Difference between revisions of "1995 AJHSME Problems/Problem 14"

(Created page with "==Problem== A team won <math>40</math> of its first <math>50</math> games. How many of the remaining <math>40</math> games must this team win so it will have won exactly <math>...")
 
(Problem)
Line 4: Line 4:
  
 
<math>\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35</math>
 
<math>\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35</math>
 +
 +
==Solution==
 +
 +
Noting that 70% is the same as <math>\frac{70}{100}=\frac{7}{10}</math>, and that, when x is the amount of wins in the last 40 games, the fraction of games won is <math>\frac{40+x}{50+40}=\frac{40+x}{90}</math>, all we have to do is set them equal: <cmath>\frac{40+x}{90}=\frac{7}{10}</cmath> <cmath>40+x=63</cmath> <cmath>x=23 \text{(B)}</cmath>

Revision as of 12:48, 5 July 2012

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: \[\frac{40+x}{90}=\frac{7}{10}\] \[40+x=63\] \[x=23 \text{(B)}\]