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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=-17<-\frac{1}{3}\) are possible values of x --> the product = 17.

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=17>\frac{2}{3}\) are possible values of x --> the product = 17.

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=17>\frac{2}{3}\) are possible values of x --> the product = 17.

Answer: C.

We CAN square an inequality if we know that the sides are non-negative, which is the case here.
_________________

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=17>\frac{2}{3}\) are possible values of x --> the product = 17.

Answer: C.

Hi Bunuel, Please correct me if i am wrong here, but dont you mean \(x=-1<-\frac{1}{3}\) and \(x=-17<\frac{-1}{3}\) are possible values of x --> the product = 17. For the last point??

Since -1 * 17 would be -17??
_________________

PS: Like my approach? Please Help me with some Kudos.

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=17>\frac{2}{3}\) are possible values of x --> the product = 17.

Answer: C.

Hi Bunuel, Please correct me if i am wrong here, but dont you mean \(x=-1<-\frac{1}{3}\) and \(x=-17<\frac{-1}{3}\) are possible values of x --> the product = 17. For the last point??

x>7/6 |12x−5|>|7−6x| (12x-5) > -(7-6x) 12x - 5 > -7 + 6x 6x > -2 x > -1/3 (if the range being tested is >7/6 and x > -1/3 is that valid or invalid?)

I think I am approaching finding x the right way, but I am not sure how I can figure out what CANNOT be the product of two possible values of x. Can anyone help? Thanks!

Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

Show Tags

23 Jul 2013, 21:08

Hi Folks,

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Plz Advise !!

Regards, TGC !

Attachments

query.JPG [ 13.53 KiB | Viewed 12278 times ]

_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Plz Advise !!

Regards, TGC !

The graphs drawn are not correct. The proper drawing is below:

Attachment:

MSP2441f260916666dabe40000498a5d50ib182823.gif [ 9.23 KiB | Viewed 12765 times ]

Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

Show Tags

23 Jul 2013, 23:20

Bunuel wrote:

targetgmatchotu wrote:

Hi Folks,

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Plz Advise !!

Regards, TGC !

The graphs drawn are not correct. The proper drawing is below:

Attachment:

MSP2441f260916666dabe40000498a5d50ib182823.gif

Hope it helps.

12x -5 , the graph touches y axis where x=0 => y = -5 taking modulus => y=5

7-6x, x=0 => y =7 .

Further, touches x axis where y=0 hence x=5/12 and x=7/6 (7/6 > 5/12)

Please correct me !!

Rgds, TGC !

So by any chance the graph cross??????
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

I have a query on this and see the attachment for the same.Please ignore the untidy drawing of mine, couldn't help due to time constraint and poor word knowledge.

I have drawn graphs for |12x+5| and |7-6x| and compared where the first modulus is greater than the second modulus. However, against the above solutions , I got only x> 2/3 solution which doesn't satisfies above solution.

The line in red is the graph for |7-6x| and the line in black is for |12x-5|. The pink line shows intersection point of the lines and has (2/3) as x - coordinate.So by the graph we can see that |12x-5| > |7-6x| only after x= 2/3.

Please tell where I am going wrong !!!

Plz Advise !!

Regards, TGC !

The graphs drawn are not correct. The proper drawing is below:

Attachment:

MSP2441f260916666dabe40000498a5d50ib182823.gif

Hope it helps.

12x -5 , the graph touches y axis where x=0 => y = -5 taking modulus => y=5

7-6x, x=0 => y =7 .

Further, touches x axis where y=0 hence x=5/12 and x=7/6 (7/6 > 5/12)

Please correct me !!

Rgds, TGC !

So by any chance the graph cross??????

I don't understand your question... As I said the graph of |12x−5|>|7−6x| is:

Re: If |12x−5|>|7−6x|, which of the following CANNOT be the prod [#permalink]

Show Tags

24 Jul 2013, 00:59

Bunuel wrote:

I don't understand your question... As I said the graph of |12x−5|>|7−6x| is:

x<-1/3 and x>2/3.

Got it,

there was a mistake in drawing the graph !!

Thanks, TGC !
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=-17<-\frac{1}{3}\) are possible values of x --> the product = 17.

Answer: C.

Dear Bunuel

Pl enlighten us in what all cases we can square the modulus

We find the points where the two distances are equal. The distance between 5/12 and 14/12 is 9/12 which gets divided into 1:2 i.e. the point where the distances will be equal will be 3/12 away from 5/12 i.e. at 8/12 = 2/3. At any point to the right of 2/3, twice the distance from 5/12 will be more than the distance from 14/12.

Another point will be 9/12 to the left of 5/12 i.e. at -4/12 = -1/3. At any left to the left of -1/3, twice the distance from 5/12 will be more than the distance from 14/12.

Re: If |12x−5|>|7−6x|, which of the following CANNOT be the [#permalink]

Show Tags

02 Apr 2014, 08:17

3

This post received KUDOS

2

This post was BOOKMARKED

Another Method \(|12x-5|>|7-6x|\) Only 2 cases can arise; the other 2 cases are the same as these ones Ist Case \(|12x-5|>|7-6x|\) \(2x-5>7-6x\) \(18x>12\) \(x>2/3\)

IInd Case \(|12x-5|>|7-6x|\) \(12x-5<6x-7\) \(x<-2/6\) \(x<-1/3\)

Now we know that\(x>2/3\) and \(x<-1/3\) So from the above we can deduce that the answer has to be negative, thus we can cross out D and E. From the next 3 options which are all negative, Option A and B both can be formed, but option C is between \(-1/3\) and \(2/3\) which is not in the range of x. Thus C is your answer.

\(x=-4<-\frac{1}{3}\) and \(x=3>\frac{2}{3}\) are possible values of x --> the product = -12; \(x=-\frac{7}{5}<-\frac{1}{3}\) and \(x=1>\frac{2}{3}\) are possible values of x --> the product = -7/5; \(x=-\frac{4}{9}<-\frac{1}{3}\) and \(x=-1<-\frac{1}{3}\) are possible values of x --> the product = 4/9; \(x=-1<-\frac{1}{3}\) and \(x=-17<-\frac{1}{3}\) are possible values of x --> the product = 17.

Answer: C.

Hi Bunnel,

I have a doubt. how you are getting 7/5, 4/9, 1/3 etc ion above post.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...