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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) \(|x|^2>1\)

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.

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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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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...

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

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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20 Dec 2011, 03:08

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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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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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21 Feb 2016, 22:14

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happyface101 wrote:

Engr2012 wrote:

happyface101 wrote:

Hi Karishma and other experts - I understand the solution but I'm still stuck on why reasoning is wrong. This is how I solved:

x/|x| < x --> x/|x| -x --> x*(1/|x|-1)<0, so:

1. x<0; (1/|x|-1)<0 -> 1/|x|<1 -> 1<|x| -> x<-1 THIS IS WHERE IT'S WRONG BUT WHY?

2. x>0; (1/|x|-1)>0 -> 1/|x|>1 -> 1>|x| -> 1>x THIS IS WHERE IT'S WRONG BUT WHY?

Thank you!

The mistake you are making is in the portion with red text above. Your 'simplified' expression is x(1/|x|-1) and not just (1/|x|-1).

Thus your cases become:

Case 1: x<0 --> |x|=-x ---> x(1/|x|-1)<0 ---> x(-1/x-1)<0 ---->x(1/x+1)>0 --->x(1+x)/x > 0--->x+1>0 --->x>-1 and this with the fact that x<0 ---> range becomes -1<x<0.

Case 2: x>0 --> |x|=x ---> x(1/|x|-1)<0 ---> x(1/x-1)<0 ---->x(1/x-1)<0 --->x(1-x)/x < 0--->1-x<0 --->x>1 and this with the fact that x>0 ---> range becomes x>1.

Hope this helps.

Thank you so much! +1 I think the knowledge gap is that I treated x*(1/|x|-1)<0 as if it's x*((1/|x|-1)=0 In an inequality problem I obviously can't solve by taking the two apart and do x<0, (1/|x|-1)<0 like I can for an equation. Thanks for helping me understand this!

Additionally, you can split the factors as you did but you made an error there.

x*(1/|x|-1)<0 implies that x*(1/|x|-1) is negative.

So either x is negative and (1/|x|-1) is positive or x is positive and (1/|x|-1) is negative

Case 1: x is negative and (1/|x|-1) is positive When x is negative, |x| = -x

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.

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 ....

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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13 Jan 2012, 12:17

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What this means is that x could be negative fraction or positive integer/fraction. Try plugging in -2, -1, -0.8, 1, 2, 3 ....

B _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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27 May 2012, 05:16

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vibhav wrote:

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. _________________

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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30 May 2012, 05:36

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pavanpuneet wrote:

Hi Shouvik,

But I am not dividing .. I first consider that x<0 then cross multiply and flip the sign. I am still not clear with where did I go wrong.

Ok,

Let me explain step by step:

Note that we have considered x<0. So anything we divide or multiply by a negative number x will change the signs of an inequality.

1. Since x<0, x/|x| < x => x/(-x) < x 2. Cross-multiplying both sides by (-x). Now since x<0, (-x)>0. So if we cross multiply it doesn't change the sign. x < -x^2 3. Now, we divide both sides by x. Since x<0, this changes the sign of the inequality. 1 > -x 4. Simplifying further, -1<x

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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12 Jun 2012, 04:31

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There was a lot of confusion between options (A) and (B). Therefore, I would like to explain why option (B) is correct using diagrams.

Forget this question for a minute. Say instead you have this question:

Question 1: x > 2 and x < 7. What integral values can x take? I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.

Attachment:

Ques3.jpg [ 4.45 KiB | Viewed 6450 times ]

You see that the overlapping area includes 3, 4, 5 and 6.

Now consider this:

Question 2: x > 2 or x > 5. What integral values can x take? Let’s draw that number line again.

Attachment:

Ques4.jpg [ 4.24 KiB | Viewed 6449 times ]

So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. overlapping). OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.

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) \(|x|^2>1\)

\(\frac{x}{|x|}\) is either 1 or -1. So x > 1 or x > -1 So which values can x take? All values that are included in at least one of the sets. Therefore, x > -1. _________________

If x/|x|<x which of the following must be true about x? [#permalink]

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15 Aug 2014, 09:28

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sri30kanth wrote:

Bunuel,

What if we square the inequality x/ |x| < x. Then we get (x^2 / x ) < x^2 which implies that x^2 < x^3. Is this correct? Please explain. Thanku

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality), which is not the case here.

Also, the second step in your solution: never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know the sign of it, we don't know the sign of x, so we cannot multiply x^2/x < x^2 by x here.

Re: If x/|x|<x which of the following must be true about x? [#permalink]

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17 Aug 2014, 03:58

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dransa 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) \(|x|^2>1\)

I disagree with OA .According to me the Answer should be A. OA is wrong for insistence take x=1/2 (since 1/2 > -1) but the inequality does not hold true for this value, so how can the answer be correct.

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