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

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?".

Think again.
Every value greater than -1 need not satisfy the inequality but every value satisfying the inequality must be greater than -1.

x/|x| can take only 2 values: 1 or -1
If x is positive, x/|x| = 1
If x is negative, x/|x| = -1
x cannot be 0.

Now let's look at the question.
x > x/|x| holds for x > 1 (x is positive) or -1 < x < 0 (x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc

Which of the following holds for every value that x can take?
(A) X > 1
(B) X > -1

I hope that you agree that X > 1 doesn't hold for every possible value of X whereas X > -1 holds for every possible value of X.

Mind you, every value greater than -1 need not be a possible value of x.

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...

x/|x|<x
if x>0, then 1/x<1 ie, x>1
if x<0, then 1/(-x)>1, ie, x>-1

So combine these two, we know that x must be greater than -1.

baner is right, (B).

Bunuel, thanks for your reply. But I have to disagree with you because:

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

What do you think about that?

OA for this question is B and it's not wrong.

Consider following:
If \(x=5\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If \(-1<x<10\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If \(-1<x<0\) or \(x>1\), then 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

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand A says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.

Hi Bunnuel,
Can it be solved as-

x/|x|<x

Hence 1/|x|<1

1/x<1 OR -1/x<1
when 1/x<1
then 1<x = x>1

when -1/x<1
then -1<x = x>-1

The two possible outcomes are x>1 or x>-1
The more restrictive is x>-1
Hence B

Thanks
Neelam

No, that't not correct.

First of all, when you are writing 1/|x|<1 from x/|x|<x you are reducing (dividing) inequality by x:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Consider a simple inequality \(4>3\) and some variable \(x\).

Now, you can't multiply (or divide) both parts of this inequality by \(x\) and write: \(4x>3x\), because if \(x=1>0\) then yes \(4*1>3*1\) but if \(x=-1\) then \(4*(-1)=-4<3*(-1)=-3\). Similarly, you can not divide an inequality by \(x\) not knowing its sign.

Next, inequality \(\frac{1}{|x|}<1\) holds true in the following ranges: \(x<-1\) and \(x>1\) (and not: x>1 or x>-1, which by the way simply means x>1).

Hope it's clear.

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.

Dear Ian,
I truly appreciate your efforts on here. 'Must be true' is a condition without exceptions. And when i plug in 0, the inequality is NOT true. That five apples are more than 0 apples is clear to me. However, the question asks for what 'Must be true'. Several ranges that make the inequality true makes this question inaccurate. Unless, if the GMAT translates 'Must be true' into 'may or may not be true' then explanation with the ranges make sense. For me, these type of questions make the GMAT into a lottery and I am not alone since many test takers have trouble with a logic that ignores exceptions. I have another example, fresh from Kaplan.

If it is true to -6<= n <= 10, which of the following must be true?

n<8
n=-6
n>-8
-10<n<7
none of the above

Same case. Official solution is n>-8, however n= -7 does not fulfill the first requirement, it therefore CAN BE TRUE - but not MUST BE TRUE

Again, I am no stranger to logic but, I am sure many will agree, these kind of questions are nonsense, especially when one considers the official (you and others) translation of the question into "Are 3 apples more than 2?" - what kind of a question is that?

I've tried to explain the logic behind this question twice, so I won't try again, but I can assure you that every mathematician in the world would agree with the answer to this question - this has nothing to do with some kind of logic exclusive to the GMAT. The same is true of the Kaplan question you quote; if n is greater than -6, it is surely true that n is greater than -8. You seem to be looking at these problems backwards: you're assuming n > -8 is true, and asking if it needs to be true that -6 < n < 10. That's the opposite of what the question is asking you to do.

I think answer should be A and not B because x cannot be equal to zero. If x equals to zero then the equation will lead to infinity.

isnt E true ? from the above 1<mod (x) which implies 1< mod(x) squared

We are told that \(x\) can take values only from these ranges.

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

Now, \(|x|^2>1\) means that \(x^2>1\) --> \(x<-1\) or \(x>1\). Since \(x<-1\) is not true about \(x\) (we know that \(-1<x<0\) and \(x>1\)), then this option is not ALWAYS true.

Hope it's clear.

Bunuel, the question doesn't mention anywhere that x is an integer. so why have we not considered the values of 0<x<1 which also doesnt satisfy the equation? If it has been considered then the solution should be A and not B. x>1 will always hold true aka must be true! while x>-1 is sometimes true.. aka can be true.

I think you don't understand the question.

Given: \(-1<x<0\) and \(x>1\). Question: which of the following must be true?

A. \(x>1\). This opinion is not always true since \(x\) can be \(-\frac{1}{2}\) which is not more than 1.
B. \(x>-1\). This option is always true since any \(x\) from \(-1<x<0\) and \(x>1\) is more than -1.