The graph shows the unemployment rates for three cities between 2000

Question

Data Insights (DI) Butler 2023-24 Click here for Details ] The graph shows the unemployment rates for three cities between 2000 and 2011. During this period, the total population of City X grew from 1.3 million to 1.7 million; of City Y grew from 2.1 million to 2.3 million; and of City Z decreased from 1.2 million to 1.1 million. https://gmatclub.com/forum/download/file.php?mode=view&id=74025 Based on the given information, fill in the blanks in each of the following statements. The unemployment rate in City X was GRAPH_1_PLACEHOLDER times greater than the average of the unemployment rates in Cities Y and Z. Among these three cities in the year 2000, GRAPH_2_PLACEHOLDER had the greatest number of unemployed people. 1.jpg

anish777 · Answer

Is there any short technique via which we can reach to the answer for the first question ? Else crude calculation method will consume a lot of time.

sassupreme · Answer

For 1- Looking at the years when the unemployment in City X is greater than Y and Z both is 6 years - 2000 ---> 2005, then 3 years when City X's rate is closer to the larger one. Combined it is 9, so 9.

For 2- In year one, City Y's population of 2.1M * 4% gives us ~84,000 which is the highest of all three.

For 2- In year one, City Y's population of 2.1M * 4% gives us ~84,000 which is the highest of all three.

Sajjad1994 · Answer

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).

Gemmie · Answer

The unemployment rate in City X was   times greater than the average of the unemployment rates in Cities Y and Z.

=> There are 9 years when City X's rate is higher than the avg of those of Cities Y and Z
 
 2000 => 2005: 6 years 
 2006 
 2007 
 2009  
 Among these three cities in the year 2000,   had the greatest number of unemployed people.

City X: 1.3 * 6%
City Y: 2.1 * 4%
City Z: 1.2 * 5%

=> City Y

Charudatta22 · Answer

Just a suggestion the word "times" created a little confusion, so how it would be: the unemployment rate in City X was greater than the average of the unemployment rates in Cities Y and Z for DROP DOWN years within given period.

The graph shows the unemployment rates for three cities between 2000

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The graph shows the unemployment rates for three cities between 2000

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