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stmnt 1 - minimum value which x can have is 7. so on arranging the numbers we have 1,3, x,8,12 where x can be b/w 3 and 8 if x=7 or x can be b/w 8 and 12 or after 12

if x = 7, avg of these numbers will be 6.2 and median is 7 so median is > avg . suff if x= 11, avg of these numbers will be 7 and median is 8 so median is > avg. suff if x= 100, avg of these numbers will be 24.8 and median is 8 so median is < avg. insuff

so stmnt1 is insuff

stmnt 2 - x is greater than median so x> 8 [ on arranging we get 1,3,8,x,12]

if x= 11, avg of these numbers will be 7 and median is 8 so median is > avg. suff if x= 100, avg of these numbers will be 24.8 and median is 8 so median is < avg. insuff

so stmnt 2 is insuff

taking together we have x>8 but still insuff. hence E

Would you mind elaborating more on your procedure? Thanks Cheers! J

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 ?

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 \(median=x\) and 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.

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