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Last month 15 homes were sold in Town X. The

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Last month 15 homes were sold in Town X. The [#permalink] New post 30 Sep 2007, 19:34
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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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 [#permalink] New post 30 Sep 2007, 20:43
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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 [#permalink] New post 01 Oct 2007, 12:25
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 10,000--it 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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 [#permalink] New post 01 Oct 2007, 12:41
The answer is I only... It is hard to do in 2 mins...
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 [#permalink] New post 01 Oct 2007, 12:52
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 10,000--it 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
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 [#permalink] New post 01 Oct 2007, 21:31
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.
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 [#permalink] New post 02 Oct 2007, 04:01
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 [15x150000-14x130000].
This knocks off st. II and III.
Only st. I will hold.
  [#permalink] 02 Oct 2007, 04:01
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