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

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

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