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20 Dec 2015, 06:25
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Difficulty:
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81% (01:34) correct
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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%
A. 5%
B. 14 2/7%
C. 16 2/3%
D. 23 1/3%
E. 60%
20 Dec 2015, 06:34
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Bunuel
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%
A. 5%
B. 14 2/7%
C. 16 2/3%
D. 23 1/3%
E. 60%
30% of 20 = 6 and 28% of 25 = 7, so percent increase \(= \frac{7-6}{6}= 16.6 %\) Answer C
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17 May 2017, 04:23
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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...
