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

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If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  25 Aug 2012, 00:47
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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?
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  25 Aug 2012, 01:46
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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.
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  25 Aug 2012, 02:50
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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 2357 times ]

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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  25 Aug 2012, 03:06
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 ..
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  25 Aug 2012, 03:15
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.
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  26 Aug 2012, 10:24
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
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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  28 Aug 2012, 20:58
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?

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/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Status: :O Joined: 11 Jul 2012 Posts: 43 GMAT 1: 670 Q48 V35 Followers: 1 Kudos [?]: 27 [0], given: 27 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] 28 Aug 2012, 21:16 WOW ! Thanks Karishma, thats one awesome collection of inequalities stuff. A kudos alone wouldnt have cut it _________________ disaster-spelled-in-670-ways-10-not-to-do-s-139230.html GMAT Demystified: gmat-demystified-great-ebook-to-get-started-with-download-140836.html Tense Tutorial: uber-awesome-resource-on-tenses-download-140837.html CR Question Bank (LSAT): critical-reasoning-question-bank-download-140838.html Manager Joined: 12 Feb 2012 Posts: 108 Followers: 1 Kudos [?]: 21 [0], given: 28 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] 12 Jul 2013, 13:15 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? Math Expert Joined: 02 Sep 2009 Posts: 29186 Followers: 4737 Kudos [?]: 50073 [0], given: 7527 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] 12 Jul 2013, 13:57 Expert's post 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. _________________ Manager Joined: 11 Jan 2011 Posts: 71 GMAT 1: 680 Q44 V39 GMAT 2: 710 Q48 V40 Followers: 0 Kudos [?]: 8 [0], given: 3 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] 24 Oct 2013, 17:15 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... Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5866 Location: Pune, India Followers: 1481 Kudos [?]: 7979 [0], given: 190 Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink] 24 Oct 2013, 19:38 Expert's post 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

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Re: If 4/x < 1/3 , what is the possible range of values of x? [#permalink]  25 Oct 2013, 10:44
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.
Re: If 4/x < 1/3 , what is the possible range of values of x?   [#permalink] 25 Oct 2013, 10:44
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