Bunuel wrote:
If 6 positive integers have a sum less than 10, what is the value of the greatest of the integers?
(1) The average (arithmetic mean) of the 6 integers is 3/2.
(2) The median of the integers is 1.
Let the integers in ascending order be a, b, c, d, e, f. We are given that a+b+c+d+e+f < 10. WE have to find 'f'.
(1) Avg = 3/2. So sum = 3/2 * 6 = 9. This means
a+b+c+d+e+f = 9. So there is a possibility that a=b=c=d=e=1, and then f=4. Or we could have a=b=c=1, and d=e=f=2. Multiple possibilities for 'f'. So
Insufficient.
(2) Median is 1. So the average of middle two integers (c &d) is 1. Which means c+d = 2. This is only possible when c=d=1. And since they are in ascending order, this also means that a=b=1 too (no integer can be 0 since they are all positive integers). So a=b=c=d=1. This means e+f < 6 (total sum has to be less than 10). But we could have e=1, f=4 Or we could have e=2, f=3. Multiple possibilities for 'f'.
Insufficient.
Combining the statements, we get that a=b=c=d=1. And now e+f=5. But again if e=1, f=4 and if e=2, f=3. WE cant uniquely determine value of 'f'.
Insufficient.
Hence
E answer