In a tournament, a player has a record of 40\% wins, out of the number of games he has played so far which in turn is $\frac{\mathrm{2}}{\mathrm{5}}$ of the total number of games he plays. What is the maximum percentage of the remaining games that the player can lose and still win 50\% of all the games played?
Answer
Answer (as printed):
Explanation
Let the total number of games played be $x$. Number of games already played $=\frac{2x}{5}$. Games already lost $=60\%$ of $\frac{2x}{5}=\frac{6x}{25}$. Number of games that the player can lose $=50\%$ of $x=\frac{x}{2}$. Number of games that the player can still lose $=\left(\frac{x}{2}-\frac{6x}{25}\right)=\frac{13x}{50}$. Remaining games to be played $=\left(x-\frac{2x}{5}\right)=\frac{3x}{5}$. Required percentage $=\left(\frac{13x}{50}\times\frac{5}{3x}\times 100\right)\%=43.3\%$.
Explanation as extracted from the printed page; notation may be imperfect.