2020 AMC 12B Problem 5

Attempt Problem 5 of the 2020 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

Teams AA and BB are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team AA has won 23\tfrac23 of its games and team BB has won 58\tfrac58 of its games. Also, team BB has won 77 more games and lost 77 more games than team A.A. How many games has team AA played?
2121
2727
4242
4848
6363
Answer: C
Concepts:system of equationsfraction
Small hint:
Let AA play aa games and BB play bb games, then write both the win and loss conditions
Big hint:
Wins: 58b=23a+7;\tfrac58 b = \tfrac23 a + 7; losses: 38b=13a+7\tfrac38 b = \tfrac13 a + 7
Solution:
Let aa be the number of games team AA played and bb the number team BB played. Team AA wins 23a\tfrac23 a and loses 13a;\tfrac13 a; team BB wins 58b\tfrac58 b and loses 38b.\tfrac38 b. The conditions give 58b=23a+7 \tfrac58 b = \tfrac23 a + 7 and 38b=13a+7. \tfrac38 b = \tfrac13 a + 7.
Subtracting the equations gives 14b=13a,\tfrac14 b = \tfrac13 a, so b=43a.b = \tfrac43 a. Substituting into the loss equation: 38⋅43a=13a+7,\tfrac38 \cdot \tfrac43 a = \tfrac13 a + 7, i.e. 12a=13a+7,\tfrac12 a = \tfrac13 a + 7, so 16a=7\tfrac16 a = 7 and a=42.a = 42.
Thus, the correct answer is C.
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