15/2 +1 makes median, hence 8 homes were sold for not more than $130,000 and 7 homes were sold for >$130,000. III) <130,000 isn't right as we know that 8 homes MUST BE not more than $130,000, BUT we don't know if they are less. II) 150,000*15-130,000*8=1,460,000 spread for 7 homes makes not less than 200,000 each home. Thus we don't know for sure if house >130,000 and <150,000. I) We know for sure that at least one house is sold for >165,000. Why actually as explained before 7 homes average beyond the median could be evenly spread with not less than $200,000. And if not 165,000 then even 200,000 could be placed in I).

BUT I sincerely think that this problem can be solved much easier and speedier way if we cut off zeros and put the revised way into solution

Last month 15 homes were sold in Town X. The average sale price of the houses was $15 and the median sales price was $13. Which must be true?

At least 1 house sold for > $16.5
At least 1 house sold for > $13 but < $15
At least 1 home sold for < $13

Akansha wrote:Last month 15 homes were sold in Town X. The average sale price of the
houses was $150,000 and the median sales price was $130,000. Which must be
true?
At least 1 house sold for > $165,000
At least 1 house sold for > $130,000 but < $150,000
At least 1 home sold for < $130,000
a. I
b. II
c. III
d. I and II
e. I and III

OA is A

Hi,
Let the 15 prices when placed in increasing order be a1,a2,...a15, where a1<=a2<=...<=a15
median = a8 = 150K
Avg = 150K
Case(I) Let's assume no house is sold above 165K. So, we shall assign maximum possible values to the sales prices.
So, a1=a2=..=a8=130
a9=a10=...=a15=165
now, average = [(130).8+(165).7]/15 = 2195K/15 <150K
so, our assumption that no house is sold above 165K is false
Hence, I must be true

Case(II)
One counter example is a1=a2=..a8 = 130K, a9=a10...=a14=170K,a15=190K
So, II need not be true

Case(III)
The example used in case II should be sufficient to disprove this
So, III need not be true
Hence, answer A

Cheers!