Vincen wrote:a, b, c, d, and e are integers. Is the median of the integers greater than the average (arithmetic mean) of the integers?
1) a<b<c<d<e
2) b-a=e-d
The OA is E.
I need some help. First, I forgot what is the median. And how can I solve this DS question?
Statement 1: a < b < c < d < e
Case 1: Say the integers are: 1 < 2 < 3 < 4 < 5
=> Median = 3;
=> Arithmetic mean = (1+2+3+4+5)/5 = 3. AM = Median. The answer is No.
Case 2: Say the integers are: 1 < 2 < 40 < 48 < 49
=> Median = 40;
=> Arithmetic mean = (1 + 2 + 40 + 48 + 49)/5 = 140/5 = 28. Median (40) > AM (28). The answer is Yes.
No unique value. Insufficient.
Statement 1: b - a = e - d
We have no information about c. Insufficient.
Statement 1 & 2 together:
Both the cases discussed in Statement 1 are applicable here too. Thus, both together are not sufficient.
The correct answer:
E
Hope this helps!
-Jay
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