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4/x < -1/3, what is the range of x

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Manager
Joined: 07 Aug 2018
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Location: United States (MA)
GMAT 1: 560 Q39 V28
GMAT 2: 670 Q48 V34
4/x < -1/3, what is the range of x  [#permalink]

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23 Sep 2018, 23:50
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$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

_________________

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8883
Location: Pune, India
Re: 4/x < -1/3, what is the range of x  [#permalink]

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24 Sep 2018, 00:18
1
1
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

Method 1:

$$\frac{4}{x}<-\frac{1}{3}$$

$$\frac{4}{x} + \frac{1}{3} < 0$$

$$\frac{x + 12}{3x} < 0$$

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Method 2:

$$\frac{4}{x}<-\frac{1}{3}$$

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

$$12 < -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x < -12$$
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

$$12 > -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x > -12$$

So here, -12 < x < 0

_________________

Karishma
Veritas Prep GMAT Instructor

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Joined: 09 Apr 2018
Posts: 31
GPA: 4
Re: 4/x < -1/3, what is the range of x  [#permalink]

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09 Oct 2018, 21:52
From the question stem we can conclude that x<0.

If we move x to the other side of the inequality, it changes sign (as we multiply with a negative):

4>$$-(\frac{1}{3})x$$

From here we can conclude that the maximum value x can be is -12, in order to make the inequality true.

Hence -12<x<0.

Intern
Joined: 09 Apr 2018
Posts: 31
GPA: 4
4/x < -1/3, what is the range of x  [#permalink]

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09 Oct 2018, 21:53
From the question stem we can conclude that x<0.

If we move x to the other side of the inequality, it changes sign (as we multiply with a negative):

4>$$-(\frac{1}{3})x$$

From here we can conclude that the maximum value x can be must be smaller than -12, in order to make the inequality true.

Hence -12<x<0.

Manager
Joined: 01 Jan 2018
Posts: 129
Re: 4/x < -1/3, what is the range of x  [#permalink]

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13 Oct 2018, 01:20
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

Method 1:

$$\frac{4}{x}<-\frac{1}{3}$$

$$\frac{4}{x} + \frac{1}{3} < 0$$

$$\frac{x + 12}{3x} < 0$$

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Method 2:

$$\frac{4}{x}<-\frac{1}{3}$$

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

$$12 < -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x < -12$$
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

$$12 > -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x > -12$$

So here, -12 < x < 0

I have doubt in second method (excuse me if very silly and basic).

Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.

Regards,
Tamal
_________________

kudos please if it helped you.

VP
Joined: 09 Mar 2016
Posts: 1285
4/x < -1/3, what is the range of x  [#permalink]

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13 Oct 2018, 05:05
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

Method 1:

$$\frac{4}{x}<-\frac{1}{3}$$

$$\frac{4}{x} + \frac{1}{3} < 0$$

$$\frac{x + 12}{3x} < 0$$

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Method 2:

$$\frac{4}{x}<-\frac{1}{3}$$

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

$$12 < -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x < -12$$
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

$$12 > -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x > -12$$

So here, -12 < x < 0

can anyone draw a wavy curvy line for ths expression ? -12 < x < 0

$$\frac{x + 12}{3x} < 0$$ in nummerator x = - 12 but what do with 3x in the denominator ?
VP
Joined: 09 Mar 2016
Posts: 1285
4/x < -1/3, what is the range of x  [#permalink]

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14 Oct 2018, 00:52
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

hello generis is my understanding correct ? took me a while to figure out how curvy method works

$$\frac{x+12}{3x} <0$$

To solve rational inequality:

- solve for when numerator equals 0

- find the values that make denominator equal 0

$$x+12=0$$
$$x = -12$$

$$3*0=0$$
$$x=0$$

So the critical points of boundaries are -12 and 0. Now to find the range draw a number line.

Attachment:

Wavy Curve Method.png [ 41.09 KiB | Viewed 673 times ]

- = $$x<0$$

+ = $$x>0$$

We need to find X less than ZERO, so I marked that range.
Manager
Joined: 07 Aug 2018
Posts: 110
Location: United States (MA)
GMAT 1: 560 Q39 V28
GMAT 2: 670 Q48 V34
4/x < -1/3, what is the range of x  [#permalink]

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Updated on: 14 Oct 2018, 01:17
tamal99 wrote:
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

Method 1:

$$\frac{4}{x}<-\frac{1}{3}$$

$$\frac{4}{x} + \frac{1}{3} < 0$$

$$\frac{x + 12}{3x} < 0$$

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Method 2:

$$\frac{4}{x}<-\frac{1}{3}$$

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

$$12 < -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x < -12$$
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

$$12 > -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x > -12$$

So here, -12 < x < 0

I have doubt in second method (excuse me if very silly and basic).

Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.

Regards,
Tamal

Hey tamal99

Eventough I am not Karishma I will try to answer your question.

In the second Case marked above we are trying to find values that are below zero (x<0), therefore x cannot be greater than 0! Maybe picking values makes it clearer:

Try to plug in numbers above 0. $$\frac{4}{x}<-\frac{1}{3}$$ No positive value will make this inequality true.

Therefore the range is -12<x<0.

Hope it makes sense!
_________________

Flashcards Quant + Verbal:https://gmatclub.com/forum/gmat-flashcards-108651.html
Thursdays with Ron:https://gmatclub.com/forum/manhattan-s-thursdays-with-ron-consolidated-video-index-223612.html#p2138753

Originally posted by T1101 on 14 Oct 2018, 01:12.
Last edited by T1101 on 14 Oct 2018, 01:17, edited 1 time in total.
VP
Joined: 09 Mar 2016
Posts: 1285
4/x < -1/3, what is the range of x  [#permalink]

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14 Oct 2018, 01:15
tamal99 wrote:
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

Method 1:

$$\frac{4}{x}<-\frac{1}{3}$$

$$\frac{4}{x} + \frac{1}{3} < 0$$

$$\frac{x + 12}{3x} < 0$$

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Method 2:

$$\frac{4}{x}<-\frac{1}{3}$$

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

$$12 < -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x < -12$$
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

$$12 > -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x > -12$$

So here, -12 < x < 0

I have doubt in second method (excuse me if very silly and basic).

Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.

Regards,
Tamal

Hi tamal99

i think we need to set denominator equal zero , pls check the link below

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8883
Location: Pune, India
Re: 4/x < -1/3, what is the range of x  [#permalink]

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15 Oct 2018, 03:41
1
tamal99 wrote:
T1101 wrote:
$$\frac{4}{x}<-\frac{1}{3}$$, what is the range of x?

A. $$-12<x$$

B. $$12>x>0$$

C. $$x>0$$

D. $$-12<x<0$$

E. $$x<0$$

Method 1:

$$\frac{4}{x}<-\frac{1}{3}$$

$$\frac{4}{x} + \frac{1}{3} < 0$$

$$\frac{x + 12}{3x} < 0$$

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Method 2:

$$\frac{4}{x}<-\frac{1}{3}$$

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

$$12 < -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x < -12$$
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

$$12 > -x$$

Multiply both sides by -1 (this will flip the inequality sign)

$$x > -12$$

So here, -12 < x < 0

I have doubt in second method (excuse me if very silly and basic).

Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.

Regards,
Tamal

In case 2, we have assumed that x is negative (hence have flipped the inequality sign later)
From the inequality, we got x > -12.
So x is negative but greater than -12. This gives us -12 < x < 0.
_________________

Karishma
Veritas Prep GMAT Instructor

Re: 4/x < -1/3, what is the range of x   [#permalink] 15 Oct 2018, 03:41
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