venmic wrote:If the average (arithmetic mean) of four different numbers is 30, how many of the
numbers are greater than 30?
(1) None of the four numbers is greater than 60.
(2) Two of the four numbers are 9 and 10, respectively.

C

Can anyone explain how best to do this one



what is the underlying concept .. I dont want to plug in numbers and check because it takes a long time

I think the assumption here is that we are not dealing with negative numbers.

It's given that a + b + c + d = 120. How many of the numbers a, b, c, and d are greater than 30?

I. The maximum possible value for only one of the four different numbers a, b, c, and d could be 60; and in that case, the remaining three could add up to 60 with only one of them being more than 30. That means we can have two numbers more than 30 in the lot if this freedom is used to the fullest. Otherwise we can also have three of the numbers more than 30, but we can absolutely not have all four of the numbers more than 30 or not a single of the numbers more than 30. This doesn't help us much in answering the objective question. Insufficient

II. If two of the four numbers are 9 and 10, then the remaining two must sum up to 101, and this can be achieved if one or both of the numbers are more than 30. Insufficient

Taking together, sum of two numbers is 101 and none of the numbers is greater than 60. What do we understand? Both of the numbers are greater than 30. Hence only two of the numbers are greater than 30. Answered

[spoiler]Hence sufficient

C[/spoiler]