What is x? (1) |x|

Question

What is x?

(1) |x| < 2
(2) |x| = 3x – 2

Bunuel · Accepted Answer

What is x?

(1) |x| < 2

This can be rephrased as  -2 < x < 2 , which is not sufficient to determine the value of x.

Not sufficient.

(2) |x| = 3x – 2

In this equation, the left-hand side (|x|) represents an absolute value, which is inherently non-negative. Thus, the right-hand side (3x – 2) must also be non-negative. This can be expressed as  3x - 2 ≥ 0 , which simplifies to  x ≥ \frac{2}{3} . Therefore, x must be a positive value, and thus |x| = x. Substituting this into our equation results in x = 3x - 2, which solves to x = 1.

Sufficient

Answer: B.

Hope it's clear.

WholeLottaLove · Answer

What is x?

(1) |x| < 2

x<2 x>-2
Invalid, x could be any number greater than 2 or less than 2.
INSUFFICIENT

(2) |x| = 3x – 2
x>=0
|x| = 3x – 2
x = 3x - 2
-2x = -2
x = 1 Valid
OR
x<0
-x = 3x - 2
-4x = -2
x = (1/2) Invalid as 1/2 is not less than zero.
SUFFICIENT

(B)

gmatter0913 · Answer

Vow! that is a very good solution :shock:

It opened my eyes on how to read modulus based problems...

rohitgoel15 · Answer

Hi Bunuel, Thanks for reply .. But this is not clear to me. Generally in modulus questions we do as below: (1) |x| < 2 I agree with u. -2

Bunuel · Answer

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true. Next, the way I'm doing is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1 . Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html Hope it helps.

shasadou · Answer

OA:

(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2 
 -2x = -2 
 x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
 1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
 x = 2 – 3x
 4x = 2
 x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
 1/2 = 3/2 – 2
 1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

The correct answer is B.

KarishmaB · Answer

We all know that if we have mods, we take two cases - positive and negative - and then solve the equation.
The point is - why do we do that?
If you remember, this is how we define mods

|x| = x if x is positive
and -x if x is negative

So basically, |x| takes different forms depending on whether x is positive or negative.
When I want to solve |x|=3x-2, I can't solve with |x|.

So I split it into two cases:
Case 1: x is positive
I get x = 3x - 2
x = 1
I accept this value of x since x has to be +ve and satisfy given equation. It does both.
Case 2: x is negative
-x = 3x - 2
x = 1/2
I reject this value since x should be negative for the equation to look like this.
So x can only take 1 value i.e. x = 1.

Remember, when you split |x| into two cases, you have to check that the value you get lies in the region in which you are expecting it to lie.

GMATinsight · Answer

Question : x = ?

Statement 1: |x| < 2 
i.e. -2 < x < 2
NOT SUFFICIENT

Statement 1: |x| = 3x – 2 
i.e. x = 3x – 2 and -x = 3x – 2
i.e. x = 1 and x = 1/2
 But 1/2 doesn't satisfy the given statement hence that is not acceptable. Therefore,
x=2 is the only acceptable solution 
SUFFICIENT

Answer: Option B

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Question

In the original condition, there is 1 variable  x  and 0 equation. So you need 1 equation.

Condition 1)
Since it is an inequality, there is no equation. Thus we can't identify the variable x and this is not sufficient.

Condition 2)
i)  x \ge 0 
 |x| = 3x - 2  is equivalent to  x = 3x - 2  or  2x - 2 = 0 .
Thus we have  x = 1 .
This is sufficient.

ii)  x < 0 
 |x| = 3x - 2  is equivalent to  -x = 3x - 2  or  4x - 2 = 0 .
Thus we have  x = 1/2 . However  x = 1/2 > 0 .
There is no negative solution.

Therefore, we have a unique solution  x = 1 .
This condition is sufficient.

The answer is B.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

BrentGMATPrepNow · Answer

Target question :    What is the value of x?

Statement 1 : |x| < 2   
There are several values of x that satisfy statement 1. Here are two: 
Case a:  x = 1  (notice that < 2)
Case b:  x = 0  (notice that |0| < 2)
Since we cannot answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : |x| = 3x - 2  
When solving equation involving ABSOLUTE VALUE, there are 3 steps:
1. Apply the rule that says:   If |x| = k, then x = k and/or x = -k  
2. Solve the resulting equations
3. Plug in the solutions to check for extraneous roots

So, first we get: 
x = 3x - 2 
Solve, to get x =  1 
To check whether this is an extraneous, plug x =  1  into the original equation to get: | 1 | = 3( 1 ) - 2
Simplify to get: |1| = 1
PERFECT, this solution checks out.

Next, we get: 
x = -(3x - 2) 
Solve, to get x =  1/2 
To check whether this is an extraneous, plug x =  1/2  into the original equation to get: | 1/2 | = 3( 1/2 ) - 2
Simplify to get: |1/2| = -1/2
This solution DOES NOT check out.

So, it MUST be the case that  x = 1 
Since we can answer the  target question  with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

davidbeckham · Answer

Bunuel , please can you explain this step? I am still not clear on how did you reach 1.

Bunuel · Answer

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

achloes · Answer

This video does a great job clarifying the concept in discussion: https://www.youtube.com/watch?v=SYB2NrzA12o

achloes · Answer

My takeaway:

For absolute value questions, plug in the positive case and negative case answers into the non-absolute value part of the original equation to make sure it doesn’t produce a negative solution. A negative solution is extraneous as absolute values are always non-negative

bumpbot · Answer

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What is x? (1) |x| < 2 (2) |x| = 3x – 2

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