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by harsh.champ » Thu Feb 04, 2010 1:25 pm
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
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by ace_gre » Thu Feb 04, 2010 1:52 pm
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
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by ajith » Thu Feb 04, 2010 3:39 pm
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
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by neelimareddym » Thu Feb 04, 2010 10:03 pm
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
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