Placerville's soccer team won one third of its first 18 games and one

Placerville's soccer team won one third of its first 18 games and one half of the rest of its games. What was the total number of games played?

So team won \(18*\frac{1}{3}+\frac{x}{2}\), where x is the remaining games.

(1) The soccer team won a total of 9 games.
So \(18*\frac{1}{3}+\frac{x}{2}=9......6+\frac{x}{2}=9........x=6\)
Total games = 18+6=24
Suff

(2) The soccer team lost at least 5/8 of its total games.
Total games = 18+x
So \((18+x)-(18*\frac{1}{3}+\frac{x}{2})\geq{\frac{5}{8}(18+x)}........8(12+\frac{x}{2})<5(18+x).......96+4x\geq{90+5x}....x\leq 6\)
x has to be EVEN as the team has won HALF of the matches. So x can be 2, 4 or 6.
However we do not require to get into calculations if we are short of time, as an inequality will generally not give us a specific value.
Insuff

A
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Bunuel need your help in this one

Wouldn't x be less than or "equal to" 6 because the statement uses the word "at least"? So, in that case, x could be 2, 4, or 6. The answer would still be A though.

There are some typos in chetan's solution above (for example in Statement 1, he solves as if the team won 8 games, instead of 9).

The team won 6 of its first 18 games. Using Statement 1, if the team won 9 games in total, it won 3 games in the remaining part of the year. Since it won half of its games in that part of the year, it played 6 games to end the year, and thus 18+6 = 24 games in total. So Statement 1 is sufficient.

For Statement 2, we know the team lost 2/3 of its games early in the year, and 1/2 of its games late in the year. So overall, it lost somewhere between 2/3 and 1/2 of its games. Overall the team lost at least 5/8 of its games. Now, 5/8 is much closer to 2/3 than it is to 1/2, so the team must have played a lot more games when it was losing 2/3 of them than when it was losing 1/2 of them. So in the second part of the year, only a small number of games were played. You could just test some small even numbers for the number of remaining games (2 and 4 say) to prove that different numbers of games are possible. But you don't even need to do that. Since the Statements must be logically consistent, the lone solution from Statement 1 absolutely must work when we use Statement 2 alone. So it's possible only 6 games were played to end the year. But then if only 4 or 2 games were played to end the year, the fraction of games the team lost will be even closer to 2/3, so those must also be possible, and Statement 2 is not sufficient.
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Thanks Ian. Yes, there were two typos. Edited
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Hey Ian, thanks for the quick trick. I have one question though, can we avoid all kinds of calculations for statement 2 as it tells us only about the games lost. And 2/3 and 1/2 games that were not won could comprise games both lost and drawn?

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