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Given the ascending set of positive integers {a, b, c, d, e,
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Updated on: 20 Jun 2011, 21:32
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Given the ascending set of positive integers {a, b, c, d, e, f}, is the median greater than the mean? (1) a + e = (3/4)(c + d) (2) b + f = (4/3)(c + d)
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Originally posted by siddhans on 19 Jun 2011, 13:36.
Last edited by siddhans on 20 Jun 2011, 21:32, edited 1 time in total.




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Re: Given the ascending set of positive integers {a, b, c, d, e,
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04 Aug 2012, 15:32
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.
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Re: Given the ascending set of positive integers {a, b, c, d, e,
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04 Aug 2012, 17:37
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.
Answer: C.
Not a good question.



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Re: Given the ascending set of positive integers {a, b, c, d, e,
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04 Aug 2012, 17:40



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Re: Given the ascending set of positive integers {a, b, c, d, e,
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12 Feb 2013, 13:29
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.



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Re: Given the ascending set of positive integers {a, b, c, d, e,
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13 Feb 2013, 00:17



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Re: Given the ascending set of positive integers {a, b, c, d, e,
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09 Mar 2014, 19:55
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,
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10 Dec 2016, 15:44
Below is the best reply that I have found on another forum. It's quite understandable. Median = (c+d)/2 Average = (a+b+c+d+e+f)/6 Median > Average (c+d)/2 > (a+b+c+d+e+f)/6 3c + 3d > a+b+c+d+e+f 2c + 2d > a+b+e+f 2(c+d) > a+b+e+f Thus, the question can be rephrased: Is 2(c+d) > a+b+e+f? Statement 1: a + e = (3/4)(c + d) No information about b+f. Insufficient. Statement 2: b + f = (4/3)(c + d) No information about a+e. Insufficient. Statement 1 and 2 together: Adding the equations, we get: a+b+e+f = (3/4)(c+d) + (4/3)(c+d) a+b+e+f = (3/4 + 4/3)(c+d) a+b+e+f = (25/12)(c+d) Sufficient. The correct answer is C.
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Re: Given the ascending set of positive integers {a, b, c, d, e,
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22 May 2017, 06:53
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: 2C + 2D > 25/12C + 25/12D? No, because 25/12 > 2. Sufficient.



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Re: Given the ascending set of positive integers {a, b, c, d, e,
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19 Dec 2017, 13:18
Hi All, This is a layered question, but it can be solved with a mix of TESTing VALUES and Algebra. We're told that the variables A, B, C, D, E, and F are ASCENDING POSITIVE INTEGERS (meaning that they are unique, A is smallest, F is largest, etc.). We're asked if the MEDIAN is greater than the MEAN. This is a YES/NO question. 1) A+E = (3/4)(C+D) Without information about B and F, your instinct is probably to call this insufficient. We can prove it rather easily though. From this equation, since all of the variables are INTEGERS, we know that A+E must sum to an integer. By extension, (C+D) will have to be a multiple of 4. IF.... C+D = 12, we could have... A=1 B=2 C=5 D=7 E=8 F=9 or higher MEDIAN = (5+7)/2 = 6 MEAN = 32/6 = 5 1/3 OR HIGHER (depending on the value of F) Thus, the median may or may not be greater than the mean (again, depending on the value of F) Fact 1 is INSUFFICIENT 2) B+F = (4/3)(C+D) Here, we have a similar situation to what we had in Fact 1, but here we do not know much about the values of A and E. Here, I'm going to TEST much larger values so that we can see the potential 'swings' in the MEAN... IF.... C+D = 120, we could have... A=148 B=49 C=50 D=70 E=71  110 F=111 MEDIAN = (50+70)/2 = 60 MEAN = 352/6  438/6 = A little less than 60 to a little more than 70. Thus, the median may or may not be greater than the mean (again, depending on the values of A and E) Fact 2 is INSUFFICIENT Combined, there's actually an interesting algebra pattern that we can take advantage of. The sum of ALL of the variables can be put in terms of (C+D).... A+E = (3/4)(C+D) C+D = (C+D) B+F = (4/3)(C+D) Thus, A+B+C+D+E+F = (3/4 + 1 + 4/3)(C+D) Sum = (9/12 + 12/12 + 16/12)(C+D) Sum = (37/12)(C+D) The MEAN of those 6 terms = (37/12)(C+D)/6 = (37/72)(C+D) The MEDIAN of those 6 terms = (C+D)/2 = (1/2)(C+D) = (36/72)(C+D) Comparing the median and the mean now, we can see that the median will ALWAYS be less than the mean. Thus, the answer to the question is ALWAYS NO. Combined, SUFFICIENT. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Given the ascending set of positive integers {a, b, c, d, e, &nbs
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