2. The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
Integer
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hard to me, it is only try, so any objections are highly appreciated
we have set S={s1,s1.....sn}, and set T={t1,t2....tk} , with s1+s2+....sn=t1+t2+....tk
and we need to estimate does n>k with given conditions
(1) mean of set S< mean of set T. in math terms:
(s1+s2+....sn)/n<(t1+t2+....tn)/k. here i think it is possible to cancel s1+s2+...sn as well as t1+t2+..tk as they are the same, and left with: 1/n<1/k ,k<n. sufficient ( i hope that i did not miss anything but not sure)
(2) for the second st i did not find any formal approach so the only way is to find possible sets, i hope that we are not restricted to the number of tems so
S={ 3}. T={1, 2} median S> median T, but n(1)<k(2) ( number of terms)
S={1,2,3,4} sum of S=10, median=2,5.T={10,10, 10} sum also 10, median 10. 2,5>10. but the number of terms in S=4,in T=3, n>k to me insuff
my pick for A
we have set S={s1,s1.....sn}, and set T={t1,t2....tk} , with s1+s2+....sn=t1+t2+....tk
and we need to estimate does n>k with given conditions
(1) mean of set S< mean of set T. in math terms:
(s1+s2+....sn)/n<(t1+t2+....tn)/k. here i think it is possible to cancel s1+s2+...sn as well as t1+t2+..tk as they are the same, and left with: 1/n<1/k ,k<n. sufficient ( i hope that i did not miss anything but not sure)
(2) for the second st i did not find any formal approach so the only way is to find possible sets, i hope that we are not restricted to the number of tems so
S={ 3}. T={1, 2} median S> median T, but n(1)<k(2) ( number of terms)
S={1,2,3,4} sum of S=10, median=2,5.T={10,10, 10} sum also 10, median 10. 2,5>10. but the number of terms in S=4,in T=3, n>k to me insuff
my pick for A

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danjuma wrote:2. The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
1. Average = Sum/total number of items in a list ; If Sum is same for two list and and average relation is known , you know the relation of number of items in a list. In this case, S has more number of items . Sufficient
2. Median is middle number or average of two middle numbers . Median can be any value as depends on what numbers are . Can not say. But I would like to have more definitive approach as well.
Also what is OA please.

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Let Ts be the total and s be the number of elements in set S
Let Tt be the toal and t be the number of elements in set T
given Ts = Tt
Is s > t ?
prerequisite : Avg = (Total)/number of elements
Stat 1:
Avg of S < Avg of T
Ts(s) < Tt(t)
but given Ts=Tt. Cancel them out .
hence s < t.
sufficient
Stat 2:
Med of S > Med of T
information doesn't help.
Let Tt be the toal and t be the number of elements in set T
given Ts = Tt
Is s > t ?
prerequisite : Avg = (Total)/number of elements
Stat 1:
Avg of S < Avg of T
Ts(s) < Tt(t)
but given Ts=Tt. Cancel them out .
hence s < t.
sufficient
Stat 2:
Med of S > Med of T
information doesn't help.
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Hi, guys.danjuma wrote:2. The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
The answer is certainly "A", letÂ´s see why:
(a) Sufficient:
Sum = Average * (Number of Terms) , therefore if the sum is the same (in S and in T) and the average is less (in S), than the number of terms must be greater (in S).
(b) Insufficient:
> Take S = {1,2,3} and T = {0,1,5} and this answers the question in the negative;
> Take S = {1,2,2,3} and T = {0,1,7} and this answers the question in the affirmative.
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This is the "safe way" to guarantee a certain statement is not enough to answer the question asked, I call it a BIFURCATION, in the sense that both EXPLICIT examples satisfy the conditions presented at the question stem AND the statement(s) considered at the time but each example answers the question asked DIFFERENTLY.fskilnik wrote: > Take S = {1,2,3} and T = {0,1,5} and this answers the question in the negative;
> Take S = {1,2,2,3} and T = {0,1,7} and this answers the question in the affirmative.
I hope you got the point!
Regards,
Fabio.
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fskilnik, How about this
Let's say the sum is 12 for both
set s has 6 integers and t=4 integers
12/6 = 2 (call this s)
12/4 = 3 (call this t)
now s>t even though the sum is same
if the sum is 12
12/4 = 3
12/6 = 2
s<t even though the sum is same, but here now, s<t
therefore A is INSUFFICIENT
If the question stem had mentioned that these are all positive/nonnegative integers then A would be sufficient.
Correct me if I am wrong. This is the first time that I am seeing what looks like a wrong OA on GMATPREP
Let's say the sum is 12 for both
set s has 6 integers and t=4 integers
12/6 = 2 (call this s)
12/4 = 3 (call this t)
now s>t even though the sum is same
if the sum is 12
12/4 = 3
12/6 = 2
s<t even though the sum is same, but here now, s<t
therefore A is INSUFFICIENT
If the question stem had mentioned that these are all positive/nonnegative integers then A would be sufficient.
Correct me if I am wrong. This is the first time that I am seeing what looks like a wrong OA on GMATPREP
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Hi, zaarathelab!
First of all, sorry for the delay. IÂ´ve not come here for a LONG time (too busy)!
You are absolutely right!!
Let me put the bifurcation explicitly, for all readers:
First case: (Please note both sums are equal.)
S = {2,2,2,2,2,2} then average(S) = 2
T = {4,4,4} then average(T) = 4
Answering in the affirmative;
Second case: (Please note both sums are equal.)
S = {3,3,3,3} then average(S) = 3
T = {2,2,2,2,2,2} then average(T) = 2
Answering in the negative;
Therefore sttm(1) is really INSUFFICIENT.
From the fact that (1) and (2) alone are insufficient, we have to consider both of them together... could you complete the problem and tell us if the answer is (C) or (E) ?
Regards (and congrats for your attention and rigour),
fskilnik.
First of all, sorry for the delay. IÂ´ve not come here for a LONG time (too busy)!
You are absolutely right!!
Let me put the bifurcation explicitly, for all readers:
First case: (Please note both sums are equal.)
S = {2,2,2,2,2,2} then average(S) = 2
T = {4,4,4} then average(T) = 4
Answering in the affirmative;
Second case: (Please note both sums are equal.)
S = {3,3,3,3} then average(S) = 3
T = {2,2,2,2,2,2} then average(T) = 2
Answering in the negative;
Therefore sttm(1) is really INSUFFICIENT.
From the fact that (1) and (2) alone are insufficient, we have to consider both of them together... could you complete the problem and tell us if the answer is (C) or (E) ?
Regards (and congrats for your attention and rigour),
fskilnik.
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You cannot use S ={0,3,0} and T={2,2,2}, because the sums of the elements in S and T are not the same...navami wrote:I feel ans should be E
Consider the below case
S T

{0 3 0 } { 2 2 2}
even after combining either set is larger for S or equal.
Obs.: the "hard part" is to find lists S and T such that:
(i) average of S < average of T < 0
(ii) the sum of elements of S = the sum of elements of T and this sum is negative
(iii) median of S > median of T ...
Hint: start with S and T found by zaarathelab, keeping averages but changing medians!
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This is quite interesting. This problem appears as a question in the GMAT PREP exam, but they mark the correct answer as A (statement 1 is sufficient). Depending on the test cases used, it seems that the answer could be A or E. Could someone please advise?fskilnik wrote:Hi, zaarathelab!
First of all, sorry for the delay. IÂ´ve not come here for a LONG time (too busy)!
You are absolutely right!!
Let me put the bifurcation explicitly, for all readers:
First case: (Please note both sums are equal.)
S = {2,2,2,2,2,2} then average(S) = 2
T = {4,4,4} then average(T) = 4
Answering in the affirmative;
Second case: (Please note both sums are equal.)
S = {3,3,3,3} then average(S) = 3
T = {2,2,2,2,2,2} then average(T) = 2
Answering in the negative;
Therefore sttm(1) is really INSUFFICIENT.
From the fact that (1) and (2) alone are insufficient, we have to consider both of them together... could you complete the problem and tell us if the answer is (C) or (E) ?
Regards (and congrats for your attention and rigour),
fskilnik.
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It's a rare goof by GMAC. The software claims that A is correct, but this would only be true if we were limited to dealing with positive integers. And we aren't. Credit to Brent for catching this...Poisson wrote:This is quite interesting. This problem appears as a question in the GMAT PREP exam, but they mark the correct answer as A (statement 1 is sufficient). Depending on the test cases used, it seems that the answer could be A or E. Could someone please advise?fskilnik wrote:Hi, zaarathelab!
First of all, sorry for the delay. IÂ´ve not come here for a LONG time (too busy)!
You are absolutely right!!
Let me put the bifurcation explicitly, for all readers:
First case: (Please note both sums are equal.)
S = {2,2,2,2,2,2} then average(S) = 2
T = {4,4,4} then average(T) = 4
Answering in the affirmative;
Second case: (Please note both sums are equal.)
S = {3,3,3,3} then average(S) = 3
T = {2,2,2,2,2,2} then average(T) = 2
Answering in the negative;
Therefore sttm(1) is really INSUFFICIENT.
From the fact that (1) and (2) alone are insufficient, we have to consider both of them together... could you complete the problem and tell us if the answer is (C) or (E) ?
Regards (and congrats for your attention and rigour),
fskilnik.