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10 Aug 2010, 03:05
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:56) correct
65%
(02:51)
wrong
based on 329
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Of the 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?
(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10
(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10
10 Aug 2010, 03:54
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amijags
Hi, and welcome to Gmat Club. Very good question, so +1 for it. Below is a solution:
Le these 5 numbers in ascending order be \(a\), \(b\), \(c\), \(d\), \(e\): \(a\leq{b}\leq{c}\leq{d}\leq{e}\).
Median would be the middle number - \(c\) and the \(mean=\frac{a+b+c+d+e}{5}\).
Given: \(e=c+4\). Question: is \(\frac{a+b+c+d+e}{5}>c\) --> is \(\frac{a+b+c+d+(c+4)}{5}>c\) --> is \(a+b+d+4>3c\)
(1) The largest number plus the median is 34 --> \(e+c=34\) --> \(c=15\) and \(e=19\) --> question becomes: is \(a+b+d+4>45\). Now, if \(a=b=15\) (max values possible for \(a\) and \(b\)) and \(d=19\) (max value possible for \(d\)) then answer would be YES but as min values of \(a\) and \(b\) are not limited at all then the answer could be NO as well. Not sufficient.
(2) The median minus the smallest number is 10 --> \(a=c-10\) --> question becomes: is \(c-10+b+d+4>3c\) --> \(b+d>2c+6\). Now, max value of \(b\) is \(c\) and max value of \(d\) is \(e=c+4\), so max value of LHS (left hand side) would be \(LHS=b+d=c+c+4=2c+4\), which is always less than right hand side: \(RHS=2c+6\). Hence the answer to the question is NO. Sufficient.
Answer: B.
Hope it's clear.
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