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MGMAT STATISTICS DS QUES

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by cramya » Mon Oct 20, 2008 6:52 pm
x is an integer greater than 7. What is the median of the set of consecutive integers from 1 to x inclusive?

(1) The average of the set of integers from 1 to x inclusive is 11.

(2) The range of the set of integers from 1 to x inclusive is 20.


[spoiler]OA:D[/spoiler]


I am not sure from 1) how u can tell the median is equal to the average of the set (this is only true if x is odd. The question stem says x>7 what if x is 8 then the average is not equal to the median

Thoughts?
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Source: — Data Sufficiency |

by aznstyle28 » Mon Oct 20, 2008 7:08 pm
The average they give is 11. With this information you can find the median by using the first number of the consecutive set, x, and the average to find x which enables you to find the median.

(1+x)/2 = 11
22 = 1+x
x=21
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by cramya » Mon Oct 20, 2008 7:18 pm
I am thinking u cant do
(1+x)/2 = 11

If x=7 then there will be 7 terms
If x=8 then there will be 8 terms etcc

1+.....+.....+last term / (x-1)+1 = 11
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by cramya » Mon Oct 20, 2008 7:20 pm
I think u r right!!! :D

Not sure what I was thinking!

Thanks Aznstyle28!
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by java_ka_jalwa » Wed Oct 22, 2008 3:49 pm
no need to calc x since, consec integers mean = median
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by kandyhot27 » Wed Oct 22, 2008 5:43 pm
IMO D

When you solve A
x (x+1)/2 = 11
x = 21....now you can find median

when you solve b
x-1=20
x =21...same as a....sufficient

hence D
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by cramya » Wed Oct 22, 2008 7:33 pm
Kandy,
Could u elaborate

stmt 1 where u did x(x+1)/2 = 11

Thnks!
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by stop@800 » Wed Oct 22, 2008 11:02 pm
cramya wrote:Kandy,
Could u elaborate

stmt 1 where u did x(x+1)/2 = 11
It should be
x(x+1)/(2x) = 11

Thnks!
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by lunarpower » Fri Oct 24, 2008 3:12 am
in any symmetric set - a description that includes all sets of consecutive integers - the following facts are true:

(1) the AVERAGE of the set is just the AVERAGE OF THE FIRST AND LAST terms.
in fact, because of symmetry, the average is equal to the average of any two terms that are symmetric about the middle number, but the first and last are usually the easiest to figure out and think about.

(2) the MEDIAN and the MEAN of the set are the same.
if you visualize the set plotted on a number line - i.e., as a set of dots that are equally spaced - this should be somewhat clear: draw a line right through the middle of the set, and that line will be at both the median and the mean.

only (2) applies here, but, since these are both useful general takeaways, it helps to list them both.
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by rohangupta83 » Fri Oct 24, 2008 4:47 am
(D)

Using statement 1

1 (1st integer)
x (last integer)

Average for the series 1,2,3,4,5,6,7.......x = 11 (all consecutive integers)

for example
average of 1,2,3,4,5 = 3

and also the average of 1,5 = 3 (1 - 1st integer and 5 - last integer)

Therefore, average for the first and last integer should also be 11

(1+x)/2 = 11
or x = 21

using statement 2

range = 20
1st digit = 1

therefore, the last digit = 1+20 = 21 (which is x)

Hence, both statements are sufficient

answer D
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by aagar2003 » Sat Aug 13, 2011 6:35 am
is there any significance of x being greater than 7 in this question? Wouldn't the answer remain D (same) even without this information?
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by lunarpower » Sat Aug 13, 2011 3:50 pm
aagar2003 wrote:is there any significance of x being greater than 7 in this question? Wouldn't the answer remain D (same) even without this information?
basically, yeah. i think the author is just hedging against the (remote) possibility that someone might allow sequences in decreasing order -- i.e., someone might come with the complaint that statement 2 could also be "the integers from 1 to -19". if you state that x > 7 then you kill that particular possibility.
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by GmatKiss » Sun Aug 14, 2011 12:53 pm
Thanks Lunarpower for your time :)
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by [email protected] » Wed Mar 14, 2012 7:56 pm
x is an integer greater than 7. What is the median of the set of consecutive integers from 1 to x inclusive?

(1) The average of the set of integers from 1 to x inclusive is 11.

(2) The range of the set of integers from 1 to x inclusive is 20.



Yes got it easily, the statement 1 can be solved by 2 formula:

1] the sum of n natural numbers in a series'

2] second already given... [n(n+1)]/2 X (1/n) inorder to get the average...
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