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Mod X. A 700-800 ques

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Mod X. A 700-800 ques

by rb90 » Fri Sep 10, 2010 3:50 am
If x is not equal to 0, is |x| less than 1?

(1) ( x/|x| ) < x

(2) |x| > x

The actual answer is C (both together are sufficient)
I need help understanding the working.
Really need the help people.
Thanks.
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by kvcpk » Fri Sep 10, 2010 4:07 am
rb90 wrote:If x is not equal to 0, is |x| less than 1?

(1) ( x/|x| ) < x

(2) |x| > x

The actual answer is C (both together are sufficient)
I need help understanding the working.
Really need the help people.
Thanks.
Question is asking whether x lies between -1 and 1

(1) ( x/|x| ) < x
there can be only 2 cases. x>0 or x<0
1) x>0
then |x|=x
x/x <x
1<x
x>1
So when x>0, x>1 [doesnot fall between -1 and 1]
2)x<0
then |x|=-x
x/(-x) <x
-1<x
x>-1
So, when x<0, -1<x<0 [falls between -1 and 1]
Contradicting results.
hence INSUFF


stmt2) |x| > x
again 2 cases.
x>0:
x>x.. Not possible.
x<0:
-x>x
2x<0
x<0 [no info if it falls between -1 and 1 only.]
Hence INSUFF

Combining:
We see that only x<0 is possible.
and for x<0, from stmt1 we know falls between -1 and 1.
Hence sufficient.

Hope this helps. Let me know if you have trouble getting this.
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by tpr-becky » Fri Sep 10, 2010 11:25 am
We need to know whether x is a fraction (the only way that an absolute value can be less than 1)

1) x/ \x\ will always be 1 or negative one (depending on the pos/neg nature of x) When x is greater than one this equation always works. BUT this also works if x is a negative fraction - a negative number over it's absolute value will always be -1 so the only time this can be less than x will be when x is a fraction. So here we can't tell if x is positive or negative and thus BCE

2) if \x\>x that means that x has to be negative because x and \x\ will always be the same distance from zero on the number line it is just a matter of positive/negative. here becuase the absolute value is greater than the original number the original number must be negative. the only problem here is that we are asking about fractions, not pos/neg so this doesn't give us enough info. thus CE

When we put them together #2 tells us that x is negative and when x is negative #1 tells us that is has to be a fraction - therefore together we know that x is a negative fraction and therefore \x\ is less than 1.
Becky
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The Princeton Review
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by anantbhatia » Fri Sep 10, 2010 11:34 pm
I think this was a very good question. Confused me a lot. But I couldn't understand the solution from the above two replies. So, I solved it in my way.

1.
|x|<1 means
-1<x<1 . Asked is if this condition is true.


2.
x/|x|<x means

either

x-x^2<0 critical pts- 0,1

or

x>-x^2 or x^2+x >0 critical pts 0, -1

Graphs for them are shown in the figure. Hence A cannot satisfy and A, D eliminated.

3.
From second condition we get
-x<x
satisfies only for x<0 and hence cannot satisfy alone. B eliminiated.

From the above equations, we get the range -1<x<0 which satisfies the above equations 1,3 and hence the answer (C)

Experts please let me know if the above understanding was correct.

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