The sum of the integers in list S is the same as the sum of the intege

Brunel,

Thank you for all your help. Brilliant stuff!

I have a concern on this one... I had marked the answer as E on GMATPrep but as per the answers on it the correct answer is A, and obviously the software will consider these answers when calculating your score. How exactly or to what extend can we rely on the score thrown up by the test if they have wrong answers on the software to begin with? It's causing a conflict. Please help.

Thank you.

Actually there aren't that many flawed questions in GMAT Prep and they remove one when spot it. I remember only 3 or 4 wrong questions in different editions, so generally their score is accurate.
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I am not understanding, because I got this same question in GMAT prep & ans indicated by them is A. And I am also unclear as why A. So is there something I am missing or the above explanation has something missing??

As indicated by Bunuel in his post here: the-sum-of-the-integers-in-list-s-is-the-same-as-the-sum-of-127755.html#p1046371
the correct answer is (E).

As always your solutions are elegant and crisp Bunuel. I liked the thought process on 1. Any other way to work on option 2 ? I am a bit wary on putting in values. Any conceptual way to negate this option ?

I am attaching the screenshot indicating OA as A from Gmat Prep test

@Bunuel : Please guide further.

PLEASE READ THE WHOLE THREAD. THE OA FOR THIS QUESTION IS WRONG IN GMAT PREP.
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My take is E.

1) S = {1,2,3,4} ; avg = 2.5
T = {100,200} ; avg = 150
avg(T) > avg(S) and {S} > {T}

now

S = {1,2,3,4} ; avg = 2.5
T = {100,200,300,400,500} ; avg = 375
avg(T) > avg(S) and {T} > {S}

Hence A is not sufficient.

2) S = {1,2,3} ; median = 2
T = {1,2} ; median = 1.5
{S} > {T}

now

S = {1,2,3} ; median = 2
T = {-5,-4,-3,-2,-1} ; median = -3
{T} > {S}

Hence B is not sufficient

(A) + (B)

S = {1,2,3} ; median = 2
T = {-5,-4,-3,-2,1000} ; median = -3
avg(S) < avg(T) and {T} > {S}

now

S = {1,2,3,4,5,10000} ; median = 4
T = {-100, -3, 1000000} ; median = -3
avg(S) < avg(T) and {S} > {T}

Hence (A)+(B) is insufficient.
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Is it possible to do this in 2 minutes with inserting numbers like in Bunuel's explanation? :S Or is there another approach?
I always prefer to insert numbers, but it seems that it just takes far too long sometimes..

The sum of integers in list S is the same as the sum of the integers in T. Does S contain more integers than T?
can we also interpret the above statement to mean that " s and t may also contain decimals along with integers but only sum of their integers is equal?" I am just trying to see how closely we should adhere to the language of the GMAC questions. Thanks

Yes, S and T may contain non-integers too. The question is only concerned about integers (Does S contain more INTEGERS that T?) and hence we don't really care about the other elements. The statements also only talk about integers.

Had the question been: "Does S contain more elements than T?"
we would have had to consider the possibility of non integer elements too.

Given, Sum of terms in S = Sum of terms in T

Is Number of terms in S > Number of terms in T?

Statement 1: Average of S < Average of T

i.e. (Sum of terms in S)/Number of terms in S < (Sum of terms in T)/Number of terms in T

i.e. (Sum of terms in S)* Number of terms in T < (Sum of terms in T)* Number of terms in S

If sum of terms in S is POSITIVE then it may be cancelled out from both sides and then
Number of terms in T < Number of terms in S

If sum of terms in S is NEGATIVE then it may be cancelled out from both sides but Inequality sign reverses i.e.
Number of terms in T > Number of terms in S

Hence NOT SUFFICIENT


Statement 2: Median of S > Median of T
But median have no relation with the sum of the terms in any set hence
NOT SUFFICIENT

Combining also doesn't give any solution as median has no relation with sum of terms and first statement is Insufficient as we don't know whether Sum of the terms of the Set S and T are positive or negative

Answer: option E

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Sum(S) = Sum(T) ==> Q: #(S) > #(T)

There are 4 variables and 1 equation. Thus the answer E is most likely.

Actually, there is no relation between average and median.
The reason A is not an answer is that the sum could be positive or negative.

Let's consider both conditions 1) and 2) together.

S = { 1, 2, 3, 4 }
T = { 1, 2, 7 }
Sum = 1 + 2 + 3 + 4 = 1 + 2 + 7 = 10
ave(S) = 10/4 and ave(T) = 10/3
med(S) = 2.5 and med(T) = 2
S has more integers than T : Yes

S = { -1, -2, -7 }
T = { -1, -2, -3, -4 }
Sum = (-1)+(-2)+(-7) = (-1)+(-2)+(-3)+(-4) = -10
ave(S) = -10/3 and ave(T) = -10/4
med(S) = -2 and med(T) = -2.5
T has more integers than S : No.

The answer is not unique.

Therefore, the answer is E as expected.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.

chetan2u niks18 gmatbusters KarishmaB Abhishek009 amanvermagmat

Can anyone help with algebraic approach to this Q? I am a bit uncomfortable with
picking numbers to prove statements are not sufficient.
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Hi adkikani,

check this solution by Bunuel. If you have any query, pls let us know -

https://gmatclub.com/forum/the-sum-of-t ... l#p1046371

Bunuel has given you the algebraic approach for statement 1 (link given by niks18 above).
When you talk of median, usually you do not have an algebraic approach. You often need to pick numbers to figure out the logic.

1)
Let S= 1,1
T=-1,0,3
Average of S > Average of T

Now let S=1,0,-3
T=-1,-1
Average of S > Average of T

Thus for Average of S > Average of T, # of integers in S can be smaller or greater than those in T.
INSUFFICIENT

2)
Again,
Let S= 1,1
T=-1,0,3
Mean of S(0.5) > Mean of T(0)

Now let S=1,0,-3
T=-1,-1
Mean of S > Mean of T

Thus for Mean of S > Mean of T, # of integers in S can be smaller or greater than those in T.
INSUFFICIENT

As the same examples have been used to show that both options each are insufficient, we can use the same examples to show that both choices together are INSUFFICIENT.

Hence correct answer is E

Why you always say "forget about conventional approach"? Most of you explanation are based on conventional approach. The way you write the explanations makes them boring, dull and eventually no takeaway for the future problems.

Posted from my mobile device
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Hi Karishma, I don't think you can consider sets with sum=0 because your DS Stem 1 provides a data point on Avg of one set being lesser than the Avg of the other. If you consider sets with Sum 0, Stem 1 info won't hold good.

However, sets with sum negative still works.