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If |12x−5|>|7−6x|, which of the following CANNOT be the

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

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New post 13 Jan 2018, 06:41
1
How did you arrive at x<-1/3 or x>2/3 from (x+1/3) ( x-2/3) >0???
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the  [#permalink]

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New post 13 Jan 2018, 06:51
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Kezia9 wrote:
How did you arrive at x<-1/3 or x>2/3 from (x+1/3) ( x-2/3) >0???


There are links on previous pages which lead to posts explaining this. Here are some more:

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



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

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New post 29 Jan 2018, 11:11
emmak 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



Let’s first solve for when (12x - 5) and (7 - 6x) are both positive.

12x - 5 > 7 - 6x

18x > 12

x > 12/18

x > 2/3

Now let’s solve for when (12x - 5) is negative and (7 - 6x) is positive.

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

-12x + 5 > 7 - 6x

-2 > 6x

-1/3 > x

So we have x < -1/3 or x > 2/3.

If x were -1/3 or 2/3, then the product would be -2/9. However, the inequalities specify that x can be neither -1/3 nor 2/3, so we know the product of two possible values of x cannot be -2/9.

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

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New post 12 May 2019, 12:05
Bunuel wrote:
emmak 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


VERITAS PREP OFFICIAL SOLUTION:

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.


Hi Bunuel,

i'm a bit confused over here. According to the question, there are 2 CP, 5/12 and 7/6.

so, the three cases are as follows:

1. x <5/12
=> -12x+5>7-6x
=> x<-1/3

2. 5/12<= x <7/6
=> 12x-5>7-6x
=> x>2/3

3. x >=7/6
=>12x-5>6x-7
=> x>-1/3

Could you please help me understand the reason for eliminating the 3rd solution?
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If |12x−5|>|7−6x|, which of the following CANNOT be the  [#permalink]

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New post 13 May 2019, 10:58
Bunuel

Hi,

Could you please resolve my query and help me understand the solution?
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the  [#permalink]

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New post 13 May 2019, 22:35
99ramanmehta wrote:
Bunuel wrote:
emmak 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


VERITAS PREP OFFICIAL SOLUTION:

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.


Hi Bunuel,

i'm a bit confused over here. According to the question, there are 2 CP, 5/12 and 7/6.

so, the three cases are as follows:

1. x <5/12
=> -12x+5>7-6x
=> x<-1/3

2. 5/12<= x <7/6
=> 12x-5>7-6x
=> x>2/3

3. x >=7/6
=>12x-5>6x-7
=> x>-1/3

Could you please help me understand the reason for eliminating the 3rd solution?


Note here that pre-condition for 3rd case is that x >= 7/6. Hence values such as -1/10, 0, 1 are not valid. So solution from third case is that x > 7/6.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the  [#permalink]

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New post 13 May 2019, 22:37
1
99ramanmehta wrote:
Bunuel wrote:
emmak 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


VERITAS PREP OFFICIAL SOLUTION:

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.


Hi Bunuel,

i'm a bit confused over here. According to the question, there are 2 CP, 5/12 and 7/6.

so, the three cases are as follows:

1. x <5/12
=> -12x+5>7-6x
=> x<-1/3

2. 5/12<= x <7/6
=> 12x-5>7-6x
=> x>2/3

3. x >=7/6
=>12x-5>6x-7
=> x>-1/3

Could you please help me understand the reason for eliminating the 3rd solution?


Also note that the most convenient way of dealing with such inequalities is squaring both sides. The absolute value sign disappears immediately and your are left with a simple quadratic.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the  [#permalink]

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New post 14 May 2019, 09:22
VeritasKarishma wrote:
99ramanmehta wrote:
Bunuel wrote:

VERITAS PREP OFFICIAL SOLUTION:

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.


Hi Bunuel,

i'm a bit confused over here. According to the question, there are 2 CP, 5/12 and 7/6.

so, the three cases are as follows:

1. x <5/12
=> -12x+5>7-6x
=> x<-1/3

2. 5/12<= x <7/6
=> 12x-5>7-6x
=> x>2/3

3. x >=7/6
=>12x-5>6x-7
=> x>-1/3

Could you please help me understand the reason for eliminating the 3rd solution?


Note here that pre-condition for 3rd case is that x >= 7/6. Hence values such as -1/10, 0, 1 are not valid. So solution from third case is that x > 7/6.


Thank you for the explanation VeritasKarishma ! Your answer is exactly what I was looking for.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the  [#permalink]

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New post 14 May 2019, 09:31
VeritasKarishma wrote:
99ramanmehta wrote:
Bunuel wrote:

VERITAS PREP OFFICIAL SOLUTION:

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.


Hi Bunuel,

i'm a bit confused over here. According to the question, there are 2 CP, 5/12 and 7/6.

so, the three cases are as follows:

1. x <5/12
=> -12x+5>7-6x
=> x<-1/3

2. 5/12<= x <7/6
=> 12x-5>7-6x
=> x>2/3

3. x >=7/6
=>12x-5>6x-7
=> x>-1/3

Could you please help me understand the reason for eliminating the 3rd solution?


Also note that the most convenient way of dealing with such inequalities is squaring both sides. The absolute value sign disappears immediately and your are left with a simple quadratic.


Absolutely, I do realise it's a better method. Thank you for pointing it out.
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Re: If |12x−5|>|7−6x|, which of the following CANNOT be the   [#permalink] 14 May 2019, 09:31

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