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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
(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. ( 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. _________________
Your attitude determines your altitude Smiling wins more friends than frowning
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. ( 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
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.
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. _________________
If you like my post, a kudos is always appreciated
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????
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.
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. _________________
I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!
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. 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)
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.
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