metallicafan wrote:
\(x, 3, 1, 12, 8\)
If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean) of the 5 numbers?
(1) \(x>6\)
(2) x is greater than the median of the 5 numbers
We have a set: {1, 3, 8, 12, x} Question: is \(median>mean=\frac{x+1+3+8+12}{5}=\frac{x+24}{5}\)? Note that as we have odd (5) # of terms in the set then the median will be the middle term when arranged in ascending (or descending) order. So:
if \(x\leq{3}\): {1, x, 3, 8, 12} then \(median=3\);
if \(3<x\leq{8}\): {1, 3, x, 8, 12} then \(median=x\);
if \(x\geq{8}\): {1, 3, 8, x, 12} then \(median=8\).
(1) \(x>6\). If \(x=7\) then the median will be 7 as well: {1, 3,
7, 8, 12} and mean will be \(mean=\frac{7+24}{5}=6.2\), so \(median=7>mean=6.2\) and the answer is YES
BUT if \(x\) is very large number then the median will be 8: {1, 3,
8, 12, x=very large number} and mean will be more than median (for example if \(x=26\) then \(mean=\frac{26+24}{5}=10\), so \(median=8<10=mean\)) and the answer will be NO. Not sufficient.
(2) x is greater than the median of the 5 numbers --> so \(median=8\): now, if \(x=11\) then \(mean=\frac{11+24}{5}=7\), so \(median=8>7=mean\) and the answer is YES. Again it's easy to get answer NO with very large \(x\). Not sufficient.
(1)+(2) Again,
x=11 and
x=very large number give two diffrent answers to the question. Not sufficeint.
Answer: E.