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by theCodeToGMAT » Thu Nov 14, 2013 10:32 pm
|12x-5|>|7-6x|

12x - 5 > 7 - 6x ==> 18x > 12 ==> x > 2/3 or 0.66
or,
5 - 12x > 7 - 6x ==> 6x < -2 ==> x < -1/3 or -0.33

{A} -12 ==> (4)(-3) YES
{B} -7/5 ==> (1)(-7/5) YES
{C} -2/9 ==> (-1/3)(2/3) => NOT POSSIBLE
{D} 4/9 ==> (-2/3)(-2/3) YES
{D} 17 ==> (17/3)(3) YES

Answer [spoiler]{C}[/spoiler]
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by Uva@90 » Fri Nov 15, 2013 7:31 am
theCodeToGMAT wrote:|12x-5|>|7-6x|

12x - 5 > 7 - 6x ==> 18x > 12 ==> x > 2/3 or 0.66
or,
5 - 12x > 7 - 6x ==> 6x < -2 ==> x < -1/3 or -0.33

{A} -12 ==> (4)(-3) YES
{B} -7/5 ==> (1)(-7/5) YES
{C} -2/9 ==> (-1/3)(2/3) => NOT POSSIBLE
{D} 4/9 ==> (-2/3)(-2/3) YES
{D} 17 ==> (17/3)(3) YES

Answer [spoiler]{C}[/spoiler]
Rahul,
I have couple of doubts with answer,
First thing is You have considered both positive and negative value for LHS, Similarly why you didn't consider for RHS.
12x - 5 > 7 - 6x
5 - 12x > 7 - 6x
I couldn't get that part.

Secondly,
In you answer you have split-ed the answer into two variables, for example
12 as (4)(-3) why not as 12 and 1 or -12 and -1?
-12 ==> (4)(-3) YES
Similarly for other options too.

could you please explain these things.

Thanks in advance

Regards,
Uva.
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by theCodeToGMAT » Fri Nov 15, 2013 8:30 am
Good to see that your lots of doubts are now only two :)

Generally, when we have inequalities both sides we simple expand the left side. This is because when we have "=" sign we will have same result; to verify that, lets consider an example:

|x-3| = |2x -2|
x-3 = 2x - 2
-x = 1 ==> x = -1

3-x = 2x -2
-3x = -5 ==> x = 5/3

2x - 2 = x - 3
x = -1

2 - 2x = x - 3
-3x = -5 ==> x = 5/3

So, results are similar.

Now, coming back to original question, we do not have "=" sign.. So, we can expect to have discrepancy with sign when we solve LHS & RHS separately.

When we solved LHS, we got x > 2/3 or 0.66 &&& x < -1/3 or -0.33

At max, sign will vary and that too for the case when we introduce "-ve" sign in LHS .. i.e. for "x < -1/3 "

Now, lets see the answer choices..
we see that
{C} ==> -2/9 which is when (2/3)(-1/3)

For any case x != -1/3 & x!= 2/3.. so we cannot have [spoiler]{C}[/spoiler] as answer.
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by theCodeToGMAT » Fri Nov 15, 2013 8:42 am
For the Second doubt:

The question says that we need to identify the not possible case.. so we need to find out ways to prove that yes the answer choice is achievable so as to reach a case which doesn't match {A} -12

We found out that x's possible values are:

x > 2/3 or 0.66

x < -1/3 or -0.33

i.e. 3,4..12,.... and -2/3, -1,-2, -3, -4.....

So, now we need to select values from above to fit in answer choices

{A} -12 ==> (4)(-3)..
Similarly for other

Is it better now?
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by Uva@90 » Fri Nov 15, 2013 10:13 am
theCodeToGMAT wrote:Good to see that your lots of doubts are now only two :)

Generally, when we have inequalities both sides we simple expand the left side. This is because when we have "=" sign we will have same result; to verify that, lets consider an example:

|x-3| = |2x -2|
x-3 = 2x - 2
-x = 1 ==> x = -1

3-x = 2x -2
-3x = -5 ==> x = 5/3

2x - 2 = x - 3
x = -1

2 - 2x = x - 3
-3x = -5 ==> x = 5/3

So, results are similar.

Now, coming back to original question, we do not have "=" sign.. So, we can expect to have discrepancy with sign when we solve LHS & RHS separately.

When we solved LHS, we got x > 2/3 or 0.66 &&& x < -1/3 or -0.33

At max, sign will vary and that too for the case when we introduce "-ve" sign in LHS .. i.e. for "x < -1/3 "

Now, lets see the answer choices..
we see that
{C} ==> -2/9 which is when (2/3)(-1/3)

For any case x != -1/3 & x!= 2/3.. so we cannot have [spoiler]{C}[/spoiler] as answer.
Rahul,
Thanks a bundle.

You made me soo clear.

Regards,
Uva.
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by vipulgoyal » Wed Dec 04, 2013 10:10 pm
|12x-5|>|7-6x|
Rahul would you mind to take another look on
5 - 12x > 7 - 6x ==> 6x < -2 ==> x < -1/3 or -0.33
if you open the LHS with -ve, the enequality will be reversed
5 - 12x < 7 - 6x
-6x < 2
x < -1/3
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by theCodeToGMAT » Wed Dec 04, 2013 10:38 pm
Hi Vipul, you made a small mistake while solving the inequality...refer the part in Red
vipulgoyal wrote:|12x-5|>|7-6x|
Rahul would you mind to take another look on
5 - 12x > 7 - 6x ==> 6x < -2 ==> x < -1/3 or -0.33
if you open the LHS with -ve, the enequality will be reversed
5 - 12x < 7 - 6x
-6x < 2
x < -1/3 ==================> Correct Version must be ::: 6x > -2 ===> x > -1/3
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by vipulgoyal » Thu Dec 05, 2013 12:07 am
yeh, but the solution require x < -1/3, could you please explain how you come to this, after opening mode with -ve the inequality you mentioned 5 - 12x > 7 - 6x seems incorrect to me
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by theCodeToGMAT » Thu Dec 05, 2013 12:21 am
vipulgoyal wrote:yeh, but the solution require x < -1/3, could you please explain how you come to this, after opening mode with -ve the inequality you mentioned 5 - 12x > 7 - 6x seems incorrect to me

The original equation is: |12x-5|>|7-6x|

Working on LHS, we have two scenarios (i.e by considering that LHS modulus is +ve & -ve)

+ve ==> 12x - 5 > 7 - 6x => 18x > 12 => x > 2/3

-ve ==> -(12x-5) > (7-6x) => 5 - 12x > 7 - 6x => 6x < -2 ==> x < -1/3

Also, we can have another possible scenario by breaking RHS modulus:

-ve ==> 12x - 5 > -(7-6x) => 12x - 5 > 6x - 7 => 6x > -2 ==> x > -1/3

For any of the cases we don't have x = -1/3

For {C} -2/9 we need (-1/3)(2/3) but we dont have x equal to -1/3 & 2/3
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by vipulgoyal » Thu Dec 05, 2013 12:47 am
Rahul, what i am trying to say while opening the ineqality with -ve at LHS, ineqality will change two times

Working on LHS, we have two scenarios (i.e by considering that LHS modulus is +ve & -ve)
|12x-5|>|7-6x|

-12x + 5 < 7 - 6x (changes first time as we opended LHS mode with -ve)
-6x < 2
6x > -2 (changes second time as it mutiplied by -1)
x > -1/3

Now breaking RHS mode with -ve the ineqality changes only once

|12x-5|>|7-6x|
12x - 5 < -7 + 6x
6x < -2
x < -1/3

though it doesnt change the solution as we got what we need x!= -1/3 & x!= 2/3
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by sanju09 » Thu Dec 05, 2013 2:43 am
[email protected] wrote:If |12x−5|>|7−6x|, which of the following CANNOT be the product of two possible values of x?

A. -12
B. -7/5
C. -2/9
D. 4/9
E. 17

Ans-C
If |12x-5|>|7-6x|, which of the following CANNOT be the product of two possible values of x?

Not always, but in this particular situation, inequality can be squared in both sides. We'll have

144x^2 - 120x + 25 > 49 - 84x + 36x^2

108x^2 - 36x - 24 > 0

9x^2 - 3x - 2 > 0

(3x - 2) (3x + 1) > 0

Either both factors are positive or both negative.

If 3x - 2 > 0 then x > 2/3, and we need not to consider the other factor. (Why?)

Or if 3x + 1 < 0 then x < -1/3, and we need not to consider the other factor. (Why?)

A. -12 = 4 (x > 2/3) times -3 (x < -1/3) hence possible
B. -7/5 = 7/5 (x > 2/3) times -1 (x < -1/3) hence possible
C. -2/9 = no possible combination according to the constraints, and we can stop testing
D. 4/9
E. 17
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by GMATGuruNY » Thu Dec 05, 2013 4:57 am
If |12x-5|>|7-6x|, which of the following CANNOT be the product of two possible values of x?

a -12
b -7/5
c -2/9
d 4/9
e 17
Determine the CRITICAL POINTS: the values where |12x-5| = |7-6x|.
There are TWO CASES to consider:
Case 1: The signs of the equation are UNCHANGED.
Case 2: For ONE SIDE of the equation, the signs are flipped.

Case 1: 12x-5 = 7-6x
18x = 12
x = 12/18 = 2/3.

Case 2: 12x-5 = -7+6x
6x = -2
x = -2/6 = -1/3.

The CRITICAL POINTS are -1/3 and 2/3.
These are the only values where |12x-5| = |7-6x|.
To determine the ranges where |12x-5| > |7-6x|, test one value to the left and right of each critical point.

x<-1/3:
Plug x = -1 into |12x-5| > |7-6x|:
|12(-1) - 5| > |7 - 6(-1)|
17 > 13.
This works.
x < -1/3 is a valid range.

-1/3 < x < 2/3:
Plug x = 0 into |12x-5| > |7-6x|:
|12*0 - 5| > |7 - 6*0|
5 > 7.
Doesn't work.
-1/3 < x < 2/3 is NOT a valid range.

x > 2/3:
Plug x = 1 into |12x-5| > |7-6x|:
|12*1 - 5| > |7 - 6*1|
7 > 1.
This works.
x > 2/3 is a valid range.

Thus, |12x-5| > |7-6x| when x < -1/3 or x > 2/3.
Eliminate answer choices that indicate a possible product.
If the two solutions are x=-1 and x=12, then their product = -1*12 = -12.
Eliminate A.
If the two solutions are x=-1 and x=7/5, then their product = -1 * 7/5 = -7/5.
Eliminate B.
If the two solutions are x=-1 and x=-4/9, then their product = -1 * -4/9 = 4/9.
Eliminate D.
If the two solutions are x=1 and x=17, then their product = 1*17 = 17.
Eliminate E.

The correct answer is C.
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by Mathsbuddy » Thu Dec 05, 2013 8:42 am
Square both sides:

(12x-5)^2 > (6x-7)^2

144x^2 -120x + 25 > 36x^2 - 84x + 49

y = 108x^2 -36x -24 > 0

9x^2 - 3x - 2 > 0

(3x - 2) (3x + 1) > 0

Roots: x = -1/3, 2/3

As the coefficient of x^2 is positive, then y > 0 lies outside these values.

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

Product P would therefore be negative and |lowest limit| > 1/3 * 2/3 = 2/9

So P < -2/9

Only answer (C)-2/9 is OUT OF RANGE
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