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Problem : Data Sufficiency - 3

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by 6983manish » Mon Mar 28, 2011 1:29 am
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
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Source: — Data Sufficiency |

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by Anurag@Gurome » Mon Mar 28, 2011 3:07 am
6983manish 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
(1) Let S = 2, T = 3, then combined set = {2, 3}; (2 + 3)/2 = 2.5 < 5; here mean of combined set is smaller than the sum of means of sets S and T.
Let S = -2, T = -3, then combined set = {-2, -3}; (-2 - 3)/2 = -2.5 > -5; here mean of combined set is greater than the sum of means of sets S and T.
No unique answer.
So, (1) is NOT SUFFICIENT.

(2) This statement only reflects that the elements of both the sets do not contain negative numbers, without stating the no. of elements in each set. So, clearly this info is NOT SUFFICIENT again.

Combining (1) and (2), If S = 2, T = 3, then combined set = {2, 3}; (2 + 3)/2 = 2.5 < 5; here mean of combined set is smaller than the sum of means of sets S and T.
If S = T = 0, then mean of combined set is equal to the sum of means of sets S and T.
No unique answer. So, combining is also NOT SUFFICIENT.

The correct answer is E.
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by preeti6 » Tue Apr 05, 2011 7:16 am
@ANURAG - I find B an answer to this. Why isnt B sufficient-

1. As rightly pointed out, in, if negative number are a part of the list, then the combined list mean is greater than the sum of the individual means. Hence not sufficient

2. In B, we are eliminating possibilities of a negative number in the sets.
We could assume for example two sets of different sizes -
a) SET X = 123, Set Y = 3,4.. Combined list mean (CLM)< Sum of indv means
b) SET X =1, SET Y= 3,4 .. CLM<SUM of indv means
c) SET X = 2,3 SET Y = 3,4 ... CLM< SUM of ind means
d) SET X = 2,3 SET Y = 0,4.. CLM< SUM of ind means!!

So B is sufficient??
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by Stuart@KaplanGMAT » Tue Apr 05, 2011 10:54 am
preeti6 wrote:@ANURAG - I find B an answer to this. Why isnt B sufficient-

1. As rightly pointed out, in, if negative number are a part of the list, then the combined list mean is greater than the sum of the individual means. Hence not sufficient

2. In B, we are eliminating possibilities of a negative number in the sets.
We could assume for example two sets of different sizes -
a) SET X = 123, Set Y = 3,4.. Combined list mean (CLM)< Sum of indv means
b) SET X =1, SET Y= 3,4 .. CLM<SUM of indv means
c) SET X = 2,3 SET Y = 3,4 ... CLM< SUM of ind means
d) SET X = 2,3 SET Y = 0,4.. CLM< SUM of ind means!!

So B is sufficient??
What about X = {0} and Y = {0}? Now the sum of the means = combined mean, giving us a "no" answer to the question.

Morals of the story:

(1)"non negative" means "0 or positive"; and

(2)unless a question specifically says that variables/items/sets must be distinct, remember that they can be identical.
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