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Last month 15 homes were sold in Town X. The average (arithmetic 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 more than $130,0000 and less than $150,000 III. At least one of the homes was sold for less than $130,000.

A. I only B. II only C. III only D. I and II E. I and III

Let's start with the first one and try to make it false.

I. At least one of the homes was sold for more than $165,000.

Worst case scenario, when \(x_{15}\) has the least value (trying to make it less than 165), would be when \(x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=max\), and \(x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}=min\): \(8*130+7x=2250\) --> \(x_{min}\approx{173}\).

So we got that I is always true: At least one of the homes was sold for more than $165,000 (as for the worst case scenario we got that least value of \(x_{15}>165\)).

But if we take the scenario which we considered: \(x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130\), and \(x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173}\) we can see that II and III with this scenario are false. So II and III are not always true.

Answer: A (I only).

what if in case 2

i take 149k as the value of 6 properties,making the total as 149*6+1040=1934 2250-1934=316k,which can be the 7th or last property value in order to maintain the average,and if in case the min value is 173k then how 165k is taken as the value of 6 properties.

Sorry, but I was not able to decipher what you were trying to say...
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Re: Last month 15 homes were sold in Town X. The average [#permalink]

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13 Aug 2017, 00:55

Bunuel wrote:

himanshukamra2711 wrote:

Bunuel wrote:

Last month 15 homes were sold in Town X. The average (arithmetic 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 more than $130,0000 and less than $150,000 III. At least one of the homes was sold for less than $130,000.

A. I only B. II only C. III only D. I and II E. I and III

Let's start with the first one and try to make it false.

I. At least one of the homes was sold for more than $165,000.

Worst case scenario, when \(x_{15}\) has the least value (trying to make it less than 165), would be when \(x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=max\), and \(x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}=min\): \(8*130+7x=2250\) --> \(x_{min}\approx{173}\).

So we got that I is always true: At least one of the homes was sold for more than $165,000 (as for the worst case scenario we got that least value of \(x_{15}>165\)).

But if we take the scenario which we considered: \(x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130\), and \(x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173}\) we can see that II and III with this scenario are false. So II and III are not always true.

Answer: A (I only).

what if in case 2

i take 149k as the value of 6 properties,making the total as 149*6+1040=1934 2250-1934=316k,which can be the 7th or last property value in order to maintain the average,and if in case the min value is 173k then how 165k is taken as the value of 6 properties.

Sorry, but I was not able to decipher what you were trying to say...

Option 2. If i take value of house as 149000 for 6 house(149*6+130*8=1934),so now the total is 1934000 and we need to have the total as 225000 so the last 7th house can have the value as 316000.Moreover,if the minimum value of the house is 173000 then how can 165000 be the value of 6 houses.(I am not able to come to the conclusion that if minimum is 173k then atleast one house is of 165k,and if this is so then why not 149k)

Let's start with the first one and try to make it false.

I. At least one of the homes was sold for more than $165,000.

Worst case scenario, when \(x_{15}\) has the least value (trying to make it less than 165), would be when \(x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130=max\), and \(x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}=min\): \(8*130+7x=2250\) --> \(x_{min}\approx{173}\).

So we got that I is always true: At least one of the homes was sold for more than $165,000 (as for the worst case scenario we got that least value of \(x_{15}>165\)).

But if we take the scenario which we considered: \(x_1=x_2=x_3=x_4=x_5=x_6=x_7=x_8=130\), and \(x_9=x_{10}=x_{11}=x_{12}=x_{13}=x_{14}=x_{15}\approx{173}\) we can see that II and III with this scenario are false. So II and III are not always true.

For stmt 1, why cant we have a scenario where houses 1 to 8 are 130 each, house 9 is 160, house 10 is 186 and remaining being the balance average i.e. around 173?

Re: Last month 15 homes were sold in Town X. The average [#permalink]

Show Tags

13 Sep 2017, 03:29

I solved for statement 1,2, and 3 correctly, but while evaluating the options forgot what I deduced - marked E, when the solution could be option A. It seems that it is a problem of attention span - I don't know how to solve such psychological problems, such as this one, lack of focus etc... Could anyone help.
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