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M28-47

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Math Expert
Joined: 02 Sep 2009
Posts: 48110

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16 Sep 2014, 02:51
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55% (hard)

Question Stats:

52% (00:48) correct 48% (01:02) wrong based on 44 sessions

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What is the value of the median of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

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Math Expert
Joined: 02 Sep 2009
Posts: 48110

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16 Sep 2014, 02:51
1
Official Solution:

(1) No number in set A is less than the average (arithmetic mean) of set A.

Since no number is less than the average, then no number is more than the average, which implies that the list contains identical elements: $$A=\{x, \ x, \ x, \ ...\}$$. From this it follows that (the average)=(the median). But we don't know the value of $$x$$, thus this statement is NOT sufficient.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

Not sufficient: if $$A=\{0, \ 0, \ 0, \ 0\}$$, then (the median)=0, but if $$A=\{1, \ 2, \ 2, \ 3\}$$, then (the median)=2.

(1)+(2) From (1) we have that the list contains identical elements. The range of all such sets is 0. Therefore, from (2) we have that (the average)=(the range)=0 and since from (1) we also know that (the average)=(the median), then (the median)=0. Sufficient.

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Joined: 14 Jul 2014
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11 Mar 2015, 10:45
Bunuel wrote:
Official Solution:

(1) No number in set A is less than the average (arithmetic mean) of set A.

Since no number is less than the average, then no number is more than the average, which implies that the list contains identical elements: $$A=\{x, \ x, \ x, \ ...\}$$. From this it follows that (the average)=(the median). But we don't know the value of $$x$$, thus this statement is NOT sufficient.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

Not sufficient: if $$A=\{0, \ 0, \ 0, \ 0\}$$, then (the median)=0, but if $$A=\{1, \ 2, \ 2, \ 3\}$$, then (the median)=2.

(1)+(2) From (1) we have that the list contains identical elements. The range of all such sets is 0. Therefore, from (2) we have that (the average)=(the range)=0 and since from (1) we also know that (the average)=(the median), then (the median)=0. Sufficient.

Hi Bunuel

How can you derive the highlighted part above? (i.e. then no number is more than the average) - I think this is possible

Eg A= { 2 } - Here, Median = Mean = 2
A = { 2, 2, 2, 3} - Here, Median = 2, Mean = 2.25 ..... What Im trying to say is that we can have a number > Mean

Either ways, I agree that this is Insufficient.

Pls clarify my doubt

Thanks
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Joined: 23 Apr 2016
Posts: 22
Location: Finland
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11 Nov 2016, 15:05
1
The question says, no number is less than average, NOT some number is less than average.
For example take a set of number 2,2,2 here no number is more than the average
now if we add say 3 to this set... to bring the average of this new set (2,2,2,3), we needs some value which will bring back the average of 3 and that number back to 2, so we will have to add a to the set.

So what Bunuel is basically saying that if something is more than average, than there must be some number less than average OR if nothing is more than average, than nothing should be less than average as well.
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Joined: 14 Oct 2012
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16 Apr 2017, 20:28
My 2 cents,
Elaborating further on what thapliya said:
Initially, set = {2,2,2} mean = median = 2
Now, set = {2,2,2,3} mean = 11/4 = 2.25 & Median = 2
To bring the mean of the set back to 2 we need to remove the additional portion of 2 (8+2, i hope this is clear. if not leave a message!!!) added to the mean by entering 3 in the set.
Set = {-1,2,2,2,3} => Mean = (11-1)/5 = 10/5 = 2.
Thus we again get Mean = Median = 2.

Coming back to the problem:
S-1 - "No number in set A is less than the average (arithmetic mean) of set A."
But in our case we CANNOT add any value which is smaller than mean. Thus we CANNOT add any new value to the set which is either greater than or less than mean of set, as we CANNOT add any smaller value to balance out higher value in our set. Thus ALL the values in the set NEED to be equal.
I hope this helps!!!
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Joined: 02 Jun 2015
Posts: 192
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28 Jun 2017, 06:49
Bunuel wrote:
What is the value of the median of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

Hi Bunuel,

Kindly help me to understand this: why can't set A be a one-element set?

Thanks
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Math Expert
Joined: 02 Sep 2009
Posts: 48110

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28 Jun 2017, 09:56
1
duahsolo wrote:
Bunuel wrote:
What is the value of the median of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

Hi Bunuel,

Kindly help me to understand this: why can't set A be a one-element set?

Thanks

The number of x's in the solution can be any. So, we can have one-element set but we'll still get the same answer. One element set for (1)+(2) will be {0}.
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Joined: 24 Jun 2017
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31 Oct 2017, 02:00
From Statement 2, cant we have single element set {1} . In this set, Range of single element set is 1 and average is 1. Is this right??? then we dont have solution for this?
Math Expert
Joined: 02 Sep 2009
Posts: 48110

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31 Oct 2017, 02:08
From Statement 2, cant we have single element set {1} . In this set, Range of single element set is 1 and average is 1. Is this right??? then we dont have solution for this?

(The range) = (Largest element) - (Smallest element).

Now, since a single-element set (Largest element) = (Smallest element), then (The range of a single-element set) = (Largest element) - (Smallest element) = 0.
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Re: M28-47 &nbs [#permalink] 31 Oct 2017, 02:08
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