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Director  Joined: 03 Sep 2006
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If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 63% (02:24) correct 37% (02:12) wrong based on 504 sessions

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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.
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Joined: 02 Sep 2009
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If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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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.

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Manager  Joined: 21 Jun 2006
Posts: 232

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Both are sufficient..
put the nos in ascending seq. 1,3,8,12 and x. We don't no where x is positioned.

X > 6 could mean that X is the median or 8 is the median.
if X =7 then mean = 6.2 and X > mean
if x= 11 then mean = 7 and X > mean

X greater than median of 5 nos. Which means that median HAS to be 8.
x> 8
Again satisfies the question for any value of x.

D
Senior Manager  Joined: 03 Jun 2007
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ArvGMAT wrote:
Both are sufficient..
put the nos in ascending seq. 1,3,8,12 and x. We don't no where x is positioned.

X > 6 could mean that X is the median or 8 is the median.
if X =7 then mean = 6.2 and X > mean
if x= 11 then mean = 7 and X > mean

X greater than median of 5 nos. Which means that median HAS to be 8.
x> 8
Again satisfies the question for any value of x.

D

what if x is 16 then mean = 8 = median which is not greater so answer is E
VP  Joined: 10 Jun 2007
Posts: 1116
Re: DS- Mean, Median  [#permalink]

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1
1
LM 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 numbes.

Got E.

(1) plug in x = 7, the lowest avg value is (7+3+1+12+8) / 5 = 31/5 =~ 6
This is lower than the median, which is 7 in this case. However, if x is really large, the avg will shoot through the roof, but the median will remain at 8. INSUFFICIENT.

(2) This tells us that x>8, so plug in x=9. We get (9+3+1+12+8) / 5 = 33/5 =~ 6.6. This still less than 8 ,which is the median. Same reason as above, INSUFFICIENT.

Together, we get x>8, INSUFFICIENT.
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$$x, 3, 1, 12, 8$$

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
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Re: Average question  [#permalink]

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Thanks Bunuel!

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If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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That moment- when you think ans. is D i.e both statements even alone are sufficient,but it turns out that both of them,even together are insufficient ie ans E. _________________
ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.
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If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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Bunuel wrote:
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 $$median=x$$ 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.

Just wanted to clarify that I have solved this question using random values. Your approach is pretty precise. Is using random values always a disadvantage on GMAT, given the time constraint? It took me arnd 3 mins to do this question , using random values. Given that it's a 700 level question, taking 3 mins on such questions is justified???

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Re: If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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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.

In the original condition, there is 1 variable(x), which should match with the number of equation. So you need 1 more equation. For 1), 1 equation, for 2) 1 equation, which is likely to make D the answer. In 1) & 2),
for 1), when x>0, 1,3,x,8,12/1,3,8,x,12/1,3,8,12,x. mean=(1+3+8+12+x)/5=(24+x)/5 and median=x,8. So, (24+x)/5>x? or (24+x)/5>8? is unknown, which is not sufficient.
For 2), in the above, median is 3, x, 8 and from x>3, x>x(impossible), x>8, it is x>8>3. So, although x>8, (24+x)/5>8?, which is x>16?, is not sufficient. Even if 1) & 2), when x>8, you cannot find out (24+x)/5>8?, x>16? from x>8>6>3. Therefore it is not sufficient and the answer is E.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x is an integer , is the median of 5 numbers shown greater than  [#permalink]

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Statement 1: For X>6, Avg<median for X=7,8,9,10,11,12,13,14,15,
Avg=median for X= 16
Avg> median for X>16

Statement 2: X>median of 5 number => X> 8
Again u can't decide if Median > Avg

Both combined X>8
Can't deduct if median> avg

So.. Option E

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Re: If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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Required to put actual values to solve this question.
Option 1: considered value of x as 7 and 50. With 7 median > mean. With value of x as 50 median < mean. Not sufficient.
Option 2: Same values can be used to eliminate this answer. With 50 median <mean and with value of x as 13, medium > mean. Hence, not sufficient.
Considering both the equation also we are not able to solve the question.

My answer is E.
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Re: If x is an integer, is the median of the 5 numbers shown gre  [#permalink]

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