swerve 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.
The OA is C.
Please, can anyone explain this DS question? I tried to solve it but I don't understand why C is the correct answer. I need help. Thanks.
Given, that the average (arithmetic mean) of four different numbers is 30, the sum of the four numbers = 30*4 = 120.
We have to find out the count of the numbers that are greater than 30.
Let's take each statement one by one.
1) None of the four numbers is greater than 60.
Say, one of the numbers is 60, thus, the sum of other three = 120 - 60 = 60.
Case 1: One of the possibilities: Other three are: 10, 20, 30. Only one number (60) is greater than 30.
Case 2: Another possibility: Other three are: 5, 15, 40. Two numbers (40 & 60) are greater than 30.
No unique answer. Insufficient.
2) Two of the four numbers are 9 and 10, respectively.
=> Sum of the other two numbers = 120 - 9 - 10 = 101.
Case 1: Say a number is 60; thus, the other number = 101 - 60 = 41. Two numbers (41 & 60) are greater than 30.
Case 2: Say a number is 30; thus, the other number = 101 - 30 = 71. Only one number (71) is greater than 30.
No unique answer. Insufficient.
(1) and (2) together
From Statement (1), we have the sum of the other two numbers = 101.
Since the average of the numbers is 30, and two numbers are less than 30 (9 and 10), at least one number must be greater than 30. Say, the number is 31; thus, the 4th number is 101 - 31 = 70 (No possible since the maximum value of a number can be 60). Say, the number is 51; thus, the 4th number is 101 - 51 = 50. Two numbers (51 & 50) are greater than 30.
Now, let's make use of the Statement (1). Say a number is 60 (maximum possible); thus, the 4th number is 101 - 60 = 41. Two numbers (41 & 60) are greater than 30.
Unique answer. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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