swerve wrote:The arithmetic mean of four integers is 30. How many of the integers are below 30?
1) Each integer is less than 60.
2) Two of the four integers are 9 and 10.
The OA is C.
Please, can any expert explain this DS question for me? I don't understand why that is the correct answer. Thanks.
If the arithmetic mean of four integers is 30, then we know that those integers sum to 4*30 = 120.
Statement 1 - Case 1: [30, 30, 30, 30] there are zero integers below 30.
Case 2: [40, 40, 20, 20 ] two integers are below 30.
This statement alone is not sufficient
Statement 2: If two of the integers are 9 and 10, the other two integers must sum to 101. Case 1: [51, 50, 9, 10] Two integers are below 30
Case 2: [100, 1, 9, 10] Three integers are below 30.
This statement alone is not sufficient
Together: We know that two integers in the set are 9 and 10, so the other two integers must sum to 101. If a new term were under 30, it could be, at most, 29, meaning the last term would have to be 101 - 29 = 72. But we know that each term is less than 60! So if the last two terms must sum to 101, and neither of them can exceed 60, we cannot have any additional terms under 30, and no matter what, we'll have exactly two integers under 30. This statement alone is sufficient. The answer is
C
