 AMC 12B Problems/Problem 8"
m (→See also) |
(→Solution: latexed answer) |
||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | There are <math>x</math> schools. This means that there are <math>3x</math> 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 <math>x</math> is an odd number. <math>x</math> cannot be less than 23, because there wouldn't be a | + | There are <math>x</math> schools. This means that there are <math>3x</math> people. Because no one's score was the same as another person's score, that means that there could only have been <math>1</math> median score. This implies that <math>x</math> is an odd number. <math>x</math> cannot be less than <math>23</math>, because there wouldn't be a <math>64</math>th place if there were. <math>x</math> cannot be greater than <math>23</math> either, because that would tie Andrea and Beth. Thus, the only possible answer is <math>23 \Rightarrow \boxed{B}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=B}} | {{AMC12 box|year=2010|num-b=7|num-a=9|ab=B}} |
Revision as of 12:51, 31 May 2011
Problem 8
Every high school in the city of Euclid sent a team of 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 th and th, respectively. How many schools are in the city?
Solution
There are schools. This means that there are people. Because no one's score was the same as another person's score, that means that there could only have been median score. This implies that is an odd number. cannot be less than , because there wouldn't be a th place if there were. cannot be greater than either, because that would tie Andrea and Beth. Thus, the only possible answer is .
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 |