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Given the ascending set of positive integers {a, b, c, d, e, f}, is the median greater than the mean?
(1) a + e = (3/4)(c + d)
(2) b + f = (4/3)(c + d)
(1) a + e = (3/4)(c + d)
(2) b + f = (4/3)(c + d)
04 Aug 2012, 16:32
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Given the ascending set of positive integers {a, b, c, d, e, f}, is the median greater than the mean?
The median of a set with even number of elements is the average of two middle elements when arranged in ascending/descending order. Thus, the median of {a, b, c, d, e, f} is \(\frac{c+d}{2}\).
So, the question asks: is \(\frac{c+d}{2}>\frac{a+b+c+d+e+f}{6}\)? --> is \(3c+3d>a+b+c+d+e+f\)? --> is \(2(c+d)>a+b+e+f\)?
(1) a + e = (3/4)(c + d) --> the question becomes: is \(2(c+d)>b+f+\frac{3}{4}(c + d)\)? --> is \(\frac{5}{4}(c + d)>b+f\)? Not sufficient.
(2) b + f = (4/3)(c + d). The same way as above you can derive that this statement is not sufficient.
(1)+(2) The question in (1) became: is \(\frac{5}{4}(c + d)>b+f\)? Since (2) says that \(b + f = \frac{4}{3}(c + d)\), then the question becomes: is \(\frac{5}{4}(c + d)>\frac{4}{3}(c + d)\)? --> is \(\frac{1}{12}(c+d)<0\)? --> is \(c+d<0\)? As given that \(c\) and \(d\) are positive numbers, then the answer to this question is definite NO. Sufficient.
Answer: C.
Not a good question.
The median of a set with even number of elements is the average of two middle elements when arranged in ascending/descending order. Thus, the median of {a, b, c, d, e, f} is \(\frac{c+d}{2}\).
So, the question asks: is \(\frac{c+d}{2}>\frac{a+b+c+d+e+f}{6}\)? --> is \(3c+3d>a+b+c+d+e+f\)? --> is \(2(c+d)>a+b+e+f\)?
(1) a + e = (3/4)(c + d) --> the question becomes: is \(2(c+d)>b+f+\frac{3}{4}(c + d)\)? --> is \(\frac{5}{4}(c + d)>b+f\)? Not sufficient.
(2) b + f = (4/3)(c + d). The same way as above you can derive that this statement is not sufficient.
(1)+(2) The question in (1) became: is \(\frac{5}{4}(c + d)>b+f\)? Since (2) says that \(b + f = \frac{4}{3}(c + d)\), then the question becomes: is \(\frac{5}{4}(c + d)>\frac{4}{3}(c + d)\)? --> is \(\frac{1}{12}(c+d)<0\)? --> is \(c+d<0\)? As given that \(c\) and \(d\) are positive numbers, then the answer to this question is definite NO. Sufficient.
Answer: C.
Not a good question.
General Discussion
04 Aug 2012, 18:37
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Bunuel Is this not a GMAT type question ?
Bunuel
Given the ascending set of positive integers {a, b, c, d, e, f}, is the median greater than the mean?
The median of a set with even number of elements is the average of two middle elements when arranged in ascending/descending order. Thus, the median of {a, b, c, d, e, f} is \(\frac{c+d}{2}\).
So, the question asks: is \(\frac{c+d}{2}>\frac{a+b+c+d+e+f}{6}\)? --> is \(3c+3d>a+b+c+d+e+f\)? --> is \(2(c+d)>a+b+e+f\)?
(1) a + e = (3/4)(c + d) --> the question becomes: is \(2(c+d)>b+f+\frac{3}{4}(c + d)\)? --> is \(\frac{5}{4}(c + d)>b+f\)? Not sufficient.
(2) b + f = (4/3)(c + d). The same way as above you can derive that this statement is not sufficient.
(1)+(2) The question in (1) became: is \(\frac{5}{4}(c + d)>b+f\)? Since (2) says that \(b + f = \frac{4}{3}(c + d)\), then the question becomes: is \(\frac{5}{4}(c + d)>\frac{4}{3}(c + d)\)? --> is \(\frac{1}{12}(c+d)<0\)? --> is \(c+d<0\)? As given that \(c\) and \(d\) are positive numbers, then the answer to this question is definite NO. Sufficient.
Answer: C.
Not a good question.
The median of a set with even number of elements is the average of two middle elements when arranged in ascending/descending order. Thus, the median of {a, b, c, d, e, f} is \(\frac{c+d}{2}\).
So, the question asks: is \(\frac{c+d}{2}>\frac{a+b+c+d+e+f}{6}\)? --> is \(3c+3d>a+b+c+d+e+f\)? --> is \(2(c+d)>a+b+e+f\)?
(1) a + e = (3/4)(c + d) --> the question becomes: is \(2(c+d)>b+f+\frac{3}{4}(c + d)\)? --> is \(\frac{5}{4}(c + d)>b+f\)? Not sufficient.
(2) b + f = (4/3)(c + d). The same way as above you can derive that this statement is not sufficient.
(1)+(2) The question in (1) became: is \(\frac{5}{4}(c + d)>b+f\)? Since (2) says that \(b + f = \frac{4}{3}(c + d)\), then the question becomes: is \(\frac{5}{4}(c + d)>\frac{4}{3}(c + d)\)? --> is \(\frac{1}{12}(c+d)<0\)? --> is \(c+d<0\)? As given that \(c\) and \(d\) are positive numbers, then the answer to this question is definite NO. Sufficient.
Answer: C.
Not a good question.
