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>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 15195 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 16021 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 _____________________________________________________________________________

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

6

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.

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.

x < -1/3 OR x > 2/3

Then proceed as given above.

I apply this distance inference. But in a different way,

The nearest the number to 5/12 must be the answer, as it yields the shortest distance to 5/12. Then answer is C.

\(x=-4<-\frac{1}{3}\) 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.

Bunuel - this is really helpful, thank you! Can you let us know if a question like this is possible on the GMAT? How is it possible to interpret the question and do all this math / testing in 2mins or 2.5mins max?

To solve the question faster, we don't have to look at all the choices. Once we know that x>2/3 and x<−1/3, we can say that x can neither be 2/3 nor -1/3 (x!=2/3 && x!=-1/3)

so the product of two possible values of x can never be 2/3*(-1/3)=-2/9, Hence the correct answer is C.

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.

x < -1/3 OR x > 2/3

Then proceed as given above.

Hi VeritasPrepKarishma,

I could not understand how did you solve it. I could not comprehend the line "Twice of the distance from 5/12 should be more than distance from 14/12." Please help !

Solution: C. Algebraically, the inequalities give you four different possibilities:

12x – 5 > 7 – 6x (which works out cleanly to 18x > 12, meaning x > 2/3) 12x – 5 < -(7 – 6x) -(12x – 5) < 7 – 6x -(12x -5) > -(7 – 6x)

With the last three all involving negatives and inequalities, it can be helpful to simply find the inequality point and then test values on either side to determine whether it's greater than or less than:

12x – 5 < -(7 – 6x) works out to:

12x - 5 < 6x - 7

6x < -2

x < −1/3, but try a value like -1 (less than −1/3) and like 0 (greater than −1/3) to ensure the - vs. + of a tricky inequality with absolute values. -1 fits with the given information and 0 does not, so it should be clear that x < −1/3.

-(12x – 5) < 7 – 6x works out to 5 - 12x < 7 - 6x, which gives you -2 < 6x, and x > −1/3. This is why testing negative/positive is so important...the four "original" inequalities allow for really all sets of possible values other than −1/3 and 2/3, so some quick trial and error can help you determine which side of the inequalities are valid and which are not. Again, it should be clear from a quick plug-in of 0 and -1 that x < −1/3.

-(12x -5) > -(7 – 6x) works out to 5 - 12x > 6x - 7, which leads you to 12 > 18x, and x < 2/3. This confirms the "break point" of 2/3, so again a quick plug in of easy numbers on either side (0 and 1) will help you determine that x must be greater than 2/3.

So you know that x is either greater than 2/3 or less than −1/3. Certain answer choices, then, are easy to pick off: -12 could be -1*12. 17 could be 1*17. 4/9 could be -1*(-4/9) and -7/5 could be -1*7/5. But -2/9 cannot be done.
_________________