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Manager  G
Joined: 07 Aug 2018
Posts: 108
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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Difficulty:   55% (hard)

Question Stats: 57% (01:28) correct 43% (01:27) wrong based on 167 sessions

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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. $$x<0$$

_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: 4/x < -1/3, what is the range of x  [#permalink]

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

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

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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 Manager  G
Joined: 01 Jan 2018
Posts: 154
Re: 4/x < -1/3, what is the range of x  [#permalink]

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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  D
Joined: 09 Mar 2016
Posts: 1230
4/x < -1/3, what is the range of x  [#permalink]

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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  D
Joined: 09 Mar 2016
Posts: 1230
4/x < -1/3, what is the range of x  [#permalink]

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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 900 times ]

- = $$x<0$$

+ = $$x>0$$

We need to find X less than ZERO, so I marked that range.
Manager  G
Joined: 07 Aug 2018
Posts: 108
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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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!
_________________

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.
VP  D
Joined: 09 Mar 2016
Posts: 1230
4/x < -1/3, what is the range of x  [#permalink]

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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

hopefully i answered your question correctly Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9704
Location: Pune, India
Re: 4/x < -1/3, what is the range of x  [#permalink]

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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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Re: 4/x < -1/3, what is the range of x   [#permalink] 15 Oct 2018, 04:41
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