Out of two -thirds of the total number of basket-ball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win three-fourths of the total number of matches, if it is true that no match can end in a tie? (1)4 (2)6 (3)5 (4)3 (5)None of the above

Let x be the total number of matches. 2/3 *x = 20( i.e 17 +3) x = 30. So 10 more matches remain. 3/4 * x = 3/4 * 30 ~ 22.5. Rounding this 23 matches out of 30 need to be won. of the remaining 10 matches 23-17 = 6 need to be won. So 4 additional matches can still be lost. (total losses can be 7) IMO A

harsh.champ wrote:Out of two -thirds of the total number of basket-ball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win three-fourths of the total number of matches, if it is true that no match can end in a tie? (1)4 (2)6 (3)5 (4)3 (5)None of the above Total matches = 30 3/4 of total matches = 22.5 or the team has to win 23 They have won 17 so far and they have to win 6 more; they can lose 4 at most

harsh.champ wrote:Out of two -thirds of the total number of basket-ball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win three-fourths of the total number of matches, if it is true that no match can end in a tie? (1)4 (2)6 (3)5 (4)3 (5)None of the above Let total matches be x. 2x/3 = 20 So x =30 Total matches the team should win = (3*30)/4 = 22.5 = 23 matches. No of matches left to win = 23 -17 = 6 matches No of matches it can lose = 10 -6 = 4