The Quest for 700: Weekly GMAT Challenge (Answer)
July 8th, 2010
Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
First, we should determine the number of games played in this competition. We can count them in at least 2 different ways:
(1) Brute force. Name the 6 teams A, B, C, D, E, and F. A plays each of the other teams once, so A plays 5 games. B also plays 5 games, but we’ve already counted 1 of those games (the game with A), so we have 4 “new” games. C also plays 5 games, but we’ve already counted 2 of those games (the games with A and with B), so we have 3 “new” games. Continuing, we get 5 + 4 + 3 + 2 + 1 = 15 games.
(2) Combinatorics. We have a pool of 6 teams, and we want to count how many different pairs of teams (to play a game) we can select, without caring about order. Using either the anagram method or the formula for combinations, we get 6!/(2!4!) = 15 games.
Now, to find the maximum and minimum total points earned by all teams in the competition, we should notice that if one team wins and the other team loses, then 3 points total are earned (3 for the win and 1 for the loss). On the other hand, if the game ends in a draw, then only 2 points total are earned (1 by each team). So the maximum total points are earned if every game ends in a win/loss, and the minimum total points are earned if every game ends in a draw.
Maximum = 3 × 15 = 45 points.
Minimum = 2 × 15 = 30 points.
The difference between the maximum and the minimum is therefore 45 – 30 = 15.
The correct answer is (A).
Posted in The Quest for 700
