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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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.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 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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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
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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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 = 4Captchar 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