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by rommysingh » Fri Sep 11, 2015 5:56 am
If list S contains nine distinct integers, at least one of which is negative, is the median of the integers in list S positive?

(1) The product of the nine integers in list S is equal to the median of list S.

(2) The sum of all nine integers in list S is equal to the median of list S.
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 one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
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Source: — Data Sufficiency |
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by GMATGuruNY » Fri Sep 11, 2015 6:04 am
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by Max@Math Revolution » Tue Sep 15, 2015 7:57 am
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.


If list S contains nine distinct integers, at least one of which is negative, is the median of the integers in list S positive?

(1) The product of the nine integers in list S is equal to the median of list S.

(2) The sum of all nine integers in list S is equal to the median of list S.

In the original condition we have 9 variables, thus we need 9 equations to match the number of variables and equations. Therefore E has high probability of being the answer.

Using both 1) & 2) together, -4,-3,-2,-1,0,1,2,3,4 is the only possible case thus the median=0 and the answer is no. therefore the conditions are sufficient. However, since the key question is about ratio and integer, we should apply common mistake type 4(A) and look at 1) again. In case of 1), we have -4,-3,-2,-1,0,1,2,3,4 thus the median=0 no and thsu the condition is sufficient. In case of 2), -4,-3,-2,-1,0,1,2,3,4 median=0, while -6,-5,-4,-3,2,3,4,5,6 median=2 therefore we have both no and yes the answer. The condition is therefore not sufficient, and thus the answer is A.

Normally 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 E has a high chance of being the answer (especially about 90% of 2by2 questions where there are more than 3 variables), which is why we attempt to solve the question using 1) and 2) together. Here, there is 80% chance that E is the answer, while C has 15% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer according to DS definition, we solve the question assuming E would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

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