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If x/|x|<x which of the following must be true about x?

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If x/|x|<x which of the following must be true about x? [#permalink] New post 15 Aug 2008, 03:06
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If \frac{x}{|x|}<x which of the following must be true about x?

(A) x>1

(B) x>-1

(C) |x|<1

(D) |x|=1

(E) |x|^2>1
[Reveal] Spoiler: OA
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Re: Inequality [#permalink] New post 15 Aug 2008, 06:27
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To condense some of what the other thread states:

If you use -1/2, it works, so at first you might think the answer is clearly b), but "must be true about X" also means "must ALWAYS be true about X". If you use +1/2 (which is > -1 as in answer B) it doesn't work.

1/2 over 1/2 = 1 < x which is 1/2 so you get 1 < 1/2 and that's not true.

So answer a MUST ALWAYS be true because only positive X over abs(x) = 1 and X > 1, it will always be greater than 1 (which is x/|x| when x is positive).

Does this condense the other thread?

nmohindru wrote:
If X/|X| < X Which of the following must be true about X ?

a) X>1

b) X>-1

c) |X| < 1

d) |X| = 1

e) |X|^2 > 1

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Re: Inequality [#permalink] New post 15 Aug 2008, 07:37
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i think OA is correct....

the question is If x / |x| < x. so we have to consider only those values of x for which this inequalty is true and what are those values
1. when x is between -1 and 0
2. when x is more than 1

lets call these conditions our universe.

Now the question is for all values of x (in our universe) which of the following is true
option B, x > -1, has both conditions 1 and 2

now you may say what about x = 1/2 .... that wasnt even part of our universe... so even if x = 1/2 is satisfying option B and not the question stem, we dont have to worry... becuase we are not supposed to take it as an example ...

for all values of x in our universe, option B is ALWAYS true...
Option A is not always true...
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Re: Inequality [#permalink] New post 16 Aug 2008, 11:50
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I believe (E) should be the answer.

we know that |x| = -x for x < 0
= x for x > 0

For x < 0,

x / |x| < x
=> x / -x < x
=> x > -1

For x > 0,

x / |x| < x
=> x / x < x
=> x > 1

Therefore |x| > 1 or |x|^2 > 1
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Re: Inequality [#permalink] New post 16 Aug 2008, 12:56
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Because the question states what MUST be true, I think the answer has to be A. We can't select only our "universe" of numbers that satisfy the selection. If the answer, like in B, says X > -1, then every single number greater than -1 must satisfy the inequality and that is not the case.

I certainly agree that the "universe" must be -1 < X < 0 and 1 < x. Answer B defines one "universe" as all numbers that are greater than -1. Answer B includes all numbers between 0 and 1 which do not satisfy the inequality; therefore, B must not always be true. The answer should be A. I disagree with the OA.

nmohindru wrote:
If X/|X| < X Which of the following must be true about X ?

X>1

X>-1

|X| < 1

|X| = 1

|X|^2 > 1

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Re: Inequality [#permalink] New post 16 Aug 2008, 13:22
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For x < 0,

x / |x| < x
=> x / -x < x
=> x > -1
x < 0 & x > -1
-1<X<0


For x > 0,
x / |x| < x
=> x / x < x
=> x > 1
x>0&x > 1
----> x>1

Question says which one must be true..

Option A ) correct.. X>1 (X>1 implies that when X>0 )
this is always true

bet on A.
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Re: Inequality [#permalink] New post 16 Aug 2008, 16:21
guys i beg to differ.. i'll try to explain myself again...

our universe is defined in this line, if x/|x| < x this is given ... x only exits if it follows this condition ... if soemthing doesnt follow it then, its not x ... its y, may be z (doesnt matter) ...

if S is a set of possible values of x that will satisfy our condition x/|x| < x
S = [ -0.9, -0.8, -0.7, ....... -0.1, 1.1, 1.2, 1.3, 1.4 ....... upto infinite]

we have to find an answer that will always be true for all values of x,

lets see A, x > 1, this statement is not true for x = -0.9, -0.8 ..... -0.1

check B x > -1, this statement will "always" be true for all x ... clearly B is the answer...

i see your point that x = 0.5 will not satisfy the question condition, but thats the whole point 0.5 is not even x, its y
but if you select A, you are leaving some valid values of x.
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Re: Inequality [#permalink] New post 16 Aug 2008, 16:37
you guys are answering a different question, as i suggested in the linked thread

"For which of the following conditions for x,
x/|x| < x is always true. "


now A will be the clear answer,
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Re: Inequality [#permalink] New post 16 Aug 2008, 16:45
nmohindru wrote:
If X/|X| < X Which of the following must be true about X ?

X>1

X>-1

|X| < 1

|X| = 1

|X|^2 > 1


For me it is disputable. I sometime go with B and sometime with A. B is broader than A but B again is not always a true case where as A is always true.

if we pay detail attention to MUST on "Which if the following must be true about X", it should be A because B is not a "must" true.
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Re: Inequality [#permalink] New post 16 Aug 2008, 16:50
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GMAT TIGER wrote:
nmohindru wrote:
If X/|X| < X Which of the following must be true about X ?

X>1

X>-1

|X| < 1

|X| = 1

|X|^2 > 1


For me it is disputable. I sometime go with B and sometime with A. B is broader than A but B again is not always a true case where as A is always true.

if we pay detail attention to MUST on "Which if the following must be true about X", it should be A because B is not a "must" true.


Please go through my earlier posts and give me a value of x for which B is not true, i dont think you can find one ....
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Re: Inequality [#permalink] New post 16 Aug 2008, 17:12
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GMAT TIGER wrote:

x = 0, 1 , or any +ve fractions.


that is not even defined as x, our x is defined in the first line ....

i'll give an example,

x is a member of set S = [-4,-3,-2,-1, 4,5,6,7]
what is always true for all x
A. x >= 4
B. x >= -4

A or B... if you select A, then i cant explain it further..
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Re: Inequality [#permalink] New post 16 Aug 2008, 18:53
durgesh79 wrote:
GMAT TIGER wrote:

x = 0, 1 , or any +ve fractions.


that is not even defined as x, our x is defined in the first line .....


Gotta. Now I believe this a classical example of gmat questions. nice one....

if x/lxl < x, x can not be 0, 1, and any +ve fraction.
x is any value > -1 except 0, 1, and any +ve fraction.

remember the condition "if x/lxl < x"

its been a looming question for me since long time i.e. why it is not A?

thanks durgesh79.
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Re: Inequality [#permalink] New post 17 Aug 2008, 07:29
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I'm pretty sure Durgesh has already explained his answer convincingly, but just in case, I'll present a different problem:

If x is positive, what must be true?

I) x > 10
II) x > -10
III) x > 0

Of course, III) says exactly 'x is positive', so III) must be true. But if x is positive, x is obviously larger than -10. II) must also be true. I) doesn't need to be true; x could be 4, or 7, for example.

The same situation applies with the question that began this thread. We know that either -1 < x < 0, or x > 1. What must be true? Well, "x > 1" does not need to be true. We know that x might be -0.5, for example. On the other hand, "x > -1" absolutely must be true: if x satisfies the inequality x/|x| < x, then x is certainly larger than -1. If we had been asked whether "x > -1,000,000" must be true, the answer would also be yes. The question did not ask "What is the solution set of "x/|x| < x"; nor did it ask "For which values of x is x/|x| < x true?".
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Re: Inequality [#permalink] New post 17 Jan 2010, 22:31
jallenmorris wrote:
To condense some of what the other thread states:

If you use -1/2, it works, so at first you might think the answer is clearly b), but "must be true about X" also means "must ALWAYS be true about X". If you use +1/2 (which is > -1 as in answer B) it doesn't work.

1/2 over 1/2 = 1 < x which is 1/2 so you get 1 < 1/2 and that's not true.

So answer a MUST ALWAYS be true because only positive X over abs(x) = 1 and X > 1, it will always be greater than 1 (which is x/|x| when x is positive).

Does this condense the other thread?

nmohindru wrote:
If X/|X| < X Which of the following must be true about X ?

a) X>1

b) X>-1

c) |X| < 1

d) |X| = 1

e) |X|^2 > 1

I think A is the right answer OA may be wrong because if it says X>-1 then the given inequality will not be true for x=0. and this is not mentioned that x=!0 so that we can ignore x=0 while considering X>-1.

Please Comment
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Re: Inequality [#permalink] New post 17 Jan 2010, 22:59
GMAT TIGER wrote:
durgesh79 wrote:
GMAT TIGER wrote:

x = 0, 1 , or any +ve fractions.


that is not even defined as x, our x is defined in the first line .....


Gotta. Now I believe this a classical example of gmat questions. nice one....

if x/lxl < x, x can not be 0, 1, and any +ve fraction.
x is any value > -1 except 0, 1, and any +ve fraction.

remember the condition "if x/lxl < x"

its been a looming question for me since long time i.e. why it is not A?

thanks durgesh79.

Hi,
A must say that a is the answer
Nothing like x=!0,1and +ve fraction is given
If these conditions have been given for x then for broader answer B could be the right answer
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Re: Inequality [#permalink] New post 18 Jan 2010, 03:11
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nmohindru wrote:
If \frac{x}{|x|}<x which of the following must be true about x?

(A) x>1

(B) x>-1

(C) |x|<1

(D) |x|=1

(E) |x|^2>1


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \frac{x}{|x|}<x:

A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0;

B. x>0 --> |x|=x --> \frac{x}{x}<x --> 1<x.

So given inequality holds true in the ranges: -1<x<0 and x>1. Which means that x can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about x. Option A can not be ALWAYS true because x can be from the range -1<x<0, eg -\frac{1}{2} and x=-\frac{1}{2}<1.

Only option which is ALWAYS true is B. ANY x from the ranges -1<x<0 and x>1 will definitely be more the -1, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.
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Re: Inequality [#permalink] New post 18 Jan 2010, 09:09
Nice n tricky question and very well explained by Bunuel

Thanks.

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Re: If x/|x|<x which of the following must be true about x? [#permalink] New post 19 Dec 2011, 14:20
Typical GMAT nonsense. A lot of people seem to talk themselves into a solution but in mathematics, there are no GMAT-truths. x>-1 must not be true, since it ignores the fact that 0 does not fulfill the requirement.
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Re: If x/|x|<x which of the following must be true about x? [#permalink] New post 20 Dec 2011, 02:08
myfish wrote:
Typical GMAT nonsense. A lot of people seem to talk themselves into a solution but in mathematics, there are no GMAT-truths. x>-1 must not be true, since it ignores the fact that 0 does not fulfill the requirement.


No one is "talking themselves into a solution" here, and there's nothing wrong with the mathematics. I explained why earlier, but I can use a simpler example. If a question reads

If x = 5, what must be true?

I) x > 0


then clearly I) must be true; if x is 5, then x is certainly positive. It makes no difference that x cannot be equal to 12, or to 1000.

The same thing is happening in this question. We know that either -1 < x < 0, or that 1 < x. If x is in either of those ranges, then certainly x must be greater than -1. It makes no difference that x cannot be equal to 1/2, or to 0.

This is an important logical point on the GMAT (even though the question in the original post is not a real GMAT question), since it comes up all the time in Data Sufficiency. If a question asks

Is x > 0?

1) x = 5


that is exactly the same question as the one I asked above, but now it's phrased as a DS question. This question is really asking, when we use Statement 1, "If x = 5, must it be true that x > 0?" Clearly the answer is yes. If you misinterpret this question, and think it's asking "can x have any positive value at all", you would make a mistake on this question and on most GMAT DS algebra questions.
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Re: Inequality [#permalink] New post 20 Dec 2011, 03:05
as bunnel said:
X>-1 will always hold condition true.

If i take value as 0.5 which is greater then -1

.5/|.5|=1 which is not less then 1

for me -1<x<0 and x>1 both holds true.

Bunuel wrote:
nmohindru wrote:
If \frac{x}{|x|}<x which of the following must be true about x?

(A) x>1

(B) x>-1

(C) |x|<1

(D) |x|=1

(E) |x|^2>1


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \frac{x}{|x|}<x:

A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0;

B. x>0 --> |x|=x --> \frac{x}{x}<x --> 1<x.

So given inequality holds true in the ranges: -1<x<0 and x>1. Which means that x can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about x. Option A can not be ALWAYS true because x can be from the range -1<x<0, eg -\frac{1}{2} and x=-\frac{1}{2}<1.

Only option which is ALWAYS true is B. ANY x from the ranges -1<x<0 and x>1 will definitely be more the -1, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.
Re: Inequality   [#permalink] 20 Dec 2011, 03:05
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