Median = 130,000

8th house = 130,000
Lets give max price to each 1st to 7th = 130,000

Sum of price from 9th to 15th house = (150,000 * 15 - 130,000 * 8) = 1210,000

Each from 9th to 15th house = 173,000 approx, If we distribute evenly

9th to 14th house each = 130,000 and 15th house = rest (1210,000 - 130,000*6), which is > 165,000

So, at least one house must be > 165,000

I - True

II - Not necessary; we can distribute it (<130,000 and >150,000)

III - Not necessary; I already given one distribution above which have equal to 130,000

[The post below is copied from this thread]

One effective approach to "must be" questions is to disprove each statement by showing that it doesn't have to be whatever it claims to be. In this case, we're asked "which statements must be true?" so our approach is to try to show that a statement could be false, thus eliminating it from the solution set

We know the average of 15 homes to be 150, so the sum is 2,250. The median (cost of home #8) is 130.

I. To disprove this, make the greatest price as small as possible, by making the bottom homes as large as possible. The greatest possible values of the bottom #8 homes is the median, 130. In this case, the bottom 8 would add up to 130*8=1,040. Consequently, to reach the known sum of 2,250, the top 7 homes must add up 2,250-1,040 = 1,210 and average 1210/7 > 172. In conclusion, if we maximize everything else, we get the minimum possible value of the greatest home. This minimum is greater than 172, so it must be true that at least one home is sold for more than 165. I must be true.

II. In I above, we saw a case that disproves this statement. we could have the bottom 8 homes @ 130 and the top 7 homes around 172. It's not necessary to have a home between 130 and 150.

III. Again, the case we used in I and II disproves this statement. The bottom 8 homes could be @ 130. It's not necessary to have a home under 130.

The answer is A.

TAKE-AWAYS:
1) Average formula has 3 components (avg, sum, # of terms). If a question ever gives you 2 of the components, it's really giving you the 3rd under the table, so you should find it.
2) When trying to minimize a specific value within a set, maximize all the others. Do the opposite when trying to maximize a specific value.
3) In median problems, always be aware that many numbers can equal the median.

If you still have trouble understanding this solution, consider the step-by-step video solution. This is GMATPrep question 1023.

You can practice similar questions from GMATPrep by using the Drill Engine to generate timed drills and by setting topic='statistics' and difficulty='700'

-Patrick