It is currently 16 Oct 2017, 18:58

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If 4/x < 1/3 , what is the possible range of values of x?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 10 Nov 2010
Posts: 247

Kudos [?]: 401 [0], given: 22

Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE: Information Technology (Computer Software)
If 4/x <1/3, what is the possible range of values for x? [#permalink]

### Show Tags

07 Apr 2011, 05:01
we can't change the inequality when we have -ve nd RHS +ve in reciprocal

10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases

Case 1 x is +ve

x>12

Case 2 when we consider x as -ve we will have Left hand side -ve but right hand side +ve so in that case we cnt flip the inequality.
But OA is showing both x>12 nd x<12

Pls comment which condition is wrong.

thanks
_________________

The proof of understanding is the ability to explain it.

Kudos [?]: 401 [0], given: 22

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1964

Kudos [?]: 2048 [0], given: 376

Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink]

### Show Tags

07 Apr 2011, 05:39
GMATD11 wrote:
we can't change the inequality when we have -ve nd RHS +ve in reciprocal

10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases

Case 1 x is +ve

x>12

Case 2 when we consider x as -ve we will have Left hand side -ve but right hand side +ve so in that case we cnt flip the inequality.
But OA is showing both x>12 nd x<12

Pls comment which condition is wrong.

thanks

4/x <1/3
12/x-1<0
(12-x)/x < 0

Means either numerator or denominators is -ve:

Case I:
If Denominator is -ve.
x<0 ------1

Numerator must be +ve
12-x > 0
-x > -12
x< 12--------------2

In equation 1 and 2, 1 is more restrictive:
x<0

Case II:
If Denominator is +ve.
x>0 ------3

Numerator must be -ve
12-x < 0
-x < -12
x > 12

In equation 3 and 4, 4 is more restrictive:
x>12

Thus;
complete Range of x:

x<0 or x>12
_________________

Kudos [?]: 2048 [0], given: 376

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1339

Kudos [?]: 1951 [0], given: 6

Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink]

### Show Tags

07 Apr 2011, 13:19
GMATD11 wrote:
we can't change the inequality when we have -ve nd RHS +ve in reciprocal

I don't understand what you mean by this?

GMATD11 wrote:
10) If 4/x <1/3, what is the possible range of values for x?

We need to consider 2 cases

You do need to consider 2 cases, but you only need to spend any time on one of them. We know:

4/x < 1/3

This will clearly be true if x is negative, since then the left side is negative, and the right side is positive, and negative numbers are certainly smaller than positive ones. So whenever x < 0, the inequality is true.

Now for the second case: if x > 0, we can multiply both sides by x without needing to worry about reversing the inequality:

4 < x/3
12 < x

So either x < 0, or 12 < x.
_________________

GMAT Tutor in Toronto

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

Kudos [?]: 1951 [0], given: 6

Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 872

Kudos [?]: 396 [0], given: 123

Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink]

### Show Tags

07 Apr 2011, 13:38
Brilliant !

And when I know the sign of x > 0 then taking the reciprocal of left hand side Vs the right hand side will reverse the direction of the inequality

4/x < 1/3 and x > 0
or x/4 > 3 (reverse the direction)
or x > 12

Kudos [?]: 396 [0], given: 123

Intern
Joined: 06 May 2011
Posts: 13

Kudos [?]: 1 [0], given: 0

Re: we can't change the inequality when we have LHS -ve nd RHS [#permalink]

### Show Tags

29 Aug 2011, 19:13
GMATD11 wrote:
10) If 4/x <1/3, what is the possible range of values for x?

4/x < 1/3 => 12/x < 1
For the fraction 12/x to be <1 ,

Scenario1 : x has to be negative.
Scenario2 : The denominator should be greater than 12

=> x < 0 or x > 12 .

Please correct me if i am wrong.

Kudos [?]: 1 [0], given: 0

Intern
Status: :O
Joined: 11 Jul 2012
Posts: 43

Kudos [?]: 49 [0], given: 27

GMAT 1: 670 Q48 V35
If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

25 Aug 2012, 01:47
3
This post was
BOOKMARKED
If $$\frac{4}{x} <\frac{1}{3}$$ , what is the possible range of values of x?

The question is from an MGMAT guide and the solution provided multiplies the inequality by x and flips the sign to arrive at the solution x < 0 or x > 12 . However the way I naturally would solve the given equation is:
$$\frac{4}{x} - \frac{1}{3} < 0$$
$$\frac{(12 - x)}{(3x)} < 0$$

Two cases:
a) $$12 - x < 0$$ &$$3x > 0$$ :: x > 12 & x > 0 therefore x > 12
b) $$12 - x > 0$$ & $$3x < 0$$ :: x < 12 & x < 0 therefore x < 12
Hence my solution: x < 12 or x > 12

Can someone kindly explain where I went wrong here?
_________________

http://gmatclub.com/forum/disaster-spelled-in-670-ways-10-not-to-do-s-139230.html

Kudos [?]: 49 [0], given: 27

Math Expert
Joined: 02 Sep 2009
Posts: 41875

Kudos [?]: 128459 [1], given: 12173

Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

25 Aug 2012, 02:46
1
KUDOS
Expert's post
harshvinayak wrote:
If $$\frac{4}{x} <\frac{1}{3}$$ , what is the possible range of values of x?

The question is from an MGMAT guide and the solution provided multiplies the inequality by x and flips the sign to arrive at the solution x < 0 or x > 12 . However the way I naturally would solve the given equation is:
$$\frac{4}{x} - \frac{1}{3} < 0$$
$$\frac{(12 - x)}{(3x)} < 0$$

Two cases:
a) $$12 - x < 0$$ &$$3x > 0$$ :: x > 12 & x > 0 therefore x > 12
b) $$12 - x > 0$$ & $$3x < 0$$ :: x < 12 & x < 0 therefore x < 12
Hence my solution: x < 12 or x > 12

Can someone kindly explain where I went wrong here?

When we consider the case when $$12 - x > 0$$ and $$3x < 0$$, we have: $$x<12$$and $$x<0$$, therefore $$x<0$$.

Solving inequalities:
inequalities-trick-91482.html (check this one first)
x2-4x-94661.html#p731476
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
_________________

Kudos [?]: 128459 [1], given: 12173

Director
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1057 [2], given: 43

WE: Science (Education)
Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

25 Aug 2012, 03:50
2
KUDOS
harshvinayak wrote:
If $$\frac{4}{x} <\frac{1}{3}$$ , what is the possible range of values of x?

The question is from an MGMAT guide and the solution provided multiplies the inequality by x and flips the sign to arrive at the solution x < 0 or x > 12 . However the way I naturally would solve the given equation is:
$$\frac{4}{x} - \frac{1}{3} < 0$$
$$\frac{(12 - x)}{(3x)} < 0$$

Two cases:
a) $$12 - x < 0$$ &$$3x > 0$$ :: x > 12 & x > 0 therefore x > 12
b) $$12 - x > 0$$ & $$3x < 0$$ :: x < 12 & x < 0 therefore x < 12
Hence my solution: x < 12 or x > 12

Can someone kindly explain where I went wrong here?

Another way to solve this type of inequality:

From $$\frac{4}{x}<\frac{1}{3}$$ it follows that $$x\neq0$$, therefore we can multiply both sides by $$3x^2$$, which is positive.
We obtain $$12x<x^2$$ or $$x^2-12x>0.$$
Now, imagine the graph of the quadratic function, $$y=12x^2-x$$, which is an upward parabola. See the attached drawing.
This parabola intercepts the X-axis at $$x=0$$ and $$x=12,$$ the two "arms" are above the X-axis, meaning the values of $$y$$ are positive when $$x<12$$ or $$x>0$$, and the values of $$y$$ are negative (graph under the X-axis) when $$0<x<12.$$

Hence the solution $$x<0$$ or $$x>12.$$

Note: Just remember the shape of the parabola, then you can easily deduce the sign of the quadratic function $$y=x^2+bx+c.$$ If the quadratic equation $$x^2+bx+c=0$$ has two roots $$x_1<x_2$$ then $$y<0$$ (negative) between the two roots and $$y>0$$ (positive) outside the roots.
Or succinctly, $$y>0$$ if $$x_1<x<x_2$$ and $$y>0$$ if $$x<x_1$$ or $$x>x_2.$$
Attachments

ParabolaUp.jpg [ 8.97 KiB | Viewed 5141 times ]

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Kudos [?]: 1057 [2], given: 43

Intern
Status: :O
Joined: 11 Jul 2012
Posts: 43

Kudos [?]: 49 [0], given: 27

GMAT 1: 670 Q48 V35
Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

25 Aug 2012, 04:06
Ah! thanks for the clarification Bunuel. I guess staying up all night took its toll. I had to reread and reread and reread and...

finally what I did was (per the first link you suggested) ::

dumping the two cases,
$$\frac{(12-x)}{(3x)} < 0$$
$$\frac{(x-12)}{(3x)} > 0$$
roots: 12 and 0
and then I could ride the waves of + - + ...... weeeeeeeeeee.... weeeeeeeeeee... and weeeeeeeeeeee

Thanks again for helping out Bunuel ..
_________________

http://gmatclub.com/forum/disaster-spelled-in-670-ways-10-not-to-do-s-139230.html

Kudos [?]: 49 [0], given: 27

Intern
Status: :O
Joined: 11 Jul 2012
Posts: 43

Kudos [?]: 49 [0], given: 27

GMAT 1: 670 Q48 V35
Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

25 Aug 2012, 04:15
Thanks EvaJager you guys are awesome \m/

PS: I have just started out with my prep and have a lot to learn (and ask ) in little less than 3 weeks.
_________________

http://gmatclub.com/forum/disaster-spelled-in-670-ways-10-not-to-do-s-139230.html

Kudos [?]: 49 [0], given: 27

Manager
Joined: 05 Jul 2012
Posts: 77

Kudos [?]: 51 [0], given: 8

Location: India
Concentration: Finance, Strategy
GMAT Date: 09-30-2012
GPA: 3.08
WE: Engineering (Energy and Utilities)
Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

26 Aug 2012, 11:24
harshvinayak wrote:
Thanks EvaJager you guys are awesome \m/

PS: I have just started out with my prep and have a lot to learn (and ask ) in little less than 3 weeks.

One way to understand this is that Sign of $$\frac{x-a}{x-b}$$ is same as the sign of (x-a)(x-b)
and for (x-a)(x-b) always remember that it will be negative between a & b and positive outside a & b

Kudos [?]: 51 [0], given: 8

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7667

Kudos [?]: 17315 [2], given: 232

Location: Pune, India
Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

28 Aug 2012, 21:58
2
KUDOS
Expert's post
2
This post was
BOOKMARKED
harshvinayak wrote:
If $$\frac{4}{x} <\frac{1}{3}$$ , what is the possible range of values of x?

The question is from an MGMAT guide and the solution provided multiplies the inequality by x and flips the sign to arrive at the solution x < 0 or x > 12 . However the way I naturally would solve the given equation is:
$$\frac{4}{x} - \frac{1}{3} < 0$$
$$\frac{(12 - x)}{(3x)} < 0$$

Two cases:
a) $$12 - x < 0$$ &$$3x > 0$$ :: x > 12 & x > 0 therefore x > 12
b) $$12 - x > 0$$ & $$3x < 0$$ :: x < 12 & x < 0 therefore x < 12
Hence my solution: x < 12 or x > 12

Can someone kindly explain where I went wrong here?

Check out these posts. They discuss how to solve inequalities efficiently (using the wave) and how to handle various complications that can arise in a question.

http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
http://www.veritasprep.com/blog/2012/07 ... qualities/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17315 [2], given: 232 Intern Status: :O Joined: 11 Jul 2012 Posts: 43 Kudos [?]: 49 [0], given: 27 GMAT 1: 670 Q48 V35 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] ### Show Tags 28 Aug 2012, 22:16 WOW ! Thanks Karishma, thats one awesome collection of inequalities stuff. A kudos alone wouldnt have cut it _________________ http://gmatclub.com/forum/disaster-spelled-in-670-ways-10-not-to-do-s-139230.html GMAT Demystified: http://gmatclub.com/forum/gmat-demystified-great-ebook-to-get-started-with-download-140836.html Tense Tutorial: http://gmatclub.com/forum/uber-awesome-resource-on-tenses-download-140837.html CR Question Bank (LSAT): http://gmatclub.com/forum/critical-reasoning-question-bank-download-140838.html Kudos [?]: 49 [0], given: 27 Manager Joined: 12 Feb 2012 Posts: 130 Kudos [?]: 59 [0], given: 28 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] ### Show Tags 12 Jul 2013, 14:15 Sorry to bring back this old post. I know how to solve this problems using the methods mentioned above. But I tried solving it another way and that method gave me a weird solution. Let say you wanted to solve the problem this way: $$\frac{4}{x} <\frac{1}{3}$$ Since x cannot be 0. Lets look at the positive/negative scenarios If x>0, then $$12 <x$$ if x<0, then $$12 >x$$ but x cannot be both negative and greater than 12. So this is a contradiction. Hence the range is $$12 <x$$. But this is incomplete. The range of x for this inquality is $$x<0$$ OR $$12 <x$$. So what am I doing wrong? Kudos [?]: 59 [0], given: 28 Math Expert Joined: 02 Sep 2009 Posts: 41875 Kudos [?]: 128459 [0], given: 12173 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] ### Show Tags 12 Jul 2013, 14:57 alphabeta1234 wrote: Sorry to bring back this old post. I know how to solve this problems using the methods mentioned above. But I tried solving it another way and that method gave me a weird solution. Let say you wanted to solve the problem this way: $$\frac{4}{x} <\frac{1}{3}$$ Since x cannot be 0. Lets look at the positive/negative scenarios If x>0, then $$12 <x$$ if x<0, then $$12 >x$$ but x cannot be both negative and greater than 12. So this is a contradiction. Hence the range is $$12 <x$$. But this is incomplete. The range of x for this inquality is $$x<0$$ OR $$12 <x$$. So what am I doing wrong? This is a correct approach but you made a mistake in the second case. If x<0, then 12>x (x is less than 12) --> intersection is x<0. So, the inequality holds true for x>12 (from the first case) and x<0 (from the second case). Hope it helps. _________________ Kudos [?]: 128459 [0], given: 12173 Manager Joined: 11 Jan 2011 Posts: 69 Kudos [?]: 25 [0], given: 3 GMAT 1: 680 Q44 V39 GMAT 2: 710 Q48 V40 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] ### Show Tags 24 Oct 2013, 18:15 Gotta bump this back up because I'm stuck on what should be a very simple inequalities problem. Given: $$\frac{4}{x}$$ < -$$\frac{1}{3}$$, I simplified the equation to: 12 < -x by cross-multiplying Now, we have 2 scenarios: 1. x > 0 : no sign changes in 12 < -x, so x < -12. However, since we know x > 0, this scenario is impossible. 2. x < 0 : 12 < -x should become 12 < -(-x) so wouldn't this just be 12 < x? However, given x < 0, this scenario doesn't appear possible either. Can someone please point out the obvious? It's driving me crazy... Kudos [?]: 25 [0], given: 3 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7667 Kudos [?]: 17315 [0], given: 232 Location: Pune, India Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] ### Show Tags 24 Oct 2013, 20:38 NvrEvrGvUp wrote: Gotta bump this back up because I'm stuck on what should be a very simple inequalities problem. Given: $$\frac{4}{x}$$ < -$$\frac{1}{3}$$, I simplified the equation to: 12 < -x by cross-multiplying Now, we have 2 scenarios: 1. x > 0 : no sign changes in 12 < -x, so x < -12. However, since we know x > 0, this scenario is impossible. 2. x < 0 : 12 < -x should become 12 < -(-x) so wouldn't this just be 12 < x? However, given x < 0, this scenario doesn't appear possible either. Can someone please point out the obvious? It's driving me crazy... First of all, the actual question is $$\frac{4}{x}$$ < $$\frac{1}{3}$$ (there is no negative with 1/3) Also, you know what you have to do but you probably do not understand why you have to do it. That is why you are facing problem in this question. Given: $$\frac{4}{x}$$ < -$$\frac{1}{3}$$, I simplified the equation to: 12 < -x by cross-multiplying There is a problem here. You don't cross multiply and then take cases. You take cases and then cross multiply. Why? Because you CANNOT cross multiply till you know (or assume) the sign of x. The result of the cross multiplication depends on whether x is positive or negative. So you need to take cases and then cross multiply. Case 1: x > 0 12 < x Case 2: x < 0 12 > x (note that the sign has flipped here because you are multiplying by a negative number) x should be less than 12 AND less than 0 so the range in x < 0. Hence, two cases: x > 12 or x < 0. Also because x is negative, you cannot just multiply it by -1 to make -x = x. That is certainly not correct. -x is positive and x is negative. They are not equal if x is not 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17315 [0], given: 232

Manager
Joined: 11 Jan 2011
Posts: 69

Kudos [?]: 25 [0], given: 3

GMAT 1: 680 Q44 V39
GMAT 2: 710 Q48 V40
Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]

### Show Tags

25 Oct 2013, 11:44
VeritasPrepKarishma wrote:
NvrEvrGvUp wrote:
Gotta bump this back up because I'm stuck on what should be a very simple inequalities problem.

Given: $$\frac{4}{x}$$ < -$$\frac{1}{3}$$, I simplified the equation to: 12 < -x by cross-multiplying

Now, we have 2 scenarios:

1. x > 0 : no sign changes in 12 < -x, so x < -12. However, since we know x > 0, this scenario is impossible.

2. x < 0 : 12 < -x should become 12 < -(-x) so wouldn't this just be 12 < x? However, given x < 0, this scenario doesn't appear possible either.

Can someone please point out the obvious? It's driving me crazy...

First of all, the actual question is $$\frac{4}{x}$$ < $$\frac{1}{3}$$ (there is no negative with 1/3)

Also, you know what you have to do but you probably do not understand why you have to do it. That is why you are facing problem in this question.

Given: $$\frac{4}{x}$$ < -$$\frac{1}{3}$$, I simplified the equation to: 12 < -x by cross-multiplying

There is a problem here. You don't cross multiply and then take cases. You take cases and then cross multiply. Why? Because you CANNOT cross multiply till you know (or assume) the sign of x. The result of the cross multiplication depends on whether x is positive or negative. So you need to take cases and then cross multiply.

Case 1: x > 0
12 < x

Case 2: x < 0
12 > x (note that the sign has flipped here because you are multiplying by a negative number)
x should be less than 12 AND less than 0 so the range in x < 0.

Hence, two cases: x > 12 or x < 0.

Also because x is negative, you cannot just multiply it by -1 to make -x = x. That is certainly not correct. -x is positive and x is negative. They are not equal if x is not 0.

Hi Karishma,

That was my mistake. There is a problem with a negative -1/3 and one without - I didn't realize this one was referencing the one with the positive 1/3.

I'll search for the -1/3 problem and explanation.

Thanks.

Kudos [?]: 25 [0], given: 3

Intern
Joined: 03 Oct 2015
Posts: 30

Kudos [?]: 7 [0], given: 95

Location: India
Concentration: Marketing, Strategy
Schools: NUS '18
GPA: 2.08
WE: Business Development (Internet and New Media)
What is the possible range of value for x? [#permalink]

### Show Tags

12 Dec 2015, 12:52
4/x < 1/3, what is the possible range of values of x??

Kudos [?]: 7 [0], given: 95

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2677

Kudos [?]: 1721 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: What is the possible range of value for x? [#permalink]

### Show Tags

12 Dec 2015, 17:05
laveen_g wrote:
4/x < 1/3, what is the possible range of values of x??

Follow posting guidelines (link in my signatures). Post complete question and all options. Do post the official answer along with 3 tags, 1 each for source, difficulty and topic discussed.

As for this question,

You are asked for the range of values for 4/x < 1/3 ---> 4/x-1/3 < 0 ---> (12-x)/3x < 0 ---> (x-12)/x > 0 ---> for this to be true, the range is x<0 or x>12.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1721 [0], given: 792

Re: What is the possible range of value for x?   [#permalink] 12 Dec 2015, 17:05

Go to page    1   2    Next  [ 33 posts ]

Display posts from previous: Sort by