Question

So far this year, Adam has played 30 games of chess and has only won 6 of them. What is the minimum number of additional games he must play, given that he is sure to lose at least one-third of them, so that for the year he will have won more games than he lost?
A) 24
B) 57
C) 87
D) It is not possible for Adam to do this.

The correct answer is B) 57, but not sure how?

Answer 1

Try adding 57 to the equation, then you can see how 57 adds into the equation and proves the answer.

Answer 2

Answer:

Step-by-step explanation:Yikes. Just so we’re all clear—this is not a real SAT question.

So far, Adam’s record is 6 wins/30 games. If he plays x more games, his record will be:

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You find the minimum x such that the fraction is greater than ½, which would mean he won more than ½ of his games.

Let’s do some math!

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So if he plays 54 more games, and wins 2/3 of them, he’ll have the same number of wins as losses. Check on that to make sure: He’ll have played 30 + 54 = 84 games, and won 6 + (2/3)54 = 42 of them. Yep, that works.

Now here is where it gets tricky (and why this question is not SAT-caliber). Remember that he’s losing 1/3 of his games. If his 52nd, 53rd, and 54th additional games were win, win, loss, then he would have had, very briefly, a winning record after his 53rd additional game.

His 55th additional game is another chance at a winning record. If he loses that game, then he’ll win the next two, and have a winning record for sure by the 57th additional game. From there, if he continues winning 2/3 of his games, he’ll have a winning record forever.