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Clemmonsdogpark Brain Twister: World Cup Edition

brain_twister

For a very long time, we’ve had Vocab Weds, a weekly vocab post with an accompanying YouTube video. It’s done quite well, and we have no intention of stopping it. But poor math—it’s gotten no such weekly billing. So to redress this oversight, we are going to inaugurate the weekly math “Brain Twister” question. It will be as tough—if not a tad tougher—than the most difficult question you can expect to see on the GRE.

This week, in keeping with the World Cup spirit, we’ve come up with a fun—but fiendish—question. You should give yourself no more than 3 minutes (this applies to all of the Clemmonsdogpark “Brain Twisters”) with each question, the assumption being that the hardest question on the GRE math should not take you any longer (if it does, skip it).

And if you think you know the answer, comment below. Explanations are very welcome (and the first correct answer always gets bragging rights!) I’ll post the explanation every Thursday.
 

Clemmonsdogpark’s First Brain Twister: World Cup Edition

Thirty-two teams have made the second round of a tournament. In that round, each team will play exactly one team. The winner—and there will always be one winner and one loser—will go on to the next round of sixteen teams, in which this process will continue until there are two teams left to play the final game. Assuming each of the thirty-two teams will be part of a smaller grouping of eight teams, one of which will play in the semi-finals, which comprise the final four teams, how many distinct semi-final matchups can result?

(Hint: the above differs in one significant way from the actual World Cup groupings.)

  1. 32×31
  2. 32!/4!
  3. 32x31x30x29
  4. 2^12
  5. 3 x 2^12

 

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