jlgdr wrote:
Nice question +1
Just to add a conceptual approach to this one
Statement 1 is insufficient since we can only find the numerical values of the median and the largest number (M=15, L=19) But we still don’t know about the other three numbers. For instance, smallest numbers could be 11 and other two numbers equally spaced giving an evenly spaced set with Median= Mean. We have extra flexibility to place the other two numbers since we are told that the numbers are not necessarily integers.
Statement 2 is a bit more tricky. It says that the smallest number is 10 less than the median. So let's say that the median is 10 and the largest and smallest numbers are 14 and 0 respectively. Now, we will have 2 numbers to the left of 10 and two to the right. Even if we have 10 and 10 to the left and 10 and 10 to the right, the mean will ALWAYS be larger than the median. Therefore B is the correct answer
Hope this helps
Cheers
J
Your conclusion for statement 2 should be the opposite. In fact, intuitively, the average is being "pulled" down more by the first number (median-10) than it is being pulled
up by the last (median+4). This conclusively tells us that the mean will always be smaller than the median. Let's try extreme examples:
0, 10, 10, 10, 14 --> mean = 8.8 < 10
0, 0, 10, 14, 14 --> mean = 7.6 < 10
So the mean is always lower than the median because of the "pull" of the first number, because it is further from the median than the last number.