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22 May 2017, 08:26
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
57% (01:31) correct
43%
(01:14)
wrong
based on 216
sessions
History
Date
Time
Result
Not Attempted Yet
Find the Median of the set of integers A?
1) Mean of the set A is 10
2) Range of the set A is 2
Source: www.GMATinsight.com
1) Mean of the set A is 10
2) Range of the set A is 2
Source: www.GMATinsight.com
22 May 2017, 08:33
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either of them individually is not sufficient. Taking them simultaneously will give only one option for such a set i.e (9,10,11) => sufficient -> Option C
22 May 2017, 10:54
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I'm going to confess to getting this one wrong when I did it without writing anything down! (Always use your scratch paper). I thought that the answer had to be E, because I know there are a lot of different sets that could have a mean of 10. We don't know how many numbers are in the set, and we don't know how big the numbers are, and that just seems like too little information.
However, the answer is C. The trick is the constraint in the problem. All of the numbers have to be integers! What I should have done, was when I thought it was insufficient, I should have tried to prove it. I would have found that the only possibilities that work are these:
(9, 11)
(9, 9, 11, 11)
(9, 9, 9, 11, 11, 11)
...
(9, 10, 11)
(9, 10, 10, 11)
(9, 9, 10, 10, 11, 11)
etc.
Either the only values are 9 and 11, in which case the median is the average of 9 and 11, or 10. Or, the values are 9, 10, and 11, and the middle value is 10.
If you try to do something tricky, it won't work. For instance, you can't do (9.5, 9.5, 9.5, 11.5), even though the average is right - because they have to be integers. And you can't do (8, 10, 10, 10, 10, 10, ...), because the average will come out to a tiny bit less than 10.
However, the answer is C. The trick is the constraint in the problem. All of the numbers have to be integers! What I should have done, was when I thought it was insufficient, I should have tried to prove it. I would have found that the only possibilities that work are these:
(9, 11)
(9, 9, 11, 11)
(9, 9, 9, 11, 11, 11)
...
(9, 10, 11)
(9, 10, 10, 11)
(9, 9, 10, 10, 11, 11)
etc.
Either the only values are 9 and 11, in which case the median is the average of 9 and 11, or 10. Or, the values are 9, 10, and 11, and the middle value is 10.
If you try to do something tricky, it won't work. For instance, you can't do (9.5, 9.5, 9.5, 11.5), even though the average is right - because they have to be integers. And you can't do (8, 10, 10, 10, 10, 10, ...), because the average will come out to a tiny bit less than 10.
