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

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4/x < -1/3, what is the range of x  [#permalink]

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New post 24 Sep 2018, 00:50
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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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Re: 4/x < -1/3, what is the range of x  [#permalink]

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New post 24 Sep 2018, 01:18
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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

Answer (D)

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

Answer (D)
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Re: 4/x < -1/3, what is the range of x  [#permalink]

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New post 09 Oct 2018, 22:52
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.

Please hit Kudos if you liked this answer :)
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4/x < -1/3, what is the range of x  [#permalink]

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New post 09 Oct 2018, 22:53
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.

Please hit Kudos if you liked this answer :)
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Re: 4/x < -1/3, what is the range of x  [#permalink]

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New post 13 Oct 2018, 02:20
VeritasKarishma 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

Answer (D)

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

Answer (D)


Hi VeritasKarishma,
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
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kudos please if it helped you.

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4/x < -1/3, what is the range of x  [#permalink]

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New post 13 Oct 2018, 06:05
VeritasKarishma 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

Answer (D)

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

Answer (D)



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 ? :?
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4/x < -1/3, what is the range of x  [#permalink]

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New post 14 Oct 2018, 01:52
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 :lol:

\(\frac{x+12}{3x} <0\)

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
Wavy Curve Method.png [ 41.09 KiB | Viewed 420 times ]



- = \(x<0\)

+ = \(x>0\)

We need to find X less than ZERO, so I marked that range.
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4/x < -1/3, what is the range of x  [#permalink]

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New post Updated on: 14 Oct 2018, 02:17
tamal99 wrote:
VeritasKarishma 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

Answer (D)

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

Answer (D)


Hi VeritasKarishma,
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!
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Originally posted by T1101 on 14 Oct 2018, 02:12.
Last edited by T1101 on 14 Oct 2018, 02:17, edited 1 time in total.
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4/x < -1/3, what is the range of x  [#permalink]

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New post 14 Oct 2018, 02:15
tamal99 wrote:
VeritasKarishma 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

Answer (D)

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

Answer (D)


Hi VeritasKarishma,
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

https://study.com/academy/lesson/solvin ... ities.html

hopefully i answered your question correctly :)
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Re: 4/x < -1/3, what is the range of x  [#permalink]

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New post 15 Oct 2018, 04:41
1
tamal99 wrote:
VeritasKarishma 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

Answer (D)

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

Answer (D)


Hi VeritasKarishma,
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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

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Re: 4/x < -1/3, what is the range of x &nbs [#permalink] 15 Oct 2018, 04:41
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