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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.

Re: Given the ascending set of positive integers {a, b, c, d, e, [#permalink]

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12 Feb 2013, 14:29

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if all of the integers are positive, then how come c+d<o ? question system contradicts with the solution... You are right Bunuel.. not an air tight question.

Re: Given the ascending set of positive integers {a, b, c, d, e, [#permalink]

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09 Mar 2014, 20:55

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We don't have to do any calculations here. For mean, we have to have the sum of the all the numbers in the set while for for the median c and d are sufficient. Since both the options together can give us the mean in terms of c+d, we can compare that against the mean which is also in terms of c+d. So C should be the right choice.
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Re: Given the ascending set of positive integers {a, b, c, d, e, [#permalink]

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04 Aug 2012, 18:37

Bunuel Is this not a GMAT type question ?

Bunuel wrote:

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.

if all of the integers are positive, then how come c+d<o ? question system contradicts with the solution... You are right Bunuel.. not an air tight question.

The question is fine in that respect.

After some manipulations the question became "is c+d<0?" So, c+d<0 is not a statement, it's a question and since we know that c and d are positive numbers, then the answer to this question is NO.

Re: Given the ascending set of positive integers {a, b, c, d, e, [#permalink]

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22 May 2017, 07:53

siddhans wrote:

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)

The problem asks if (C+D)/2 > (A+B+C+D+E+F)/6?

In other words, is 2C + 2D > A+B+E+F?

Statement 1: A+E = 3/4*(C+D)

We can quickly realize that B and F are unknown, so we cannot answer the question provided (is 2C+2D>A+B+E+F?). Insufficient.

Statement 2: Here we are given B+F = 4/3(C+D). Similarly, we do not know the values of A or E, so this statement is insufficient.

Statements 1+2: When combined, we know that B+F and A+E can be expressed in terms of C and D, so via substitution we can do algebra to arrive at this answer: