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M24 Q 33

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M24 Q 33 [#permalink] New post 12 Mar 2009, 13:59
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If sets S and T are merged into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

(1) S and T are one-element sets
(2) Neither set S nor set T contains negative numbers

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SOLUTION: m24-q-76516.html#p1340646

Last edited by Bunuel on 06 Mar 2014, 06:30, edited 1 time in total.
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Re: M24 Q 33 [#permalink] New post 12 Mar 2009, 18:06
icandy wrote:
If sets S and T are merged into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

1. S and T are one-element sets
2. Neither set S nor set T contains negative numbers

(C) 2008 GMAT Club - m24#33

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

I will let you guys discuss this. How ever, The two Set opeartions I remember are union and intersection. How are we supposed to interpret merge.



I believe two operations and "union" and "union all" --> if you know ORACLE sql LANGUAGE. :lol: :lol: :lol:
( I may be wrong.. but thats what I interpret.. as IT .. nerd)

e.g
A ={1,3} B={1,5}

Merge = {1,3,5} ( all distinct elements) avg = 3
Sum = {1,1,3,5} avg= 10/4 = 2.5


another e.g.
A ={1,5} B={3,5}
Merge = {1,3,5} ( all distinct elements) avg = 3
Sum = {1,3,5,5} avg= 14/4 = 3.66



I will go with E.
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Re: M24 Q 33 [#permalink] New post 27 Mar 2009, 09:47
Couldnt get this!!
It is sum of the means of A and B.
e.g
A ={1,3} B={1,5}

Merge = {1,3,5} ( all distinct elements) avg = 3
Sum = {1,1,3,5} avg= 10/4 = 2.5 >> cant understand this.

Sum of means should be 2 + 3 = 6.May be my understanding is wrong. Please explain.

x2suresh wrote:
icandy wrote:
If sets S and T are merged into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

1. S and T are one-element sets
2. Neither set S nor set T contains negative numbers

(C) 2008 GMAT Club - m24#33

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

I will let you guys discuss this. How ever, The two Set opeartions I remember are union and intersection. How are we supposed to interpret merge.



I believe two operations and "union" and "union all" --> if you know ORACLE sql LANGUAGE. :lol: :lol: :lol:
( I may be wrong.. but thats what I interpret.. as IT .. nerd)

e.g
A ={1,3} B={1,5}

Merge = {1,3,5} ( all distinct elements) avg = 3
Sum = {1,1,3,5} avg= 10/4 = 2.5


another e.g.
A ={1,5} B={3,5}
Merge = {1,3,5} ( all distinct elements) avg = 3
Sum = {1,3,5,5} avg= 14/4 = 3.66



I will go with E.
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Re: M24 Q 33 [#permalink] New post 29 Mar 2009, 19:37
icandy wrote:
If sets S and T are merged into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

1. S and T are one-element sets
2. Neither set S nor set T contains negative numbers

I will let you guys discuss this. How ever, The two Set opeartions I remember are union and intersection. How are we supposed to interpret merge.


For me merged = combined.

1: S and T are one-element sets
If S = T = 2, mean of the merged set (2, 2) = 2 and sum of the means of sets S and T is 4.
If S = 2, and T = -2, mean of the merged set (2, -2) = 0 and sum of the means of sets S and T is 0.
If S = 0, and T = 0, mean of the merged set (2, 2) = 0 and sum of the means of sets S and T is 0. NSF..

2: Neither set S nor set T contains negative numbers
If S = T = 2, mean of the merged set (2, 2) = 2 and sum of the means of sets S and T is 4.
If S = 0, and T = 0, mean of the merged set (2, 2) = 0 and sum of the means of sets S and T is 0. NSF.

Togather alos nsf.
E.
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Re: M24 Q 33 [#permalink] New post 11 Mar 2010, 06:22
Merge of sets means you need to combine.

S= {1}
T= {1,2}

Merge = {1,1,2}
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Re: M24 Q 33 [#permalink] New post 11 Mar 2010, 07:07
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Both solution 1 and 2 fail the zero test.

S1: S = {0} and T = {0} => Mean of S = 0 and Mean of T = 0
Set ST = {0, 0} => Mean of ST = 0
Thus, S1 is not sufficient.

S2. Neither S nor T contain negative number.
Take the same sets: S = {0} and T = {0}
S2 is not sufficient.

Answer is E.
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Re: M24 Q 33 [#permalink] New post 11 Mar 2010, 07:35
icandy wrote:
If sets S and T are united into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

1. S and T are one-element sets
2. Neither set S nor set T contains negative numbers

[Reveal] Spoiler: OA
E

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s= {1,2,3}
mean1 = 2
t= {2,3,4}
mean2 = 3
sUt= {1,2,3,4}
mean3= 2.5 < mean1 + mean2

s={-1,-2,-3}
mean1 = -2
t={-2,-3,-4}
mean2 = -3
sUt= {-1,-2,-3,-4}
mean3 = -2.5 > mean1 + mean2

s={0}
mean1 = 0
t = {0}
mean2 = 0
sUt = {0}
mean3 = 0 = mean1+mean2

hence none of the two stmts solve this questions. Hence E.
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Re: M24 Q 33 [#permalink] New post 11 Mar 2010, 15:26
vshrivastava wrote:
Both solution 1 and 2 fail the zero test.

S1: S = {0} and T = {0} => Mean of S = 0 and Mean of T = 0
Set ST = {0, 0} => Mean of ST = 0
Thus, S1 is not sufficient.

S2. Neither S nor T contain negative number.
Take the same sets: S = {0} and T = {0}
S2 is not sufficient.

Answer is E.



You only showed that you can answer the question with a definite NO.
The question being "will the mean of this set be smaller than the sum of means of sets S and T ?"

This sufficiently answers the question as a NO for both cases.

In order for the cases to be insufficient, you need to be able answer the question with both a yes and a no.
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Re: M24 Q 33 [#permalink] New post 14 Mar 2010, 01:53
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i understand that both options failed the zero test but why do we need a zero in set

ex- a set of 3 pen, 2 pencil and 0 flower
we can simply say it a set of 3 pen and 2 pencils, why do we consider something like zero in a set

Please explain
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Re: M24 Q 33 [#permalink] New post 19 Mar 2010, 07:18
hardnstrong wrote:
friends
i understand that both options failed the zero test but why do we need a zero in set

ex- a set of 3 pen, 2 pencil and 0 flower
we can simply say it a set of 3 pen and 2 pencils, why do we consider something like zero in a set

Please explain


Zero is considered as an entity here.
like A has 0 sets of readings 2m , 0m, 1m
so zero is the value.
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Re: M24 Q 33 [#permalink] New post 20 Mar 2010, 05:23
hardnstrong wrote:
friends
i understand that both options failed the zero test but why do we need a zero in set

ex- a set of 3 pen, 2 pencil and 0 flower
we can simply say it a set of 3 pen and 2 pencils, why do we consider something like zero in a set

Please explain

"zero" in this case does not imply an empty set; rather, it it a set consisting of an element-digit 0.
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Re: M24 Q 33 [#permalink] New post 08 Dec 2010, 02:45
I think the answer should be C.
(1) and (2) says S = {s} (mean = s) and T = {t} (mean = t), united = {s, t} (mean = (s+t)/2)
s, t = non negative.
Question: s+t < (s+t)/2 ????
If s = t = 0, the answer to the question "will the mean of this set be smaller than the sum of means of sets S and T?" is NO.
If s = 0, t > 0 (and vice versa), the answer to question "will the mean of this set be smaller than the sum of means of sets S and T?" is NO.
If s, t > 0, the answer to the question "will the mean of the set be smaller than the sum of means of sets S and T?" is still NO.
In any case, the answer is still NO. Then (1) and (2) together are sufficient enough.
Why is E the correct answer????
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Re: M24 Q 33 [#permalink] New post 15 Mar 2011, 05:19
The answer is E.

Let A = {2}, B = {3},

So the mean of merge is 2.5, which is < 5

Let A = {1}, B = {1}

So the mean of merge is 1, which is equal to mean of either set.
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Re: M24 Q 33 [#permalink] New post 15 Mar 2011, 21:48
firasath wrote:
vshrivastava wrote:
Both solution 1 and 2 fail the zero test.

S1: S = {0} and T = {0} => Mean of S = 0 and Mean of T = 0
Set ST = {0, 0} => Mean of ST = 0
Thus, S1 is not sufficient.

S2. Neither S nor T contain negative number.
Take the same sets: S = {0} and T = {0}
S2 is not sufficient.

Answer is E.



You only showed that you can answer the question with a definite NO.
The question being "will the mean of this set be smaller than the sum of means of sets S and T ?"

This sufficiently answers the question as a NO for both cases.

In order for the cases to be insufficient, you need to be able answer the question with both a yes and a no.


I think, vshrivastava has shown that both statement put together also we are getting two answer: one time > and other time =. This much is enough to tell that answer is E.
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Re: M24 Q 33 [#permalink] New post 16 Mar 2011, 12:51
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icandy wrote:
If sets S and T are united into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

1. S and T are one-element sets
2. Neither set S nor set T contains negative numbers

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I will let you guys discuss this. How ever, The two Set opeartions I remember are union and intersection. How are we supposed to interpret merge.


Statement 1)

S and T are one element sets so let's have Set S = 1 and set T = 2

If we unite set S and T we will get {1,2} and the average of this set is 1.5
If we add the average of set S which equals 1 and the average of set T which equals 2 we get a total of 3
1.5 < 3 YES

Now let's assign different values for S and T. Like other people have mentioned if we have set S = 0 and T = 0 we will get an average of 0

Average of united S and T {0,0} = 0
Average of set S {0} = 0; the average of set T {0} = 0
0<0 NO
Which means this statement is insufficient

Statement 2)
If no values are negative we can recycle the values we used for statement 1 and get Insufficient as well

Statement 1 & 2 together are also Insufficient

The answer is E.
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Re: M24 Q 33 [#permalink] New post 20 Mar 2013, 04:37
If S = {a,b} and T = {b,c}, then the mean of the set that results after the merge is x= (a+b+c)/3 and the SUM of the two separate means is y = (a+b)/2 + (b+c)/2.
Since we need to verify whether x<y, it has to be a+4b+c>0. Since a,b and c can be thought as the means of the terms that in T and S are and are not common between the two sets, then the second condition itself is enough to guarantee an answer.
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Re: M24 Q 33 [#permalink] New post 20 Mar 2013, 08:59
B

( 1,2,3) (2,3,4)
it says that there is no - ive numbers in either of the sets then the the average of the united set will always be lesser than the sum of the average of individual sets in any case if the united set has repeated numbers ( 1,2,3,2,3,4) or just the distinct numbers ( 1,2,3,4)
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Re: M24 Q 33 [#permalink] New post 09 Apr 2013, 04:26
Answer is 'E'.

And the only condition which causes the answer to be 'E' is if both sets contain only '0' i.e. S = {0}, T = {0}, which is perfectly allowable.

Even though the above is a very specific case, it is enough to cause some ambiguity in the solution, thus answer is 'E'.
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Re: M24 Q 33 [#permalink] New post 06 Mar 2014, 06:30
Expert's post
If sets S and T are merged into a single set, will the mean of this set be smaller than the sum of means of sets S and T ?

(1) S and T are one-element sets
(2) Neither set S nor set T contains negative numbers

Even when we consider both statements together, we cannot answer the question. If S={0} and T={0} then the answer is NO but if S={0} and T={1} then the answer is YES. Not sufficient.

Answer: E.
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Re: M24 Q 33 [#permalink] New post 27 Mar 2014, 05:17
:) no need for such complex calculations, just see both the statements left the possibility of 0 as an element. => {0} + {0} = {0,0} Answer is definitely E
Re: M24 Q 33   [#permalink] 27 Mar 2014, 05:17
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