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### Integer

by danjuma » Mon Sep 27, 2010 9:23 am
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.

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by clock60 » Mon Sep 27, 2010 11:11 am
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

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by clock60 » Tue Sep 28, 2010 10:07 am
hi again Danjuma
it will be very kind of you to post OA!!!!
as i want to know the answer and the way of my solving, or you are waiting for more replies?

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by GMATMadeEasy » Tue Sep 28, 2010 11:24 am
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.

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by cipherstrength128bits » Sat Oct 16, 2010 2:12 am
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 ?

pre-requisite : 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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by [email protected] » Mon Oct 18, 2010 5:32 am
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.
Hi, guys.

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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by [email protected] » Mon Oct 18, 2010 5:35 am
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.
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.

I hope you got the point!

Regards,
Fabio.
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by zaarathelab » Fri Aug 19, 2011 12:50 pm

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/non-negative 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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by [email protected] » Fri Aug 26, 2011 1:00 pm
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

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

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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by navami » Tue Sep 06, 2011 5:02 pm
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.
This time no looking back!!!
Navami

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by [email protected] » Wed Sep 07, 2011 8:08 am
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.
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...

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! Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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by Poisson » Tue Aug 30, 2016 10:11 am
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

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

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.
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?

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by [email protected] » Tue Aug 30, 2016 10:24 am
Poisson wrote:
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

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

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.
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?
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...
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