problem9

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Captchar: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sun Dec 16, 2012 4:04 am

by Captchar » Sun Dec 23, 2012 10:56 am

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

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Source: Beat The GMAT — Problem Solving | Sun Dec 23, 2012 10:56 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sun Dec 23, 2012 11:04 am

Captchar 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?

The problem cannot be solved without the assumption that there can be no ties or draws in any game. The team either wins or loose.

The team has played (17 + 3) = 20 games till now.
Hence, 2/3 of total number of games = 20
Hence, total number of games = 20*(3/2) = 30 and number of remaining games = 20/2 = 10

If the team has to win at least 3/4 of the games, then team cannot loose more than (1/4)*30 = 7.5 games
Hence, the team cannot loose more than 7 games.

Hence, the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games = (7 - 3) = 4

The correct answer is D.

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puneetkhurana2000: Master | Next Rank: 500 Posts; Posts: 131; Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members; Thanked: 39 times

by puneetkhurana2000 » Sun Dec 23, 2012 4:06 pm

Assuming no match ends in a draw.

WE can solve this using mixtures as:-

Winning ratio in first twenty games is 17/20.
Winning ratio overall is 3/4 , for simplification lets make this 3/4 equivalent to 15/20.
Lets say Winning ratio in rest of ten games is x/10, for simplification lets make this x/10 equivalent to 2x/20.

Now ((17/20)- (15/20))/((15/20)- (2x/20)) = 2/1, solving we get x = 5.5. But x can only be an integer so we can safely assume x to be 6.

Now 10 - 6 = 4 is the number of games allowed to be lost to acquire a minimum winning %age of 75%.

Answer D.

Hope this helps!!!

Thanks

Puneet

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puneetkhurana2000: Master | Next Rank: 500 Posts; Posts: 131; Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members; Thanked: 39 times

by puneetkhurana2000 » Sun Dec 23, 2012 4:09 pm

Back-solving through options is also a good way in these kind of questions.

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The Iceman: Master | Next Rank: 500 Posts; Posts: 194; Joined: Mon Oct 15, 2012 7:14 pmFollowed by: 6 members; Location: India; Thanked: 47 times

by The Iceman » Mon Dec 24, 2012 2:23 am

Captchar 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

To maximize loss we need to minimize wins given we have total wins atleast = greatest integer function of (3*30/4) = 23. To peg the wins at minimum, therefore, the team must lose 7 matches in all. Hence additional losses = 7-3 = 4

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