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In 2010, a basketball team won 30 percent of its 20 basketball games. In 2011, the team won 28 percent of its 25 basketball games. What was the percent increase from 2010 to 2011 in the number of basketball games the team won?
A. 5%
B. 14 2/7%
C. 16 2/3%
D. 23 1/3%
E. 60%
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A. 5%
B. 14 2/7%
C. 16 2/3%
D. 23 1/3%
E. 60%
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Bunuel
30% of 20 = 6 and 28% of 25 = 7, so percent increase \(= \frac{7-6}{6}= 16.6 %\) Answer C
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In 2010, a basketball team won 30 percent of its 20 basketball games. In 2011, the team won 28 percent of its 25 basketball games. What was the percent increase from 2010 to 2011 in the number of basketball games the team won?
In 2010 Team won 30% of 20 games = \(\frac{30}{100}\) x 20 = 6
In 2011 Team won 28% of 25 games = \(\frac{28}{100}\) x 25 = 7
Increase in games won from 2010 to 2011 = 7-6 = 1
Percent increase = Change in games won / Initial games won x 100 = \(\frac{1}{6}\) x 100 = \(\frac{50}{3}\) = 16\(\frac{2}{3}\)
Answer C...
In 2010 Team won 30% of 20 games = \(\frac{30}{100}\) x 20 = 6
In 2011 Team won 28% of 25 games = \(\frac{28}{100}\) x 25 = 7
Increase in games won from 2010 to 2011 = 7-6 = 1
Percent increase = Change in games won / Initial games won x 100 = \(\frac{1}{6}\) x 100 = \(\frac{50}{3}\) = 16\(\frac{2}{3}\)
Answer C...
Bunuel
We can apply a few simple formulas to solve this
(.3)(20)= 6
25(.28)= 7
7-6/6 = 1/6 x 100 = 16 2/3
Thus
"C"
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Bunuel
Games won in 2010 is 6
Games won in 2011 is 7
Percentage increase is \(\frac{1}{6}*100 = 16 \frac{2}{3}\) %
Thus, answer will be (C)
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