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4 June, 05:59

A nonprofit group organizes its local fundraisers in teams, with each of its L team leaders responsible for D group directors, and each of those D group directors responsible for F fundraisers. If each person on a team is either a team leader, group director, or fundraiser, and there are more group directors than team leaders, how many team leaders are on the Dallas team? (1) There are 81 total members on the Dallas team (2) There are 5 group directors on the Dallas team

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  1. 4 June, 06:07
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    Each of L team leaders has D group directors, making the total number of group directors equal to (L) (D). And each of those group directors has F fundraisers, again requiring multiplication: that total is (L) (D) (F). (You can try this by plugging in small numbers - if each of 2 leaders has 3 directors, you know there would be 6 directors)

    So while statement 1 is not sufficient (there are multiple combinations that could get you to 81, such as L = 1, D = 2, and F = 39; or L = 1, D = 5, and F = 15), statement 2 guarantees that there is only one team leader. This is because 5 is a prime number, and you know that the number of group directors = LD. The only possible way for LD to equal 5 is if L is 1 and D is 5, or if D is 1 and L is 5. And since the stimulus tells you that there are more directors than leaders, the combination must be 5 directors and 1 leader. Accordingly, statement 2 is sufficient.
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