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The city x that never sleeps

Expert replies
by pkw209 » Mon May 03, 2010 3:20 pm
In city x last April, was the average daily high temperature greater than the median daily high temperature?

a) In city x last April, the sum of the 30 daily high temperatures was 2,160 degrees.

b) In city x last April, 60 percent of the daily high temperatures were less than the average daily high temperature.
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Source: — Data Sufficiency |

by Stuart@KaplanGMAT » Mon May 03, 2010 4:10 pm
pkw209 wrote:In city x last April, was the average daily high temperature greater than the median daily high temperature?

1) In city x last April, the sum of the 30 daily high temperatures was 2,160 degrees.

2) In city x last April, 60 percent of the daily high temperatures were less than the average daily high temperature.
Step 1 of the Kaplan Method for DS: Analyze the Stem

We're asked if the average (sum/# of terms) is greater than the median (the middle term (or average of the 2 middle terms in an even-termed set) of an ordered set).

So, if we know the average and the median, we can answer the question. Information about the weighting of the set may also be sufficient.

Step 2 of the Kaplan Method for DS: Evaluation the Statements

From 1, we can compute the average temperature; however, we have no information about the median: insufficient.

From 2, we know that 60% of the terms are less than the average. So:

if there are an odd numbers of terms, the middle term must be below the average; and
if there are an even number of terms, the two middle terms must be below the average.

Accordingly, the median will always be below the average: sufficient, choose (B).

Examples from (2), in case you're not convinced:

5 term set: the breakdown is 60/40, or 3 terms below the average and 2 terms above. The median is the 3rd term, which is below the average.

10 term set: the breakdown is 60/40, or 6 terms below the average and 4 above. The median is the average of the 5th and 6th terms, both of which are below the average, so the average of those terms is also definitely below the average.

Since exactly 60% of the terms are below the average, the number of terms must be a multiple of either 5 or 10.
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by pkw209 » Mon May 03, 2010 4:32 pm
Thank you, Stuart.

1) Is it safe to assume that if more than 50% of the terms within a set are BELOW the mean, then the mean is GREATER than the median?

2) Is it safe to assume that if more than 50% of the terms within a set are ABOVE the mean, then the mean is LESS than the median?
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