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Last month 15 homes were sold in Town X. The [#permalink]
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30 Sep 2007, 20:34
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Last month 15 homes were sold in Town X. The average(aritmetic mean) sale price of the homes was 150,000 and the median sale price was 130,000. Which of the following statements must be true ?
I. At least one of the homes was sold for more than $165,000
II. At least one of the homes was sold for 130,000 and less than 150,000
III. At least one of the homes was sold for less than 13000



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Joined: 13 Feb 2007
Posts: 63

this is hard to do in 2 min.
i used a calculator, and statement I seems to be true.
statement II seems to be true
i think statement III can be false.
there are some problems with this question though
II says one house should be sold for 130,000 and less than 150,000. if one was sold for 130k then it was certainly sold for less than 150k.
i think that III should be 130,000.



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Joined: 01 Oct 2007
Posts: 87

I'm assuming II is (greater than 130,000 and less than 150,000), and III is (less than 130,000). To make calculations easier, divide everything by a factor of 10,000it won't change any of the relationships.
If the mean of the houses is 15, then their total value is 225.
If the median is 13, that means that 7 values are >= 13, and 7 are <= 13.
Suppose that all the houses up to the median value are the same value: 13. What's their total value? 13*8 = 104. What's the total value of the 7 houses over the median value, then? 121. Therefore, the average value of these 7 houses is 121/7 = 17 2/7.
Now suppose the actual values of those 7 houses is equal to that mean of 17 2/7. We have a counterexample to conditions II and III. There are 8 houses = 13, and 7 houses = 17 2/7. No houses are less than 13, and no houses are between 13 and 16.5.
That might be enough to answer the question, depending on the solutions. But what about condition I? Given that the median's 13, our example assigns the largest possible value to the houses in the bottom half of the distribution. That means that 17 2/7 is the least possible value for the mean of the houses in the top half of the distribution. That means that at least one of the houses must have a value > = 17 2/7. So I must be true.



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Joined: 18 Jun 2007
Posts: 67

The answer is I only... It is hard to do in 2 mins...



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Joined: 02 Aug 2007
Posts: 346
Location: Greater New York City area
Schools: Tuck, Ross (R1), Duke, Tepper, ISB (R2), Kenan Flagler (R2)

johnrb wrote: I'm assuming II is (greater than 130,000 and less than 150,000), and III is (less than 130,000). To make calculations easier, divide everything by a factor of 10,000it won't change any of the relationships.
If the mean of the houses is 15, then their total value is 225.
If the median is 13, that means that 7 values are >= 13, and 7 are <= 13.
Suppose that all the houses up to the median value are the same value: 13. What's their total value? 13*8 = 104. What's the total value of the 7 houses over the median value, then? 121. Therefore, the average value of these 7 houses is 121/7 = 17 2/7.
Now suppose the actual values of those 7 houses is equal to that mean of 17 2/7. We have a counterexample to conditions II and III. There are 8 houses = 13, and 7 houses = 17 2/7. No houses are less than 13, and no houses are between 13 and 16.5.
That might be enough to answer the question, depending on the solutions. But what about condition I? Given that the median's 13, our example assigns the largest possible value to the houses in the bottom half of the distribution. That means that 17 2/7 is the least possible value for the mean of the houses in the top half of the distribution. That means that at least one of the houses must have a value > = 17 2/7. So I must be true.
Johnrb, hats off..Excellent explanation



Senior Manager
Joined: 27 Jul 2006
Posts: 296

Maybe my view is wrong, but I think that I and II are the answer.
See we know that the 8th house has a value of 130,000 and that the seven below it have a value of 130,000 or less.
We also know that the average is 150,000 which makes the least possible value of the 7 above the median about 173,000 dollars (I got this figure by taking the 8*20000 needed for the first homes to get to 150,000 and divided it by the seven homes and added this to the 150,000 needed.



Manager
Status: Post MBA, working in the area of Development Finance
Joined: 09 Oct 2006
Posts: 169
Location: Africa

For a quick POE, it would be a good idea to have a skewed disribution
Imagine 14 houses of 130000 and one house of 430000 [15x15000014x130000].
This knocks off st. II and III.
Only st. I will hold.










