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22 Oct 2012, 06:03
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
48% (01:53) correct
52%
(02:02)
wrong
based on 293
sessions
History
Date
Time
Result
Not Attempted Yet
If the average (arithmetic mean) of 4 different numbers is 30, how many of the numbers are less than 30?
(1) The median of the numbers is 30.
(2) The largest and smallest of the numbers are odd, and the two middle numbers are even.
Is there an alternative method instead of picking numbers?
(1) The median of the numbers is 30.
(2) The largest and smallest of the numbers are odd, and the two middle numbers are even.
Is there an alternative method instead of picking numbers?
22 Oct 2012, 08:04
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1) There are four numbers and the median is 30. So the average of the 2nd highest number and the 3rd highest number is 30. Since the numbers are all different, they cannot both be 30. So one has to be higher than 30 while the other has to be lower. So two numbers are lesser than 30. The fact that the average is 30 does not even have to be used here. Sufficient.
2) This statement does not give any valuable information. I think we can easily form two different sets that satisfy the given conditions although I have no idea how to do it without picking numbers. Insufficient.
Hence answer is A.
Kudos Please.... If my post helped
2) This statement does not give any valuable information. I think we can easily form two different sets that satisfy the given conditions although I have no idea how to do it without picking numbers. Insufficient.
Hence answer is A.
Kudos Please.... If my post helped
General Discussion
19 Nov 2013, 15:15
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If the 4 numbers are different, "at least one is <30 and two are >30" or "at least one is >30 and two are <30"
1) Tells you that none of them are = 30, so two will be >30 and two will be <30.
2) Tells you nothing.
IMO A
1) Tells you that none of them are = 30, so two will be >30 and two will be <30.
2) Tells you nothing.
IMO A
