Is the average of a set of 5 distinct positive integers {a, b, 4, 6, 2

statement 1 -the highest number in the set is 6

1 possible set can be (1,2,3,4,6) Then avg =3.2 and median 3 , so avg > median .
Another set can be (2,3,4,5,6) then avg =4 and median 4, so avg =median
so statement 1 insufficient.

statement 2 -the lowest number in the set is 2
1possible set can be (2,3,4,5,6) then avg =4 and median 4, so avg =median
Another possible set can be (2,3,4,6,10) avg =6 and median 4 , so avg > median
so statement 2 insufficient.

combining 2 statements, only set possible is (2,3,4,5,6). avg = median. we get a definite answer 'No' average is not greater than median. Answer C.

Hey, here's how I would answer this question. I'm going to try and write it out as clearly as I can. Minus the explanation this can be solved well within 2 minutes.

1. Highest number in the set is 6

Since they're all integers and distinct, the mean will be highest when a and b equal 3 & 5 (or vice versa, doesn't matter).

=> {2,3,4,5,6}
Mean = 4
Median = 4

Mean will be at its lowest when a & b equal 1 & 3.

=> {1,2,3,4,6}
Mean = 3.2
Median = 3

This is all we need to know to conclude that the first statement is INSUFFICIENT. In the first case the mean is not greater than the median, but in the second case it is.

2. The lowest number in the set is 2

Since the numbers are positive and there is now no upper limit on the integers, it is important to first find the lowest possible value of the mean since adding 300 or any huge number will ensure that the mean is greater than the median.

a & b equal 3 & 5

=> {2,3,4,5,6}
Mean = 4, Median = 4.

Again we have a case where the mean is not greater and one where it is (i.e. adding a huge number to the set). INSUFFICIENT.

Combining the two we know that 2 is the lowest number and 6 is the highest, and from our calculations it is clear that while the mean can equal the median, it CAN NOT be greater.

Answer: C

Hope this was helpful.

Thanks Guys, i know what i was missing ...
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I have a question w.r.t Statement 2. Can we get a 'Yes' case. The cases that I can see in the official solution and the ones that are posted here produce a 'No' in case of Statement 2 (precisely Mean is not greater than Median, and Mean=Median) I cannot locate a case where Mean > Median and I believe that probably can make Statement 2 sufficient.

Bunuel - I would like to understand where am I going wrong or may be there is an altered version of this question that's present in the database?

Thank You in advance

The average of {2, 4, 6, 100, 200}, which is ~62, is greater than the median of {2, 4, 6, 100, 200}, which is 6.
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Thanks Bunuel

Is the average of a set of 5 distinct positive integers {\(a\), \(b\), 4, 6, 2} greater than the median?

First of all, we must remember that the set contains distinct positive integers.

(1) The largest number in the set is 6.

Since 6 is the largest number in the set, both \(a\) and \(b\) must be less than 6. Therefore, \(a\) and \(b\) can be any two from 1, 3, and 5. If \(a = 1\) and \(b = 3\), the median is 3 and the average is 3.2. However, if \(a = 3\) and \(b = 5\), the average and the median are equal, at 4. Not sufficient.

(2) The smallest number in the set is 2.

Since 2 is the smallest number in the set, both \(a\) and \(b\) must be greater than 2. If \(a\) and \(b\) are very large numbers, say 10 and 20, the average will be greater than the median, which will be 6. However, if \(a = 3\) and \(b = 5\), the average and the median are equal, at 4. Not sufficient.

(1)+(2) From the information given, we know that both \(a\) and \(b\) must be greater than 2, and less than 6. Thus, \(a=3\) and \(b=5\), or vice-versa. In either case, the average and the median are equal, at 4. Sufficient.


Answer: C
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