If x is an integer, is the median of the 5 numbers shown

[BH]

x, 3, 1, 12, 8
If x is an integer, is the median of the 5 numbers greater than the average (arithmetic mean) of the 5 numbers?

(1) x>6
(2) x is greater than the median of the 5 numbers

[RonPurewal]

x, 3, 1, 12, 8
If x is an integer, is the median of the 5 numbers greater than the average (arithmetic mean) of the 5 numbers?

(1) x>6
(2) x is greater than the median of the 5 numbers

(1)
x must be greater than 1 or 3. if it is 8 or less, then it will be the median.
therefore:
if x = 7, then the list is 1, 3, 7, 8, 12, and median = 7. average = 31/5 = 6.2; answer to question is YES
if x = 8, then the list is 1, 3, 8, 8, 12, and median = 8. average = 32/5 = 6.4; answer to question is YES
if x = REALLY BIG, then the list is 1, 3, 8, 12, x, and median = 8. average is REALLY BIG, so the answer to the question is NO.
insufficient

(2)
this means that x is at least 9, from the observations above.
the list is either 1, 3, 8, x, 12, or 1, 3, 8, 12, x, depending on the size of x.
regardless of which one is the proper ordering, the expressions for the median and the mean are the same:
median = 8
mean (average) = (1 + 3 + 8 + 12 + x)/5 = (24 + x)/5
since x is at least 9, the mean is at least (24 + 9)/5 = 33/5 = 6.6.
this is inconclusive, because the mean could be 6.6 (if x = 9) or could be in the millions (if x is huge).
insufficient

(together)
this doesn't help, because statement (2) by itself already implies that x > 6
so, still insufficient

answer = e

[gmatwork]

For statement (1): I don't understand when you say:

"x must be greater than 1 or 3. if it is 8 or less, then it will be the median."

I understand the test cases but how did you reach the above conclusion?

For (2)I don't get it when you say:

"this means that x is at least 9, from the observations above.
the list is either 1, 3, 8, x, 12, or 1, 3, 8, 12, x, depending on the size of x."
regardless of which one is the proper ordering, the expressions for the median and the mean are the same:

Why can't be x be between 1 and 3?

Can you please explain?

[RonPurewal]

For statement (1): I don't understand when you say:

"x must be greater than 1 or 3. if it is 8 or less, then it will be the median."

if x is between 6 and 8, then the ordered list will be
1, 3, x, 8, 12
thereby making x the median.

For (2)I don't get it when you say:

"this means that x is at least 9, from the observations above.
the list is either 1, 3, 8, x, 12, or 1, 3, 8, 12, x, depending on the size of x."
regardless of which one is the proper ordering, the expressions for the median and the mean are the same:

Why can't be x be between 1 and 3?

if x is between 1 and 3, then the ordered list is
1, x, 3, 8, 12
and so x is smaller than the median, which is 3.

[namnam123]

It takes me 1 hour to find out the best and simplest method for this.

IF THE DISTANCE BETWEEN MEDIAN AND THE NUMBERS TO THE RIGHT IS FURTHER THAN THE DISTANCE BETWEEN MEDIAN AND THE NUMBER TO THE LEFT, MEDIAN <AVERAGE

example

1,3,5,7, 9

mean=average,

1,3, 4, 7 , 9
the distance between median and the number to the right is 7-4=3 , 9-4=5, the distance is 3+5=8
the distance between median and the numbers to the left is
4-1=3, 4-3=1 , distance =4
this case, median <average because median is closer the the left numbers than the right numbers.

[RonPurewal]

It takes me 1 hour to find out the best and simplest method for this.

i hope that you didn't actually sit there for an hour working on a single gmat problem.

even if you eventually *did* find this approach -- you are building very, very bad test-taking habits. the most important skill on this entire test is the ability to QUIT after a couple of minutes if you are stuck.

also -- there is no such thing as the "best" method!
any method that WORKS is just as good as any other method that WORKS!

you should not care at all about what is "best" or "simplest". you should only be concerned with finding as many ways as possible to solve the problems.

[asharma8080]

In the spirit of finding another method. Here is an alternate method

Problem: Median (M) > Average (A) ?

A = (24+x) / 5
M = Median

First statement
x = GT6
Then:
A = (24+GT6) / 5 = GT6
Possible Median if x > 6 => 7, 8

Average can be ANYTHING. N.S

Second statement
x > M; this means M is 8 (1, 3, 8, 12, x or 1,3, 8,x,12)
or x = GT8

Thus, A = (24+GT8) / 5 = GT6
Again M = 8 BUT Average can be anything. N.S

Together: x > 6 and x > 8 or x > 8 but Average can be anything when x is GT8. N.S

E

[jlucero]

In the spirit of finding another method. Here is an alternate method

Problem: Median (M) > Average (A) ?

A = (24+x) / 5
M = Median

First statement
x = GT6
Then:
A = (24+GT6) / 5 = GT6
Possible Median if x > 6 => 7, 8

Average can be ANYTHING. N.S

Second statement
x > M; this means M is 8 (1, 3, 8, 12, x or 1,3, 8,x,12)
or x = GT8

Thus, A = (24+GT8) / 5 = GT6
Again M = 8 BUT Average can be anything. N.S

Together: x > 6 and x > 8 or x > 8 but Average can be anything when x is GT8. N.S

E

While your method makes sense, there's something in your explanation that is subtle but important to recognize. For each of the three statements, you say something along the lines of "Average can be ANYTHING. N.S" Let me emphasize that this does NOT make something insufficient. This isn't a value question where you're trying to find what the average is. This is a Y/N question where you're trying to find a scenario where the average is less than the median AND a scenario where the average is greater than the median, to prove that the statement is insufficient.

Instead of saying, the median could be 7 or 8 and the average could be a number greater than 6, you should say, WHEN the average is 6.4 the median is 8 (median is larger), or WHEN the average is 100 the median is 8 (average is larger).

[NinaP494]

Mean has a greater flexibility in terms of how big it can get. So for the median to be greater than the mean we need to limit the upward movement of the mean by restricting values of x. Thus we need a constraint such as x<(certain number) and since neither of the choices does so answer is E.

For instance, even if we were given x>0
then say x=1 {1 1 3 8 12} 3>25/5? NO
now say x=8 {1 3 7 8 12} 8>(24+8)/5? Yes

on the other hand, say if we were given x<6, the answer would be a definitive No

Is my approach correct? thnaks

[RonPurewal]

no. for instance, if you were given a statement that said x > 17, that statement would be sufficient.