himu wrote:What is the value of x?

(1) 4 < x < 6

(2) |x|=4x−15

Statement 1: 4 < x < 6
It's possible that x=5 or that x=5.5.
INSUFFICIENT.

Statement 2: |x|=4x-15
Case 1: No signs are changed
x = 4x-15
15 = 3x
x=5.
To confirm that x=5 is a valid solution, plug x=5 into the original equation:
|5| = 4*5 - 15
5 = 5.
Since the two sides of the equation are equal, x=5 is a valid solution.

Case 2: Flip the signs for one side of the equation
-x = 4x-15
15 = 5x
x=3.
To confirm that x=3 is a valid solution, plug x=3 into the original equation:
|3| = 4*3 - 15
3 = -3.
Since the two sides of the equation are NOT equal, x=3 is NOT a valid solution.

Thus, x=5.
SUFFICIENT.

The correct answer is B.

GMATGuruNY wrote:
Statement 2: |x|=4x-15

Case 1: No signs are changed
x = 4x-15
15 = 3x
x=5.
To confirm that x=5 is a valid solution, plug x=5 into the original equation:
|5| = 4*5 - 15
5 = 5.
Since the two sides of the equation are equal, x=5 is a valid solution.

Case 2: Flip the signs for one side of the equation
-x = 4x-15
15 = 5x
x=3.
To confirm that x=3 is a valid solution, plug x=3 into the original equation:
|3| = 4*3 - 15
3 = -3.
Since the two sides of the equation are NOT equal, x=3 is NOT a valid solution.

Thus, x=5.
SUFFICIENT.

The correct answer is B.

This is an interesting case! So if I don't check the two values of x again, I'll automatically think that (2) is insufficient.

My question is: when I see |a| = b, how do I know a has only 1 value without testing all values again? Thank you.

feedrom wrote:

GMATGuruNY wrote:
Statement 2: |x|=4x-15

Case 1: No signs are changed
x = 4x-15
15 = 3x
x=5.
To confirm that x=5 is a valid solution, plug x=5 into the original equation:
|5| = 4*5 - 15
5 = 5.
Since the two sides of the equation are equal, x=5 is a valid solution.

Case 2: Flip the signs for one side of the equation
-x = 4x-15
15 = 5x
x=3.
To confirm that x=3 is a valid solution, plug x=3 into the original equation:
|3| = 4*3 - 15
3 = -3.
Since the two sides of the equation are NOT equal, x=3 is NOT a valid solution.

Thus, x=5.
SUFFICIENT.

The correct answer is B.

This is an interesting case! So if I don't check the two values of x again, I'll automatically think that (2) is insufficient.

My question is: when I see |a| = b, how do I know a has only 1 value without testing all values again? Thank you.

|x|=4x-15.
As Ceilidh has pointed out, whenever we have the same variable on both sides but absolute value on only one side, we need to confirm that both solutions are valid.
The easiest approach is to plug the two solutions back into the equation, as I did in my post above.

Alternate approach:
The absolute value notation on the lefthand side implies that the lefthand side represents a NONNEGATIVE VALUE.
Thus, the righthand side -- 4x-15 -- must also represent a nonnegative value.
This restricts the range of x as follows:
4x-15 ≥ 0
4x ≥ 15
x ≥ 3.75.
Thus, for the righthand side to be nonnegative, x must be greater than or equal to 3.75.
Implication:
A solution is valid only if it is greater than or equal to 3.75.

In my post above, algebra yielded two possible values for x:
3 and 5.
But x=3 is invalid because it is less than 3.75.
Thus, the only valid solution is x=5.

Wow !!!

This is sheer enlightenment !

Thanks so much GMATGuruNY & Ceilidh for sharing your knowledge.That was more than helpful.
With absolute values, when you have a variable inside the absolute value and only numbers outside of it, you will have 2 solutions (and you don't need to retest them).

If you have the same variable both inside and outside of the absolute value sign, as in |x|=4x-15, you have to plug your solutions back in to see if they work.

The absolute value notation on the lefthand side implies that the lefthand side represents a NONNEGATIVE VALUE.
Thus, the righthand side -- 4x-15 -- must also represent a nonnegative value.

Implication:
A solution is valid only if it is greater than or equal to 3.75.

In my post above, algebra yielded two possible values for x:
3 and 5.
But x=3 is invalid because it is less than 3.75.
Thus, the only valid solution is x=5.

I think I should rename the topic.What I think is in the REAL GMAT we students get such TRICKY questions wrong .At the outset these seem very simple so our feeling is we got it right, but alas....the GMAT algorithm is working in the background ((

thanks again for that inequality stuff...I'll implement that & keep it at the back of my mind !

Regards,
~Himu.

Amazing! Thank so much, Guru and Ceilidh!