Take statement 2..
What if there are like 130K, 130K.. .. (14 times 130K).. and last one has to be greater than 150K to make mean 150K.
So second out.. B and D gone..

Take statement 3.
Take the same example as in Statement 2..
B and E out..

Answer = A.

The above poster's approach is great.

If we understand medians, and have good technique, then we can answer this question very quickly.

Median just refers to the middle number in a sequence of ordered numbers. So, the median here: {3, 3, 3} is 3 even though all the numbers are 3.

We are told that the median is 130k. So, the 8th house sold for 130k. But the 1st through 7th houses may also have sold for 130k. Eliminate III; eliminate C and E.

We can also easily eliminate II. We could have 8 houses that sold for 130k or less; the rest can sell for well above 150k. Eliminate B and D.

The answer must be choice A, and there is no need to evaluate I.

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

Correct

I did it at other place. Some has opened a new thread.
https://www.beatthegmat.com/last-month-t58988.html

My turn to burn up my fatigue lazy brain...

Median = 130,000
Mean = 150,000

Since the Mean is 150,000 and the median is 130,000, we can determine that at least on house must be sold over 165,000

because

Mean > Median

150,000 - 130,000 = 20,000

Since Mean is greater than Median we can assume that at least one house sold 170,000