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In the first half of the season a football team participatin [#permalink]

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23 Jun 2013, 15:30

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In the first half of the season a football team participating in the local league won 40% of its matches. It then went on to win all its matches except two in the second half of the season. If it is known that for the entire season the percentage of matches won by the team was 66.67% and that the team played double the number of matches in the second half of the season as compared to the first half of the season then how many matches did the team play in the first half of the season?

Re: In the first half of the season a football team participatin [#permalink]

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23 Jun 2013, 16:13

The way I figured out this question was by creating a simple formula to figure out the win % in the 2nd half of the season. Since they played twice as many games, I weighted it twice as much.

1st half of season 2nd half 1/3 (win % or .4) + 2/3 ( x) = 2/3 2/15 + 2/3 x = 2/3 2/3x = 8/15 x= 4/5 or 80% wins in 2nd half of the season.

Also, the question stem says that it only lost 2 games in the second half of the season, so they must have won 8 games. (Let's say you use 4 games out of 5, it's still 80%, but then they only lost 1 game. So you must use 8 wins and 2 losses .)

Since there was 10 games in the 2nd half of the season and it was twice as many as the first, the answer is 5 games played in the first half of the season.

Re: In the first half of the season a football team participatin [#permalink]

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24 Jun 2013, 00:38

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guerrero25 wrote:

In the first half of the season a football team participating in the local league won 40% of its matches. It then went on to win all its matches except two in the second half of the season. If it is known that for the entire season the percentage of matches won by the team was 66.67% and that the team played double the number of matches in the second half of the season as compared to the first half of the season then how many matches did the team play in the first half of the season?

(A) 2 (B) 5 (C) 10 (D) 15 (E) 20

1st half ==> number of games: X ==> win: 0.4X 2nd half ==> number of games: 2x ==> win: 2X - 2 (win all except 2)

==> Win entire season = [0.4X + 2X -2] / 3x = 0.667 ==> X is approximate 5

B is correct.
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Re: In the first half of the season a football team participatin [#permalink]

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20 May 2014, 08:22

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It would help GMAT aspirants to remember a few common fraction- percentage equivalents:

1/3 = 0.333 2/3 = 0.667 1/4 = 0.25 1/5 = 0.20 1/6 = 0.167 (don't learn it by rote; 1/6 = half of 1/3; so quickly estimate the half of 0.33) 1/8= 0.125 (using the same technique as above; it's the half of 1/4)

Knowing that 66.7% = 2/3 would have saved time in solving this equation.
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Re: In the first half of the season a football team participatin [#permalink]

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29 Aug 2015, 07:36

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Re: In the first half of the season a football team participatin [#permalink]

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03 Jul 2016, 22:24

Try plugin some numbers. answer choice a) 2 ...should not consider this because we are looking for integer value for games won. b) first half = 5, second half = 10. won = 2 and 8 win ratio = 10/15 = 0.666 close enough

c) first half = 10 , second half = 20 won = 4 and 18 win ratio = 22/30.. notice that this is greater than 2/3rd.

d) first half = 15 , second half = 30 won = 6 , 28 ..win ratio = 34/45. = 30/45+4/45..again greater than 2/3.

e) first half = 20, second half = 40 won=8 and 38...win ratio = 46/60 = 40/60+1/10..

Notice that as the number of games increase second half winning ratios get better and better and this leads to improvement in overall win ratios. We can solve this using algebra

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