Of the 5 numbers, the largest number is 4 greater than the median. Is

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.