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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. $3$
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The team played 20 matches, won in 17 of them and lost in 3.

Let the total number of matches be 'm'

So, $\frac{2}{3}$ * m = 20  $\Rightarrow$ m = 30

Now, the team is yet to play 10 matches and has to win 3/4th of total matches

i.e. $\frac{3}{4}$ * 30 = 22.5 $\approx$ 23 out of 30 matches. $\therefore$ Total matches lost = 7

So, maximum matches the team can lose to win 3/4th of total matches =  7 - 3 = 4

(A) is correct.

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