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A certain basketball team that has played 2/3 of its games

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Manager
Joined: 05 Sep 2007
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Location: New York
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A certain basketball team that has played 2/3 of its games [#permalink]  13 Apr 2008, 15:00
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A certain basketball team that has played 2/3 of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games?
(A) 7
(B) 6
(C) 5
(D) 4
(E) 3
Current Student
Joined: 27 Mar 2008
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Schools: Kellogg Class of 2011
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Kudos [?]: 40 [0], given: 1

Total games = x

(2/3)x = 17 +3
x = 30

Remaining games = 10

3/4 x = 22.5

For the team to win 23 games, it will have 7 losses or 4 more losses

C
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Joined: 01 Apr 2008
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Actually I believe the answer is A (7).

Re-read the question, it is asking for the number of remaining games that the team can lose. Of they remaining 10 games, they can lose 7 and still win 3/4 of their total games.

Cool?
Current Student
Joined: 27 Mar 2008
Posts: 416
Schools: Kellogg Class of 2011
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Kudos [?]: 40 [0], given: 1

No. The question says

What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games?

If the team lost 7 games (from this point forward i.e. after 20 games and a 17-3 record), it would end up with a total of 20 wins and 10 losses. In this scenario is would only win 2/3 of its games and not 3/4 as the question asks.
Director
Joined: 14 Oct 2007
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I disagree with A.

The answer is D as posted originally by yellowjacket.

atleast 3/4 wins of ALL its 30 games = 23 wins.
the team has won 17 games so far out of 20. that means they need to win 6 games out of the next 10 to guarantee a 3/4 ratio. This means they can afford to lose 4 games.

please let me know if i am missing something.
VP
Joined: 09 Jul 2007
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Location: London
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el1981 wrote:
A certain basketball team that has played 2/3 of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games?
(A) 7
(B) 6
(C) 5
(D) 4
(E) 3

D for me

(2/3)*X = 20
X=30(total games to be played)

so far played 20=17(won)+3(lost)

the number of games it wins (3/4)*30=~23
Left 7 games. 3 already lost so it cannot loose any more/less than 4
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