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\(x=-4<-\frac{1}{3}\) 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, 22:08

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 12971 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 13520 times ]

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

Show Tags

24 Jul 2013, 00:20

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, 01:59

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, 09:17

5

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.

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