Question

Out of two-thirds of the total number of basketball 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 more than three fourths of the total number of matches, if it is true that no match can end in a tie?

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

Always solve inequalities when calculating maximum losses to maintain a given ratio.

Answer 1

The Correct Option is B

Let the total number of matches be \( x \). According to the problem, two-thirds of the total matches are considered, i.e., \( \frac{2}{3}x \) matches. The team has won 17 matches and lost 3, so it has played $17 + 3 = 20$ matches. Let the maximum number of matches the team can lose be \( L \). Thus, the total number of matches won will be \( 17 \), and the total matches played will be \( 20 + L \). The team wins more than three-fourths of the total matches, so: \[ \frac{17}{20 + L}>\frac{3}{4} \] Step 1: Solve the inequality: \[ \frac{17}{20 + L}>\frac{3}{4} \Rightarrow 4 \times 17>3 \times (20 + L) \Rightarrow 68>60 + 3L \] Step 2: Simplify the inequality: \[ 68 - 60>3L \Rightarrow 8>3L \Rightarrow L<\frac{8}{3} \approx 2.67 \] Thus, the team can lose at most 2 matches to maintain the winning ratio above 3/4. Therefore, the correct answer is 8.