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26 May 2017, 04:43
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Set S has 5 numbers such that its average (arithmetic mean) is greater than its median. Also, set T has 7 numbers such that its average (arithmetic mean) is greater than its median. If all the elements in set S and set T are different, is the average (arithmetic mean) of set S and set T combined greater than the median of set S and set T?
(1) The average (arithmetic mean) of set T is greater than that of set S.
(2) The median of set T is greater than that of set S.
(1) The average (arithmetic mean) of set T is greater than that of set S.
(2) The median of set T is greater than that of set S.
26 May 2017, 04:45
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Official Solution:
In general, If you take the 1st step of the variable approach and modify the original condition and the question, "relation" of median, mean, or standard deviation mostly has E as the answer. Thus,
by solving con 1) & con 2),
\(S = \{0.1,0.2,0.3,0.4,0.51\}, T = \{1000,1001,1002,1003,1004,1005,1006.5\},\)
and \(Z = \{0.1,0.2,0.3,0.4,0.51, 1000,1001,1002,1003,1004,1005,1006.5\}\)
then, median \(= 1000.5 >\) mean\(=585\ldots \ldots\) NO
\(S = \{0.1,0.2,0.3,0.4,0.52\}, T = \{0.01,0.11,0.21,0.31,0.41,0.51,0.62\},\)
and \(Z = \{0.01,0.1,0.11,0.21,0.3,0.31,0.4,0.41,0.51,0.52,0.62\}\)
then, median\(=\frac{0.3+0.31}{2} = 0.305 < \frac{0.3(5)+0.31(7)}{5+7}=0.3058\ldots ..\)=mean YES
For this question, the two answers"yes" and "no" coexist, hence it is not sufficient. The answer is E. As such, you need to remember that when median, mean, or standard deviation is"relation", E is most likely to be the answer.
Answer: E
In general, If you take the 1st step of the variable approach and modify the original condition and the question, "relation" of median, mean, or standard deviation mostly has E as the answer. Thus,
by solving con 1) & con 2),
\(S = \{0.1,0.2,0.3,0.4,0.51\}, T = \{1000,1001,1002,1003,1004,1005,1006.5\},\)
and \(Z = \{0.1,0.2,0.3,0.4,0.51, 1000,1001,1002,1003,1004,1005,1006.5\}\)
then, median \(= 1000.5 >\) mean\(=585\ldots \ldots\) NO
\(S = \{0.1,0.2,0.3,0.4,0.52\}, T = \{0.01,0.11,0.21,0.31,0.41,0.51,0.62\},\)
and \(Z = \{0.01,0.1,0.11,0.21,0.3,0.31,0.4,0.41,0.51,0.52,0.62\}\)
then, median\(=\frac{0.3+0.31}{2} = 0.305 < \frac{0.3(5)+0.31(7)}{5+7}=0.3058\ldots ..\)=mean YES
For this question, the two answers"yes" and "no" coexist, hence it is not sufficient. The answer is E. As such, you need to remember that when median, mean, or standard deviation is"relation", E is most likely to be the answer.
Answer: E
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