For what range of values of 'x' will the inequality 15x - 2/x > 1? : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 13:22

### 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

# For what range of values of 'x' will the inequality 15x - 2/x > 1?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 83
Followers: 2

Kudos [?]: 37 [3] , given: 2

For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Sep 2010, 06:02
3
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

44% (02:47) correct 56% (01:58) wrong based on 363 sessions

### HideShow timer Statistics

For what range of values of 'x' will the inequality 15x - 2/x > 1?

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

[Reveal] Spoiler:
D
[Reveal] Spoiler: OA
Manager
Joined: 25 Jun 2010
Posts: 91
Followers: 1

Kudos [?]: 35 [2] , given: 0

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Sep 2010, 07:17
2
KUDOS
15x - 2/x > 1

Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5

Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.
Manager
Joined: 16 Jun 2010
Posts: 188
Followers: 2

Kudos [?]: 85 [0], given: 5

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Oct 2010, 12:15
I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?
_________________

Please give me kudos, if you like the above post.
Thanks.

Manager
Joined: 25 Jun 2010
Posts: 91
Followers: 1

Kudos [?]: 35 [1] , given: 0

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Oct 2010, 12:23
1
KUDOS
Whenever you multiply an inequality by X , you should check for the -ve and +ve values of X, as it inverts the sign.
Manager
Joined: 16 Jun 2010
Posts: 188
Followers: 2

Kudos [?]: 85 [0], given: 5

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Oct 2010, 12:26
Thanks for the lightning quick reply, I was not aware of this.
_________________

Please give me kudos, if you like the above post.
Thanks.

Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93659 [4] , given: 10583

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Oct 2010, 12:36
4
KUDOS
Expert's post
3
This post was
BOOKMARKED
Eden wrote:
For what range of values of 'x' will the inequality 15x - 2/x > 1?

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

[Reveal] Spoiler:
D

devashish wrote:
I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?

anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so.

One could also do as follows:

$$15x-\frac{2}{x}>1$$ --> $$\frac{15x^2-x-2}{x}>0$$ --> $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$:

Both denominator and nominator positive --> $$x>\frac{2}{5}$$;
Both denominator and nominator negative --> $$-\frac{1}{3}<x<0$$;

So inequality holds true in the ranges: $$-\frac{1}{3}<x<0$$ and $$x>\frac{2}{5}$$.

_________________
Manager
Joined: 16 Jun 2010
Posts: 188
Followers: 2

Kudos [?]: 85 [0], given: 5

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Oct 2010, 12:45
Bunuel wrote:
One could also do as follows:

$$15x-\frac{2}{x}>1$$ --> $$\frac{15x^2-x-2}{x}>0$$ --> $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$:

Both denominator and nominator positive --> $$x>\frac{2}{5}$$;
Both denominator and nominator negative --> $$-\frac{1}{3}<x<0$$;

So inequality holds true in the ranges: $$-\frac{1}{3}<x<0$$ and $$x>\frac{2}{5}$$.

Bunuel you are awesome
_________________

Please give me kudos, if you like the above post.
Thanks.

Manager
Joined: 25 Jun 2010
Posts: 91
Followers: 1

Kudos [?]: 35 [0], given: 0

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

03 Oct 2010, 18:00
Nice alternate solution.

$$15x^2-x-2 -> x^2 - x/15 - 2/15 -> x^2 + x/3 - 2x/5 - 2/15 -> (x+1/3)(x-2/5)$$

This works well, however I would suggest to rely on quadratic equation formula: $$-b(+-)\sqrt{D}/2a$$, as you don't need to think, how to factorize an equation?

Try yourself to factorize the equation above vs using the formula (Unfortunately, if the equation doesn't has a fractional value (which I expect rarely could be the case in GMAT) , for example when D in above formula is irrational, then you might be trying to factorize the fraction forever.)
Current Student
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 991
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Followers: 165

Kudos [?]: 1475 [0], given: 227

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

12 Aug 2013, 10:48
What I have learnt, if x is in denom, take LCM after bringing all element at one side.

If we have more than 2 roots, then we can use following method to know the range.

Attachment:

Solution.jpg [ 16.75 KiB | Viewed 3787 times ]

_________________

Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Manager
Joined: 30 May 2013
Posts: 190
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 64 [0], given: 72

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

14 Aug 2013, 22:13
hi,

can any one explain this with graphical explanation??

Rrsnathan.
Senior Manager
Joined: 10 Mar 2013
Posts: 290
GMAT 1: 620 Q44 V31
GMAT 2: 690 Q47 V37
GMAT 3: 610 Q47 V28
GMAT 4: 700 Q50 V34
GMAT 5: 700 Q49 V36
GMAT 6: 690 Q48 V35
GMAT 7: 750 Q49 V42
GMAT 8: 730 Q50 V39
Followers: 11

Kudos [?]: 100 [0], given: 2405

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

26 Nov 2013, 19:46
anshumishra wrote:
15x - 2/x > 1

Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5

Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.
Intern
Joined: 24 Feb 2014
Posts: 4
Followers: 0

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

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

02 Mar 2014, 00:22
TooLong150 wrote:
anshumishra wrote:
15x - 2/x > 1

Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5

Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

I have the same doubt. Shouldn't the answer be D?
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93659 [0], given: 10583

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

02 Mar 2014, 04:03
siriusblack1106 wrote:
TooLong150 wrote:
anshumishra wrote:
15x - 2/x > 1

Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5

Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

I have the same doubt. Shouldn't the answer be D?

____________
_________________
Moderator
Joined: 20 Dec 2013
Posts: 189
Location: United States (NY)
GMAT 1: 640 Q44 V34
GMAT 2: 710 Q48 V40
GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)
Followers: 7

Kudos [?]: 66 [0], given: 71

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

02 Mar 2014, 17:02
TooLong150 wrote:
anshumishra wrote:
15x - 2/x > 1

Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5

Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

After you determine boundary points, you still need to test the potential +ve/-ve ranges per the inequality
_________________
Intern
Joined: 13 Dec 2013
Posts: 40
Schools: Fuqua (I), AGSM '16
GMAT 1: 620 Q42 V33
Followers: 2

Kudos [?]: 17 [0], given: 10

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

07 Jun 2014, 16:21
Bunuel wrote:
Eden wrote:
For what range of values of 'x' will the inequality 15x - 2/x > 1?

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

[Reveal] Spoiler:
D

devashish wrote:
I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?

anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so.

One could also do as follows:

$$15x-\frac{2}{x}>1$$ --> $$\frac{15x^2-x-2}{x}>0$$ --> $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$:

Both denominator and nominator positive --> $$x>\frac{2}{5}$$;
Both denominator and nominator negative --> $$-\frac{1}{3}<x<0$$;

So inequality holds true in the ranges: $$-\frac{1}{3}<x<0$$ and $$x>\frac{2}{5}$$.

Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or -ive.
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93659 [0], given: 10583

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

08 Jun 2014, 02:35
Enael wrote:
Bunuel wrote:
Eden wrote:
For what range of values of 'x' will the inequality 15x - 2/x > 1?

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

[Reveal] Spoiler:
D

devashish wrote:
I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?

anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so.

One could also do as follows:

$$15x-\frac{2}{x}>1$$ --> $$\frac{15x^2-x-2}{x}>0$$ --> $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$:

Both denominator and nominator positive --> $$x>\frac{2}{5}$$;
Both denominator and nominator negative --> $$-\frac{1}{3}<x<0$$;

So inequality holds true in the ranges: $$-\frac{1}{3}<x<0$$ and $$x>\frac{2}{5}$$.

Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or -ive.

I didn't multiply.

$$15x-\frac{2}{x}>1$$ --> subtract 1: $$15x-\frac{2}{x}-1>0$$ --> put in common denominator: $$\frac{15x^2-x-2}{x}>0$$ --> factorize: $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$.

Hope it's clear.
_________________
Manager
Joined: 23 May 2013
Posts: 167
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
Followers: 2

Kudos [?]: 70 [0], given: 39

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

10 Jun 2014, 07:51
anshumishra wrote:
15x - 2/x > 1

Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5

Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

A graphical approach supplements this solution quite well: The roots are the blue dots, the yellow line separates the two cases, and the red lines are where the inequality holds true.

We assume two cases: x>0 and x<0.

For the first case, x>0, we want to see where 15x^2 -x-2 >0. Solving for the roots using the quadratic formula, we find roots when x = 2/5 and x = -1/3. Thus, for x>0, 15x^2 -x-2 >0 when x>2/5.

For the second case, x<0, we want to see where 15x^2 -x-2 <0. Again, using the roots, we know that the 15x^2 -x-2 < 0 when x > -1/3. We also assumed x<0 so we must include that in our solution.

Thus we have x>2/5 and -1/3<x<0. Answer: D
Attachments

solution.png [ 36.79 KiB | Viewed 2630 times ]

Intern
Joined: 22 Feb 2014
Posts: 30
Followers: 0

Kudos [?]: 10 [0], given: 14

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

15 Jul 2014, 04:31
Bunuel wrote:
Eden wrote:
For what range of values of 'x' will the inequality 15x - 2/x > 1?

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

[Reveal] Spoiler:
D

devashish wrote:
I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?

anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so.

One could also do as follows:

$$15x-\frac{2}{x}>1$$ --> $$\frac{15x^2-x-2}{x}>0$$ --> $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$:

Both denominator and nominator positive --> $$x>\frac{2}{5}$$;
Both denominator and nominator negative --> $$-\frac{1}{3}<x<0$$;

So inequality holds true in the ranges: $$-\frac{1}{3}<x<0$$ and $$x>\frac{2}{5}$$.

Hi
Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ?
I don't know how to solve this????
Math Expert
Joined: 02 Sep 2009
Posts: 36638
Followers: 7106

Kudos [?]: 93659 [0], given: 10583

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

15 Jul 2014, 05:26
GGMAT760 wrote:
Bunuel wrote:
Eden wrote:
For what range of values of 'x' will the inequality 15x - 2/x > 1?

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

[Reveal] Spoiler:
D

devashish wrote:
I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?

anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so.

One could also do as follows:

$$15x-\frac{2}{x}>1$$ --> $$\frac{15x^2-x-2}{x}>0$$ --> $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$:

Both denominator and nominator positive --> $$x>\frac{2}{5}$$;
Both denominator and nominator negative --> $$-\frac{1}{3}<x<0$$;

So inequality holds true in the ranges: $$-\frac{1}{3}<x<0$$ and $$x>\frac{2}{5}$$.

Hi
Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ?
I don't know how to solve this????

Hope this helps.
_________________
Senior Manager
Joined: 07 Apr 2012
Posts: 464
Followers: 2

Kudos [?]: 53 [0], given: 58

Re: For what range of values of 'x' will the inequality 15x - 2/x > 1? [#permalink]

### Show Tags

23 Sep 2014, 01:50
PiyushK wrote:
What I have learnt, if x is in denom, take LCM after bringing all element at one side.

If we have more than 2 roots, then we can use following method to know the range.

Attachment:
Solution.jpg

Can you explain why before and after the "0" you change the sign?
What was the equation that got you to draw this?
Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?   [#permalink] 23 Sep 2014, 01:50

Go to page    1   2    Next  [ 32 posts ]

Similar topics Replies Last post
Similar
Topics:
16 If x > 3,000, then the value of x/(2x+ 1) is closest to 15 16 Jan 2014, 03:48
10 For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What 7 08 Mar 2012, 14:48
1 If y^2 = 2, what is the value of 2^(x-y)(x+y)/2^(x-1)(x+1) 2 26 Feb 2012, 14:40
45 What range of values of x will satisfy the inequality |2x + 19 25 May 2011, 06:04
61 For what range of values of x will the inequality 22 16 Feb 2011, 23:20
Display posts from previous: Sort by