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| DS an easier approaches other than substitution? |
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vanD
Guest
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Is |n| < 1 ?
(1) nx – n < 0 (2) x–1 = –2 from a posting in scoretop.com by reach_sat |
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Guest
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Stcy,
Thanks for the follow up. I found out that this questions is from his Kaplan class work. |
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Guest
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Is the answer E ?
Starting with #2: x = -1 , which is not suff, so cross of BD statement #1: N (x-1) < 0 , so this says N could be pos or neg interger or fraction, depending on what X is, so insufficient Together, N(-2)<0 , so N has to be positive, but could be over 1 or a fraction, so insuff, and I got E |
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boo
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Dan Bernstein
MGMAT STAFF
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The above explanation is correct. This problem offers great practice on both rephrasing Data Sufficiency questions and deducing the easier statement to begin your work.
Rephrase: Is -1 < n < 1? Since statement (2) offers no informtion about n, begin with a BD/ACE grid and eliminate BD. Statement (1) tells us n(x-1) < 0. In this case, if n>0, x<1; if n<0, x>1. Eliminate A. Together, we are dealing with the first scenario from statement (1): x<0 since it equals -1. Thus, n(-2) < 0 means n>0, but n could still be an integer or fraction. The correct answer is E. |
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Guest
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Dan, in the process of cutpaste, i overlooked that the ^ mark had dropped off... appreciate your help
Is |n| < 1 ? (1) n^x – n < 0 (2) x^–1 = –2 |
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Emily Sledge
MGMAT STAFF
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Can you clarify Statement (2)?
Is it: n^(x – n) < 0 or is it: (n^x) – n < 0 For solvability, I am guessing it is the latter, but just want to confirm... |
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Guest
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latter.. you are right.
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Saurabh Malpani
Guest
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C is the answer
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Saurabh Malpani
Guest
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n x (n^x) < n True/False
0.5 (1/2) 0.5^(1/2) < .5 FALSE 0.5 (1/2) 0.5^(2) < .5 TRUE 0.5 (-1/2) 0.5^(-1/2) < .5 FALSE 3 (1/2) Sqrt(3) < 3 FALSE So A is insufficient. B in itself insufficient. Taking A and B--we get that only the last statement is true. Hence mod(n) NOT < 1 I always like to proceed in tabular form as that helps me to keep my DS organized and visually see the answer. I hope that helps Saurabh Malpani
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Guest
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I am looking for a more formula based logically deductive approach than substit...Can somebody please make an attempt in that direction?
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Stacey Koprince
MGMAT STAFF
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Happy to provide the more strictly "math-based" approach - but be aware that many GMAT questions are designed to trip you up if you use the official math way. Alternate methods, such as substituting numbers, are often preferable (as I think is true in this case).
Is |n| < 1 ? (1) (n^x) – n < 0 (2) x–1 = –2 From the question, we're asked if -1<n<1. As Dan mentioned earlier, it's important to rephrase the question to make sure you fully understand what they're asking. Easiest to start with (2) - this tells us nothing about "n" so cross off B and D. (We can determine that x = -1, though.) (1) really says (n^x) – (n^1) < 0. I can factor out an n to get n[n^(x-1) - 1] < 0, or n < 0 and n^(x-1) - 1 < 0. The first solution, n<0 is not sufficient to answer my question - it could be between -1 and 1 but it might not be. I can't solve the second one at all because I have two variables and only one inequality. (1) and (2) together: Now I can say that (n^-1) - n < 0 or (1/n) - n < 0. I can multiply every term by n but now I have to split the problem into two parts (depending upon whether n is positive or negative). If n is pos, then n - n^2 < 0 or n(1-n) < 0 If n is neg, then n - n^2 > 0 or n(1-n) > 0 Following through on these statements shows that they are insufficient - n could be between -1 and 1 but doesn't have to be. Real numbers are easier: n could be 2 b/c 1/2 - 2 < 0; in this case, |n| is NOT less than 1. n could be -1/2 b/c 1/(-1/2) - (-1/2) = -2 + 1/2 which is less than zero; in this case |n| IS less than 1. |
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| DS an easier approaches other than substitution? |
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