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

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20 Dec 2011, 07:29

IanStewart wrote:

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.

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?

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

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13 Jan 2012, 12:17

1

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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
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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.

Answer: B.

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

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.

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.

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

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27 May 2012, 05:05

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.

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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29 May 2012, 02:30

Hi bunuel, Nice explanation..however, I was trying to do find the ranges and got confused! Your way is clear to me when you take x<0 then x/-x<x =>-1<x ...however, if I say....(if I cross multiply) x>-x2 then x2+x>0 => x(x+1) >0...thus, given then it is >; x<-1 and x>0...since the x<0...x<-1 holds... what am I doing wrong?

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

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30 May 2012, 03:21

Bunuel wrote:

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.

bunuel...i have the same question as vibhav ... what if the value of x lies in the range of 0<x<1 where the function is not valid. i am not getting the solution. any expert comment please..
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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30 May 2012, 04:54

321kumarsushant wrote:

Bunuel wrote:

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.

bunuel...i have the same question as vibhav ... what if the value of x lies in the range of 0<x<1 where the function is not valid. i am not getting the solution. any expert comment please..

Hi 321kumarsushant,

The solution states that x is only valid for -1<x<0 and x>1. So, we already know that x is not defined at all in the range 0<x<1.

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

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30 May 2012, 05:10

pavanpuneet wrote:

Hi bunuel, Nice explanation..however, I was trying to do find the ranges and got confused! Your way is clear to me when you take x<0 then x/-x<x =>-1<x ...however, if I say....(if I cross multiply) x>-x2 then x2+x>0 => x(x+1) >0...thus, given then it is >; x<-1 and x>0...since the x<0...x<-1 holds... what am I doing wrong?

Hi pavanpuneet,

We cannot divide both sides both sides of the statement by x, as we don't know whether a is positive or negative. Hence, we have to take the other approach.

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

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 7612 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 7608 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.
_________________

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

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13 Jun 2012, 13:04

VeritasPrepKarishma wrote:

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

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

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.

But wat if x will be zero?? that will be infinitive..!!.. I think answer a is correct.. m still confused .
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