Luke has played 20 games and has a $95 \%$ winning percentage. Without losing any more games, how many more games in a row must he win to reach exactly a $96 \%$ winning percentage?
(A) 1
(B) 3
(C) 4
(D) 5
(E) 10
Solution 1

Since Luke has a $95 \%$ winning percentage, then he hasn't won $5 \%$, or $\frac{1}{20}$, of his games to date. Since he has played only 20 games, there has only been 1 game that he has not won.

For Luke to have exactly a $96 \%$ winning percentage, he must not have won $4 \%$, or $\frac{1}{25}$, of his games. Since he wins every game between these two positions, when he has the $96 \%$ winning percentage, he has still not won only 1 game. Therefore, he must have played 25 games in total, or 5 more than initially.

## Solution 2

Since Luke has played 20 games and has a $95 \%$ winning percentage, then he has won

$\frac{95}{100} \times 20=\frac{95}{5}=19$ games.

Let the number of games in a row he wins before reaching the $96 \%$ winning percentage be $x$. Then

Winning $\%=\frac{\text { Games Won }}{\text { Games Played }}=\frac{19+x}{20+x}=\frac{96}{100}$

Cross-multiplying,

$$
\begin{aligned}
100(19+x) & =96(20+x) \\
1900+100 x & =1920+96 x \\
4 x & =20 \\
x & =5
\end{aligned}
$$

Therefore, we wins 5 more games in a row.

ANswer: (D)