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Help - Tired of missing this type of question

This topic has 1 expert reply and 6 member replies
mguerreiro001 Junior | Next Rank: 30 Posts Default Avatar
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Help - Tired of missing this type of question

Post Sat Jan 19, 2008 12:22 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Stuart Kovinsky GMAT Instructor
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    Post 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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    mguerreiro001 Junior | Next Rank: 30 Posts Default Avatar
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    Post Sat Jan 19, 2008 2:02 pm
    Thank you. I think I got the idea.

    mido362 Newbie | Next Rank: 10 Posts Default Avatar
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    Post 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)

    sam2304 Legendary Member
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    Post 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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    Thanked by: mido362
    mido362 Newbie | Next Rank: 10 Posts Default Avatar
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    6 messages
    Post 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

    mido362 Newbie | Next Rank: 10 Posts Default Avatar
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    6 messages
    Post 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

    mido362 Newbie | Next Rank: 10 Posts Default Avatar
    Joined
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    6 messages
    Post Wed Apr 11, 2012 4:10 am
    Thanks Sam...now i realized my mistake

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