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Re: GMAT quant DS question from GMAT club tests [#permalink]

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01 May 2013, 22:37

doe007 wrote:

vinaymimani wrote:

The question at NO point of time has asked "to find the rage in which all values of x would satisfy the inequality x / |x| < x", It just says which of the following MUST be TRUE. You don't assume that the given options encompass all the valid ranges. You find the valid ranges, then look for a common thread which binds them together and MUST BE TRUE, irrespective of the range(s).

When the question is on is MUST BE TRUE, all value in the range MUST satisfy the inequality.

I don't think so. If x>1, then which of the following must be true?

A.x=3 B.x is not equal to 2 C.x>2 D.x>9 E.x>-1

What will be your answer to the above question? BTW it is a question I just made now, so no Source. _________________

Re: GMAT quant DS question from GMAT club tests [#permalink]

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01 May 2013, 23:41

doe007 wrote:

vinaymimani wrote:

I don't think so. If x>1, then which of the following must be true?

A.x=3 B.x is not equal to 2 C.x>2 D.x>9 E.x>-1

What will be your answer to the above question?

Here you gave an example which has only one correct answer, that is option E. For all other options, we cannot prove the question from the given information.

But, in the original question, for the ranges shown in A and E, we cannot say that there is a value for which the inequality will not hold true. Thus, option A and option E are "MUST BE TRUE" conditions. There is no logic to say that there is a possibility for those ranges not satisfying the condition.

So this example is not analogous to the question in consideration.

What's the point here? Clearly that question has one and only one correct answer and there is no ambiguity or controversy in the OA of that question.

No question from GMAC will have more than one correct answer. From that point itself, the question of this topic has noncompliance.

By extending the same logic for agreeing with x > -1, I can as well say x > -5 which will MUST BE TRUE for all values satisfying the inequality in the question (all x's satisfying the given inequality MUST BE satisfying the condition x > -5). But, that cannot be the objective of any question.

I gave that question to demonstrate that a question can have both the correct options, which subscribe to the valid ranges and still not be the correct MUST-be-true answer. The question is ONLY asking which condition is MUST be true for x. BY no means, assume this condition to be a solution for the inequality. All it is asking is for the given inequality, which attribute of the variable x, MUST be TRUE. Also, the very reason that we are having this conversation/debate is the objective of the question.

I think carrying this conversation further will not do any more value addition. Ergo, I won't be posting anymore on this topic. We can both agree to disagree. _________________

Re: GMAT quant DS question from GMAT club tests [#permalink]

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02 May 2013, 01:03

I have deleted my earlier posts as there is point to agree with the OA -- ALL the values satisfying the given inequality also satisfies the range in option B. All other options are wrong as there are some x's which will satisfy x / |x| < x but will not be in that range. Examples are: A) x>1 is not true for x = -0.5 C) |x|<1 is not true for x = 2 D) |x|>1 is not true for x = -0.5 E) −1<x<0 is not true for x = 2 I admit that I read the question in incorrect way and from that I went to a different direction.

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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13 Jun 2013, 14:31

Can you tell me what I am doing wrong here?

There are two cases for x/|x|<x

Negative: x/-x<x -1<x x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x x is positive when x>0, so:

x>0, x>1

So why aren't the values considered: -1<x<1? Why do we not consider the values of x>0?

Thanks!

Bunuel wrote:

Marcab wrote:

If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

Explanations required for this one. Not convinced at all with the OA.

My range is -1<x<0 and x>1.

Notice that we are asked to find which of the options MUST be true, not COULD be true.

Let's see what ranges does \(\frac{x}{|x|}< x\) give for \(x\). Two cases:

If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);

If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).

So, \(\frac{x}{|x|}< x\) means that \(-1<x<0\) or \(x>1\).

Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\).

Answer: B.

As for other options:

A. \(x>1\). Not necessarily true since \(x\) could be -0.5; C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2; D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5; E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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13 Jun 2013, 14:35

Expert's post

WholeLottaLove wrote:

Can you tell me what I am doing wrong here?

There are two cases for x/|x|<x

Negative: x/-x<x -1<x x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x x is positive when x>0, so:

0<x<1

So why aren't the values considered: -1<x<1?

Thanks!

Bunuel wrote:

Marcab wrote:

If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

Explanations required for this one. Not convinced at all with the OA.

My range is -1<x<0 and x>1.

Notice that we are asked to find which of the options MUST be true, not COULD be true.

Let's see what ranges does \(\frac{x}{|x|}< x\) give for \(x\). Two cases:

If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);

If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).

So, \(\frac{x}{|x|}< x\) means that \(-1<x<0\) or \(x>1\).

Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\).

Answer: B.

As for other options:

A. \(x>1\). Not necessarily true since \(x\) could be -0.5; C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2; D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5; E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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14 Jun 2013, 07:31

Thank you for that link. Unfortunately I am still a bit lost.

When we take the negative case (x<0) we have -1<x therefore we join x<0 and -1<x thus: -1<x<0

When we take the positive case (x>0) we have 1<x I would assume we join the two cases thus: x>0, x>1

In your link you made reference to the AND/OR concepts and I am vaguely familiar with them but I am still a bit confused as to why we count the range of the negative cases (which makes sense because we are given that x falls within -1 and 0) but we don't count any # between zero and one but we do count numbers greater than one (is it because only values that fall within both x>0 and X>1 count? And if so, why?)

Thanks!

VeritasPrepKarishma wrote:

WholeLottaLove wrote:

Hi - I just fixed the question.

There are two cases for x/|x|<x

Negative: x/-x<x -1<x x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x x is positive when x>0, so:

x>0, x>1

So why aren't the values considered: -1<x<1? Why do we not consider the values of x>0?

Check out this link which discusses this question and this issue in detail:

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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14 Jun 2013, 08:32

Expert's post

WholeLottaLove wrote:

Thank you for that link. Unfortunately I am still a bit lost.

When we take the negative case (x<0) we have -1<x therefore we join x<0 and -1<x thus: -1<x<0

When we take the positive case (x>0) we have 1<x I would assume we join the two cases thus: x>0, x>1

In your link you made reference to the AND/OR concepts and I am vaguely familiar with them but I am still a bit confused as to why we count the range of the negative cases (which makes sense because we are given that x falls within -1 and 0) but we don't count any # between zero and one but we do count numbers greater than one (is it because only values that fall within both x>0 and X>1 count? And if so, why?)

Thanks!

VeritasPrepKarishma wrote:

WholeLottaLove wrote:

Hi - I just fixed the question.

There are two cases for x/|x|<x

Negative: x/-x<x -1<x x is negative when x<0, so:

-1<x<0

Positive: x/x<x 1<x x is positive when x>0, so:

x>0, x>1

So why aren't the values considered: -1<x<1? Why do we not consider the values of x>0?

Check out this link which discusses this question and this issue in detail:

In simple terms, this is what the concept is all about:

You have 5 numbers in your list: 1, 4, 7, 8, 11

Which of the following is true for all the numbers in your list? (A) Every number > 0 (B) Every number > 2 (C) Every number > 13

I hope you agree that answer is (A)

Every number is greater than 0. Do you have a problem that 2 is not a part of the numbers you have so how can (A) be the answer? Every number greater than 0 needn't be in your list. The question was what is true for all the numbers in the list.

This is the same concept.

You got the following ranges for x: -1 < x < 0, x > 1

Which of the following must be true? x > 1 x > -1

Do you see that for all values of x that you got, x must be greater than -1? _________________

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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01 Jul 2013, 13:55

If x doesn't equal 0 and (x)/|x|<x, which of the following must be true?

(I made the mistake of assuming that all values of x would make the statement true [i.e. what is the value of (x)/|x|<x] as opposed to looking what values of x are possible given the constraint of (x)/|x|<x)

So: (x)/|x|<x

x/x<x 1<x OR x/-x<x -1<x

(A) x>1 x is greater than 1. This lies entirely outside of the desired range. INVALID

(B) x>-1 As long as x>-1 it will fall within EITHER desired range of x>-1 or x>1 VALID

(C) |x|<1 x<1 OR x<-1 X may fall in the range (less than one) but it may also fall out of it (x<-1). INVALID

(D) |x|>1 x>1 OR x<-1 Both values of x fall entirely out of the desired range. INVALID

(E) -1<x<0

I'm still a little unsure as to why (E) isn't valid...I think it's because (E) states that x MUST be between -1 and 0 when in fact x simply has to be greater than -1.

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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02 Jul 2013, 10:20

Marcab wrote:

If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

m09 q22

Explanations required for this one. Not convinced at all with the OA.

My range is -1<x<0 and x>1.

can you please, explain me what does the must be true clause mean in this question ??

0.1 to 0.9999999999 none satisfy this relation and more over they are greater than -1 and when again we have an option called > 1, why do we choose this to be wrng ? do we have any value > 1 but still don't satisfy this question ??

please explain, i did understand from -1 to 0 there are values which accept this relation but accpeting this doesn't mean we can omit from 0 to -9 ...

Im confused, im out of nuts . please help me for this

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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02 Jul 2013, 10:47

1

This post received KUDOS

Expert's post

krrish wrote:

Marcab wrote:

If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|>1\)

(E) \(-1<x<0\)

m09 q22

Explanations required for this one. Not convinced at all with the OA.

My range is -1<x<0 and x>1.

can you please, explain me what does the must be true clause mean in this question ??

0.1 to 0.9999999999 none satisfy this relation and more over they are greater than -1 and when again we have an option called > 1, why do we choose this to be wrng ? do we have any value > 1 but still don't satisfy this question ??

please explain, i did understand from -1 to 0 there are values which accept this relation but accpeting this doesn't mean we can omit from 0 to -9 ...

Im confused, im out of nuts . please help me for this

Question: if \(-1<x<0\) or \(x>1\), then which of the following must be true? Notice that \(-1<x<0\) or \(x>1\) is given to be true: x is either from {-1, 0} or from {1, +infinity}

Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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15 Dec 2014, 22:51

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Re: If x#0 and x/|x|<x, which of the following must be true? [#permalink]

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08 Jul 2015, 11:02

Bunuel wrote:

Marcab wrote:

That's what I am saying. Since x cannot take this value then how can B be answer. How can x>-1 when 0<x<1 is not accepted?

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. \(-1<x<0\)

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.

Hope it's clear.

Hi Bunuel,

I am doing an analogy with data sufficiency problems, in this case I think that MUST BE problems are the exact opposite of DS problemas... For instance If you change the question stem to the place of a statement and also the statement to the place of the question stem (-1<x<0 and x>1 is converted in a statement; and x>-1 is converted in the question stem...) Something like this:

Is x>-1?

(1) -1<x<0 and x>1

Then it would be sufficient.

I do not know if I am wrong with this analogy of MUST BE problems with DS PROBLEMS. Could you help me?

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