Difference between revisions of "2014 AMC 8 Problems/Problem 16"
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| − | + | ==Problem== | |
| + | The "Middle School Eight" basketball conference has <math>8</math> teams. Every season, each team plays every other conference team twice (home and away), and each team also plays <math>4</math> games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams? | ||
| − | + | <math>\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160</math> | |
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| + | ==Solution== | ||
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| + | Within the conference, there are 8 teams, so there are <math>\dbinom{8}{2}=28</math> pairings of teams, and each pair must play two games, for a total of <math>28\cdot 2=56</math> games within the conference. | ||
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| + | Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of <math>4\cdot 8 =32</math> games outside the conference. | ||
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| + | Therefore, the total number of games is <math>56 + 32 = \boxed{\text{(B) }88}</math>. | ||
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| + | ==Solution 2== | ||
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| + | We can also figure out the number of games the 8 teams play within the conference with the formula n(n - 1)/2, which is used for these kind of match questions, where Team A playing Team B is the same as Team B playing Team A. In this formula, n is the number of teams. So we will do (8 X 7 / 2) X 2, as each team plays the other team twice. The 2s cancel out, to yield 8 X 7, which is 56 games. Then, we can just proceed as above to find the number of games the 8 teams play against 4 teams outside the conference, which is 8 X 4 = 32. Summing these values up yields 88, which is answer choice B. | ||
| + | |||
| + | ==Video Solution 1 by OmegaLearn== | ||
| + | https://youtu.be/Zhsb5lv6jCI?t=991 | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/G2Akf39uwcE | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | ==Video Solution 3== | ||
| + | https://youtu.be/Zhsb5lv6jCI?t=991 | ||
| + | |||
| + | https://youtu.be/w7Y-iq_kEaY ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=15|num-a=17}} | {{AMC8 box|year=2014|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:17, 7 January 2025
Contents
Problem
The "Middle School Eight" basketball conference has 
teams. Every season, each team plays every other conference team twice (home and away), and each team also plays 
games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
Solution
Within the conference, there are 8 teams, so there are 
pairings of teams, and each pair must play two games, for a total of 
games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of 
games outside the conference.
Therefore, the total number of games is 
.
Solution 2
We can also figure out the number of games the 8 teams play within the conference with the formula n(n - 1)/2, which is used for these kind of match questions, where Team A playing Team B is the same as Team B playing Team A. In this formula, n is the number of teams. So we will do (8 X 7 / 2) X 2, as each team plays the other team twice. The 2s cancel out, to yield 8 X 7, which is 56 games. Then, we can just proceed as above to find the number of games the 8 teams play against 4 teams outside the conference, which is 8 X 4 = 32. Summing these values up yields 88, which is answer choice B.
Video Solution 1 by OmegaLearn
https://youtu.be/Zhsb5lv6jCI?t=991
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution 3
https://youtu.be/Zhsb5lv6jCI?t=991
https://youtu.be/w7Y-iq_kEaY ~savannahsolver
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
