# 1995 AJHSME Problems/Problem 14

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## Problem

A team won $40$ of its first $50$ games. How many of the remaining $40$ games must this team win so it will have won exactly $70 \%$ of its games for the season? $\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

## Solution

Noting that 70% is the same as $\frac{70}{100}=\frac{7}{10}$, and that, when x is the amount of wins in the last 40 games, the fraction of games won is $\frac{40+x}{50+40}=\frac{40+x}{90}$, all we have to do is set them equal: $$\frac{40+x}{90}=\frac{7}{10}$$ $$40+x=63$$ $$x=\boxed{\text{(B)}\ 23}$$

## Solution2

Alternatively we can note that they will play a total of $40+50=90$ games and must win $0.7(90)=63$ games. Since they won $40$ games already they need $63-40=\boxed{(B)23}$ more games.

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