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Set range based on median and arithmetic mean

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Set range based on median and arithmetic mean

by nikhilsrl » Sun Jan 23, 2011 7:36 am
If set Z has a median of 19, what is the range of set Z?

1. Z = {18, 28, 11, x, 15, y}
2. The average (arithmetic mean) of the members of set Z is 20.

This is a Kaplan Gmat Premier question (Pg 480, Qn 49). I did not quite understand the solution provided. Any help is appreciated.

The ans is C.
Last edited by nikhilsrl on Wed Jan 26, 2011 1:24 am, edited 2 times in total.
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Source: — Data Sufficiency |
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by hrkrishna » Sun Jan 23, 2011 7:57 am
'C'
Here is my understanding

A: Insufficeinte becasue there are 2 variables and we cannot find the range.

B. Insufficent as it only talks about mean.

Combining these two

since the mean is 20. The sum of all these numbers have to be 120.

=> 11 + 15 + 18 + 28 + x + y = 120.
= 62 + X + Y = 120
=> X+ Y = 58.

but from question stem we know that the median is 19, and as there are '6' terms in the set, the sum of 3rd and 4th term has to be 38 (19*2) . and if you see no two known terms in the set make the sum as 38. so either X or Y has to be one of the term that make a median.
Lets consider X is a part of median.

if we consider the sequence as
11 X 15 18 28 and Y = 120.

The best answers could be X as 14 and Y as 42. leaving us 15 and 18 as the median number which is not true.

COnsider X and 18 as the median numbers giving us X as '20' there by Y = 38.

So, we finally know the sequence as
11 15 18 20 28 40.

leaving us the range as '29'.

Hope i'm not missing anything.
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by VivianKerr » Mon Jan 24, 2011 5:38 pm
Hey Nikhil,

This is a "Value" DS, so we need a specific number for the range. "Range" is found by subtracting the smallest number in a set from the biggest number in the set. So right away, we know sufficiency will only be achieved if we have the smallest AND the largest number.

1. We need to know the missing x and y to determine the smallest and the biggest numbers. INSUFFICIENT alone.

2. If the average is 20, we are still not able to determine the smallest and biggest numbers. INSUFFICIENT alone.

Let's try to combine them:

{11, 15, 18, 28, x, y} is the set. If we can choose values for x and y such that the average is 20, and get more than one outcome for the range, we will be able to choose (E).

Average = sum of terms/# of terms

20 = (11 + 15 + 18 + 28 + x + y) / 6

120 = 11 + 15 + 18 + 28 + x + y

120 = 72 + x + y

48 = x + y

We are told that the MEDIAN has to be 19. Let's arrange the data in numerical order: {11, 15, 18, 28} and somewhere in there x and y must be added. Because there is an even number of items in the set (6), we can only get a median of 19 if one of the missing numbers is 20. (This is because we take the average of the middle two numbers to find the median in a set with an even number of terms). (18 + 20) / 2 = 19.


48 = 20 + y

y = 38

The answer is (C).
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by tomada » Fri Jan 28, 2011 11:05 am
At the risk of being picayune, y=28. I know it has no bearing on the answer, but it was bothering me :-)
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