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Answer is C because combined we come conclude that neither 0,-1 and 1 can be X and therefore [x]>1

My question: What is the most effiecient way of answering this question? It it best to just plug in numbers and by doing so, conclude that between S1 and S2 X is neither 0, 1, or -1?

My route was:

1) Simplifying S1 to.... 2X>1 so X>1/2 OR X<-1 2) Simplifying S2 to....X>1 OR 2X<-1 so X>-1/2

Is that incorrect? And if so, how would you combine S1 and S2 and exclude numbers? Thank you.

1: gives x>1/2 and x<-1

so we get both YES and NO

2: gives x> 1 and X<-1/2 so both YES and NO

so A B D are knocked off

together we get that the values are between 1/2 and -1/2, hence the ANSWER is NO..hence C _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

I have read your post on this topic in detail. But the reason why the inequality sign was flipped (from the original question) for the first part of the expression (1-2x)>0 and not for the second part of the expression (1 + x)<0 is not clear.

I have read your post on this topic in detail. But the reason why the inequality sign was flipped (from the original question) for the first part of the expression (1-2x)>0 and not for the second part of the expression (1 + x)<0 is not clear.

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is |x| > 1 means is x<-1 (-2, -3, -4, ...) or x>1 (2, 3, 4, ...), so for YES answer x can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as (2x-1)(x+1)>0 (so that the coefficient of x^2 to be positive after expanding): roots are x=-1 and x=\frac{1}{2} --> ">" sign means that the given inequality holds true for: x<-1 and x>\frac{1}{2}. x could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as (x-1)(2x+1)>0: roots are x=-\frac{1}{2} and x=1 --> ">" sign means that the given inequality holds true for: x<-\frac{1}{2} and x>1. x could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x<-1 and x>1. Sufficient.

Neither condition 1 nor condition 2 suffice. But i guess drawing the possible outcomes of condition 1 and 2 on no. line will get the solution. Answer is C

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is |x| > 1 means is x<-1 (-2, -3, -4, ...) or x>1 (2, 3, 4, ...), so for YES answer x can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as (2x-1)(x+1)>0 (so that the coefficient of x^2 to be positive after expanding): roots are x=-1 and x=\frac{1}{2} --> ">" sign means that the given inequality holds true for: x<-1 and x>\frac{1}{2}. x could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as (x-1)(2x+1)>0: roots are x=-\frac{1}{2} and x=1 --> ">" sign means that the given inequality holds true for: x<-\frac{1}{2} and x>1.x could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x<-1 and x>1. Sufficient.

Answer: C.

Hi Bunuel,

I understood the question and got the answer right by finding the intersection for inequalities. The approach I normally use for such questions is as follows:

(1 - 2x)(1 + x) < 0

for above statement to be true the 2 expressions should have opposite signs, giving us (x > 1/2 and x > -1) OR (x < 1/2 and x < -1) Now, finding intersection gives us x > 1/2 OR x < -1. As a result, x < -1 is fine but x > 1/2 includes 1 as an option - making the statement insufficient.

Now my question is, how did you use the logic below to skip some of the steps above and directly go to the conclusion: x<-1 and x> 1/2. . What's the rule if instead of greater than (>) sign it was less than (<).

If I can understand the logic, I believe it will help save some valuable time.

">" sign means that the given inequality holds true for: x<-1 and x> 1/2.