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Re: The sum of the integers in list S is the same as the sum of [#permalink]

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15 Apr 2014, 00:21

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.

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.
_________________

Re: The sum of the integers in list S is the same as the sum of [#permalink]

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01 Jul 2014, 00:26

Bunuel wrote:

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?

Given: \(sum(S)=sum(T)\). Question: is \(t<s\), where \(s\) and \(t\) are # of integers in lists S and T respectively.

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T --> \(\frac{sum}{s}<\frac{sum}{t}\) --> cross multiply: \(sum*t<sum*s\). Now, if \(sum<0\) then \(t>s\) (when reducing by negative flip the sign) but if \(sum>0\) then \(t<s\). not sufficient.

(2) The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.

(1)+(2): If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.

If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.

Not sufficient.

Answer: E.

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??

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?

Given: \(sum(S)=sum(T)\). Question: is \(t<s\), where \(s\) and \(t\) are # of integers in lists S and T respectively.

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T --> \(\frac{sum}{s}<\frac{sum}{t}\) --> cross multiply: \(sum*t<sum*s\). Now, if \(sum<0\) then \(t>s\) (when reducing by negative flip the sign) but if \(sum>0\) then \(t<s\). not sufficient.

(2) The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.

(1)+(2): If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.

If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.

Not sufficient.

Answer: E.

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??

Re: The sum of the integers in list S is the same as the sum of [#permalink]

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01 Jul 2014, 21:39

Bunuel wrote:

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?

Given: \(sum(S)=sum(T)\). Question: is \(t<s\), where \(s\) and \(t\) are # of integers in lists S and T respectively.

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T --> \(\frac{sum}{s}<\frac{sum}{t}\) --> cross multiply: \(sum*t<sum*s\). Now, if \(sum<0\) then \(t>s\) (when reducing by negative flip the sign) but if \(sum>0\) then \(t<s\). not sufficient.

(2) The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.

(1)+(2): If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.

If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.

Not sufficient.

Answer: 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 ?

Re: The sum of the integers in list S is the same as the sum of [#permalink]

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03 Jul 2014, 02:57

VeritasPrepKarishma wrote:

patternpandora wrote:

Bunuel wrote:

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?

Given: \(sum(S)=sum(T)\). Question: is \(t<s\), where \(s\) and \(t\) are # of integers in lists S and T respectively.

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T --> \(\frac{sum}{s}<\frac{sum}{t}\) --> cross multiply: \(sum*t<sum*s\). Now, if \(sum<0\) then \(t>s\) (when reducing by negative flip the sign) but if \(sum>0\) then \(t<s\). not sufficient.

(2) The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.

(1)+(2): If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.

If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.

Not sufficient.

Answer: E.

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??

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??

Re: The sum of the integers in list S is the same as the sum of [#permalink]

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06 Jul 2014, 08:37

Apex231 wrote:

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?

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T. (2) The median of the integers in S is greater than the median of the integers in T.

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}

Re: The sum of the integers in list S is the same as the sum of [#permalink]

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30 Aug 2014, 06:30

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..

Re: The sum of the integers in list S is the same as the sum of [#permalink]

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06 Jan 2016, 06:22

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
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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.
_________________

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?

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T. (2) The median of the integers in S is greater than the median of the integers in T.

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
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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?

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T. (2) The median of the integers in S is greater than the median of the integers in T.

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.
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