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DS an easier approaches other than substitution?
vanD
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Is |n| < 1 ?
(1) nx – n < 0
(2) x–1 = –2

from a posting in scoretop.com by reach_sat
Stacey Koprince
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Joined: 06 Mar 2007
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Hi, VanD

We actually need to know the author of problems posted here, not just the username of someone who posted the problem on another site. We need to ensure that we are not allowing posts from official administrations of the GMAT - it is illegal to reproduce questions from the official test. Can you go back to that site and see if you can find out the source from the user reach_sat? Sorry for the hassle, but we can't risk being shut down.

Thanks!
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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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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
Dan Bernstein
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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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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
Emily Sledge
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Joined: 06 Mar 2007
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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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latter.. you are right.
Saurabh Malpani
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C is the answer

esledge wrote:
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...
Saurabh Malpani
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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




Saurabh Malpani wrote:
C is the answer

esledge wrote:
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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I am looking for a more formula based logically deductive approach than substit...Can somebody please make an attempt in that direction?
Stacey Koprince
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Posts: 2630
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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.
DS an easier approaches other than substitution?
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