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Help - Tired of missing this type of question

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Help - Tired of missing this type of question

by mguerreiro001 » Sat Jan 19, 2008 12:22 pm
In City X last April, was the average (arith mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high tempratures was 2,160o.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

ANS: B
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Re: Help - Tired of missing this type of question

by Stuart@KaplanGMAT » Sat Jan 19, 2008 12:52 pm
mguerreiro001 wrote:In City X last April, was the average (arith mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high tempratures was 2,160o.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.
Average is the sum of the terms/# of terms.
Median is the middle term of a set of numbers (or, if there are an even number of terms, the average of the two middle terms).

Q: was average > median?

(1) We have the sum and the number of terms, so we can calculate the average, but we have no information about the individual terms, so we can't calculate the median.

Insufficient

(2) if 60% of the daily highs were below the average, then 40% of the highs were at or above the average. Since the median is the middle term (which occurs at the 50th percentile), the median will fall within the 60% of the highs that are below the average.

Therefore, median is < average: sufficient.

(2) is suff, (1) is insuff - choose (B)
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by mguerreiro001 » Sat Jan 19, 2008 2:02 pm
Thank you. I think I got the idea.
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by mido362 » Tue Apr 10, 2012 10:37 pm
Why are we considering that the median (15th and 16th day) will fall in the 60%?

In other word maybe the 60 % is distributed as follows:
30% from day 1 to day 9 and 30% from day 22 to day 30.In that case the median is not fall in the 60% and maybe is still higher than the average.

Is my logic right or wrong?


Stuart Kovinsky wrote:
mguerreiro001 wrote:In City X last April, was the average (arith mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high tempratures was 2,160o.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.
Average is the sum of the terms/# of terms.
Median is the middle term of a set of numbers (or, if there are an even number of terms, the average of the two middle terms).

Q: was average > median?

(1) We have the sum and the number of terms, so we can calculate the average, but we have no information about the individual terms, so we can't calculate the median.

Insufficient

(2) if 60% of the daily highs were below the average, then 40% of the highs were at or above the average. Since the median is the middle term (which occurs at the 50th percentile), the median will fall within the 60% of the highs that are below the average.

Therefore, median is < average: sufficient.

(2) is suff, (1) is insuff - choose (B)
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by sam2304 » Tue Apr 10, 2012 10:47 pm
mido362 wrote:Why are we considering that the median (15th and 16th day) will fall in the 60%?

In other word maybe the 60 % is distributed as follows:
30% from day 1 to day 9 and 30% from day 22 to day 30.In that case the median is not fall in the 60% and maybe is still higher than the average.

Is my logic right or wrong?
Your question is right. The values can be anywhere but you go wrong in calculation. When you are finding the median, the values should be in increasing order. So the values may be on any day as said by you but we have to arrange it in increasing order to find the median. Median will be 50th percentile and average will fall in the 60th percentile. Sufficient to answer our question.
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by mido362 » Wed Apr 11, 2012 4:10 am
Thanks Sam...now i realized my mistake
Last edited by mido362 on Wed Apr 11, 2012 4:57 am, edited 1 time in total.
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by mido362 » Wed Apr 11, 2012 4:10 am
Thanks Sam...now i realized my mistake
Last edited by mido362 on Wed Apr 11, 2012 4:57 am, edited 1 time in total.
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by mido362 » Wed Apr 11, 2012 4:10 am
Thanks Sam...now i realized my mistake
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by [email protected] » Mon Apr 09, 2018 3:06 pm
Hi All,

We're asked if, in City X last April, the AVERAGE (arith mean) daily high temperature was GREATER than the MEDIAN daily high temperature. This is a YES/NO question and is a great example of a 'concept' question (meaning that you don't have to do much math to answer it IF you recognize the concepts involved).

1) In City X last April, the sum of the 30 daily high tempratures was 2,160o.

With the information in Fact 1, we can calculate the AVERAGE high temperature, but we have no way of knowing how the MEDIAN actually relates to this average (the MEDIAN could be higher, lower or equal to the AVERAGE).
Fact 1 is INSUFFICIENT

2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

By definition, when we have 30 individual temperatures, the MEDIAN will equal the average of the "middle two" temperatures (after we've put the temperatures in order from least to greatest). Fact 2 tells us that 60 PERCENT of the daily high temperatures were LESS than the average - and since 60% is far more than half of the numbers, the two numbers that would be used to calculate the MEDIAN would be in that group. Thus, we know that the MEDIAN would ALWAYS be less than the AVERAGE - and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer: B

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