In a set of 24 positive integers

David Recine

This is basically a matter of testing cases. With the data we have, could we get different possible answers to this inequality?

The answer is "yes." Why? Well, recall that the median is defined one of two ways. In an odd list of numbers (five different numbers, twenty-seven different numbers, etc...), the median is the single number that is exactly in the middle of the set, with an equal amount of numbers on either side. But for an even set of numbers, such as a set of eight numbers, or the set of twenty-four numbers seen in this problem, there is no one middle number. Instead, the median is an average of the two numbers in the middle.

Revisiting the problem, we know that the two middle numbers consist of the last, highest number in the first twelve numbers, and the first, lowest number in the second twelve numbers. And we know that every number in the first twelve numbers is less than 50, while every number int he last twelve numbers is greater than 50.

What we don't know is HOW MUCH less than 50 the first 12 numbers are, or how much greater than 50 the second half of the numbers are.

If the first half of the numbers are all a lot less than 50, and the second half of the numbers are all only a little bit more than 50, then the median will be less than 50. Let's take a look at such a case. Suppose that the numbers are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, (12, 51), 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 63

The median will be the average of the two middle numbers, which I've put in parentheses. Here, we have a median of (12+51)/2, or 63/2, which is 31.5, much less than 50. In that case, Quantity A < Quantity B.

So if the smaller middle value is as far from 50, and the larger middle value is close to 50, A < B. But what if both middle values are as close to 50 as possible? In that case, we'd have 49 and 51 for those middle values, and a median of (49+51)/2 = 100/2 = 50. In that case, we have A = B.

By just testing those two cases (middle values of 12 and 51 vs. middle values of 49 and 51), we know we have two possible answers, and thus don't have enough information to determine a fixed inequality between A and B. This leads us to answer choice D.

But just for the sake of fully analyzing this problem, what if the lower middle value was as close to 50 as possible, but the higher middle value was very far above 50. Let's say, for instance, that the middle values were 49 and 1001. (49+1001)/2 = 1050/2 = 525. In that and similar cases, A > B.

Hopefully that clears things up, but let me know if you still have any doubts. :)

Aug 21, 2018 • Reply

Abigail Stricker

Sorry for the delay, thank you David! I got it completely.

Sep 4, 2018 • Reply