Question

Our basketball team has finished $80\%$ of its season, during which we won $40\%$ of the games we played. What percent of the remainder of our games must we win in order to finish the season with the same number of wins as losses?

Answer 1

Answer:

90%

Explanation:

Let x be the total number of matches,

Given,

80% of games were finished,

So, the number of finished games = 80% of x = 0.8x

∴ Remained games= x - 0.8x = 0.2x

Now, 40% of 80% of games were won,

So, the winning games in 0.8x matches = 40% of 0.8x

= 0.4 × 0.8x

= 0.32x

In order to finish games with the same number of wins as losses,

Winning percentage in all games must be 50%,

Thus, the total winning games = 50% of x = 0.5x

Let y be the winning percentage in 0.2x games,

So, the total winning games in 0.2x games = $\frac{y\times 0.2x}{100}$

∵ Number of winning matches in 80% games + number of winning matches in 20% = total winning matches

⇒ $0.32x + \frac{0.2xy}{100}= 0.5x$

$\frac{0.2xy}{100}=0.18x$

$0.2y=18$

$\implies y=90$

Hence, the percent of winning the in the remaining games must be 90%.