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GMATPrep DS Obstacles ... One more

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Source: — Data Sufficiency |
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by rijul007 » Sat Mar 03, 2012 8:44 pm
myfish wrote:I tried this many times but I can't figure this one out, any help is appreciated. Thanks so much.
IMO: E

1,3,8,12,x

If 3<=x<8, Median = x
If x>=8, Median = 8

Average = (1+3+8+12+x)/5 = (24+x)/5 = 4.8 + x/5

QUes. Is Median>Average?

Statement 1
x>6

If x = 7
1,3,7,8,12

Median = 7
Average = 4.8 + 7/5 = 6.2
Med > Avg? Yes

If x = 50
1,3,8,12,50
Median = 8
Average = 4.8 + 10 = 14.8
Median > Average? No

Insuff


Statement 2
x is greater than median of the 5 numbers

Median can only be either x or 8.
Acc to this statment, 8 is the median

x>8

If x = 9
1,3,8,9,12
Median = 8
average = 33/5 = 6.6
Median> average? Yes

If x = 50
1,3,8,12,50
Median = 8
Average = 74/5 = 14.8
Median> Average? NO

Insuff


Combing two staements, we get x>8 which is same as statement 2

Hence, option E

What is the OA?
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by myfish » Sat Mar 03, 2012 8:59 pm
You are a magician. E is the correct answer. Thank you!!


[quote="rijul007"][quote="myfish"]I tried this many times but I can't figure this one out, any help is appreciated. Thanks so much.[/quote]

IMO: [spoiler]E[/spoiler]

1,3,8,12,x

If 3<=x<8, Median = x
If x>=8, Median = 8

Average = (1+3+8+12+x)/5 = (24+x)/5 = 4.8 + x/5

QUes. Is Median>Average?

[b]Statement 1[/b]
x>6

If x = 7
1,3,7,8,12

Median = 7
Average = 4.8 + 7/5 = 6.2
Med > Avg? Yes

If x = 50
1,3,8,12,50
Median = 8
Average = 4.8 + 10 = 14.8
Median > Average? No

Insuff


[b]Statement 2[/b]
x is greater than median of the 5 numbers

Median can only be either x or 8.
Acc to this statment, 8 is the median

x>8

If x = 9
1,3,8,9,12
Median = 8
average = 33/5 = 6.6
Median> average? Yes

If x = 50
1,3,8,12,50
Median = 8
Average = 74/5 = 14.8
Median> Average? NO

Insuff


Combing two staements, we get x>8 which is same as statement 2

Hence, option [spoiler]E[/spoiler]

What is the OA?[/quote]
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by krusta80 » Sun Mar 04, 2012 12:17 am
myfish wrote:I tried this many times but I can't figure this one out, any help is appreciated. Thanks so much.
average = (x+24)/5

Part (1)

x > 6

Let's look at it for all possible values of x, until we know that it'll be the same result as x increases...

x = 7 -> median is 7 -> mean is 31/5 MEDIAN IS GREATER
x = 8 -> median is 8 -> mean is 32/5 MEDIAN IS GREATER

Every time we add one to x going forward, the mean will increase by 1/5, while the median remains constant at 8. Therefore, eventually the MEAN will be greater. INSUFFICIENT.


Part (2)
x > median

For x to be greater than the median, it will have to be at least 9.

x = 9 -> median is 8 -> mean is 33/5 MEDIAN IS GREATER

As before, the mean will eventually overtake the median. INSUFFICIENT


(1) and (2)
Since part (2) is really just a subset of part 1's values for x, we can conclude that it is still INSUFFICIENT

E
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