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by sk818020 » Wed Jul 28, 2010 3:39 pm
I believe this one is going to be A.

The other two are possible but not necessary. Logically solved, for any average, if there is a point in your set of data that is above or below the average there must be a point(s) that is/are on the other side of the average. This is a property of averages.

The average price of 15 homes was 150 and that the median was 130. From this we can conclude two thing;

The sum of all the values of homes was;

15*150=2250

At most half the houses sold for 130. You would use at most because you are trying to figure out what is the very least the top half of the average is.

7*150=1050

So, the other six houses have to account for at least 1200 (2250-1050) of the average. To minimize the cost of the rest of the homes you would simply make them all the same price. So the minimum the highest price house could of cost was 1200/6=200. If you decreased the price of any of the other houses the home that cost the most would increase. Thus, you can conclude that there must have been at least one house that sold for greater than 165.

Hope this helps.

Thanks,

Jared

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by Patrick_GMATFix » Wed Jul 28, 2010 4:51 pm
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
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