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26 May 2017, 04:43
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In a certain company, the average (arithmetic mean) salary of finance executives is \(\)x\(and the average (arithmetic mean) salary of marketing executives is\)\(y\). Is the average (arithmetic mean) salary of finance executives and marketing executives combined greater than \(\)\frac{x+y}{2}\(?\\
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(1) The number of the finance executives is less than the number of the marketing executives.\\
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(2)\)x\(is\)10,000 less than \(y\).
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(1) The number of the finance executives is less than the number of the marketing executives.\\
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(2)\)x\(is\)10,000 less than \(y\).
26 May 2017, 04:45
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Official Solution:
If you take the 1st step of the variable approach and modify the original condition and the question, you can see that it is a "2 by 2 question" which frequently appears in the current Gmat math as shown in the table below.
finance marketing Number f m Average salary x y
Question: \(\frac{fx+my}{f+m} > \frac{x+y}{2}\) ? and if you expand this, you get \(2(fx+my)>(f+m)(x+y)\)?, or \(2fx + 2my>fx + fy + mx + my\)?. If you move this to one side, you get \(fx + my - fy - mx > 0\)?, \(f(x-y) - m(x-y)>0\)?, or \((f-m)(x-y)>0\)?.
In the original condition, there are 4 variables \((f,m,x,y)\). In order to match the number of variables to the number of equations, there must be 4 equations. Therefore, E is most likely to be the answer.By solving con 1) & con 2), you get from \(f < m\) (by con 2) and \(x = y - 10,000\) to \(f - m < 0\) and \(x - y = -10,000 < 0\), then finally to \((f- m)(x - y) = ( - )( - ) > 0\), hence always yes and sufficient.Therefore, the answer is C.
Answer: C
If you take the 1st step of the variable approach and modify the original condition and the question, you can see that it is a "2 by 2 question" which frequently appears in the current Gmat math as shown in the table below.
finance marketing Number f m Average salary x y
Question: \(\frac{fx+my}{f+m} > \frac{x+y}{2}\) ? and if you expand this, you get \(2(fx+my)>(f+m)(x+y)\)?, or \(2fx + 2my>fx + fy + mx + my\)?. If you move this to one side, you get \(fx + my - fy - mx > 0\)?, \(f(x-y) - m(x-y)>0\)?, or \((f-m)(x-y)>0\)?.
In the original condition, there are 4 variables \((f,m,x,y)\). In order to match the number of variables to the number of equations, there must be 4 equations. Therefore, E is most likely to be the answer.By solving con 1) & con 2), you get from \(f < m\) (by con 2) and \(x = y - 10,000\) to \(f - m < 0\) and \(x - y = -10,000 < 0\), then finally to \((f- m)(x - y) = ( - )( - ) > 0\), hence always yes and sufficient.Therefore, the answer is C.
Answer: C
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