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Gmat prep# DS AM

Expert replies
Source: — Data Sufficiency |

by jahnaviyeruva » Wed Oct 22, 2008 12:38 am
Answer is E
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Re: Gmat prep# DS AM

by drpawan » Thu Oct 23, 2008 8:32 am
aditi_bc wrote:pl solve
Here is the solution:

AM = (x+2+7+11+16)/5 = (x+36)/5
given AM(x,2,7,11,16) = Median

so lets think the possibilities of x from the (x+36)/5:
x (x+36)/5
-- ----------
4 8 --8 is not there in the set
9 9 --9 can be included in place of x which can also be the median
14 10 --10 is not there in the set and so on..
19 11 --19 can be included in the place of x as 11 can be median

Statement 1 --> 7<x<11

Only 9 is fits in x(as per above possibilities). Hence this statement is sufficient.

Statement 2 --> x is the median of 5 numbers.

From the above possibilities only 9 and 19 can be included in the set. But as x is also a median..19 cannot be median. Because if 19 is included in the set, then 2,7,11,16,19 --> 19 is AM and 11 is median which fails. If you consider 9, then 2,7,9,11,16 --> 9 is both AM and median
hence this statement is sufficient.

So OA is D
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by cramya » Thu Oct 23, 2008 4:44 pm
Given

The average of x,2,7,11,16 is equal to the median.

1) 7<x<11 from this we know x is the median

Total count of numbers given is odd then after arranging them in ascending order the middle value is the median

2,7,x,11,16

2+7+x+11+16/5 = x

We can solve for x
SUFFICIENT

2)x is the median so we know the order is again
2,7,x,11,16

2+7+x+11+16/5 = x

We can solve for x
SUFFICIENT

Hence D)
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by rohangupta83 » Thu Oct 23, 2008 5:00 pm
agreed imo D
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