The graph shows the unemployment rates for three cities between 2000

Official Explanation

Don’t waste time trying to calculate exact averages. Rather, think about the different cases in which City X’s unemployment rate would have been greater than the average unemployment rate of the other two cities. One such case is when City X’s rate is greater than either of the other two rates. This happened every year from 2000 to 2005. Next, City X’s unemployment rate will be greater than the average unemployment rate of the other two cities when City X’s rate is between the two other rates, but closer to the larger one of the two. This happened in 2006, 2007, and 2009. In 2011, City X’s rate was between the other two but closer to the lower of the two, while in 2008 and 2010 City X’s rate was the lowest of the three. Thus, in all but these three years (that is, in a total of 9 years), City X’s rate was greater than the average of the other two rates.

The correct answer is (C).

In 2000, City X had the highest unemployment rate, but City Y was significantly more populous. The 4% of City Y’s 2.1 million residents who were unemployed are more than the 5.8% of City X’s 1.3 million residents or the 5.2% of City Z’s 1.2 million residents. Thus, the correct answer is choice (B).

The correct answer is (B).