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Modulus Question

by Mo2men » Thu Oct 20, 2016 6:21 pm
What is the value of x?

(1) |6 - 3x| = x - 2

(2) |5x + 3| = 2x + 9

OA: A

Source: Magoosh

Is it the only solution is by using the rule: if |a|= b then a=b or -a=b. Is other way to solve this question?
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Source: — Data Sufficiency |

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by ceilidh.erickson » Thu Oct 20, 2016 7:24 pm
Absolute value questions can be tricky. We know that the rule (stating it a little bit differently than you did, but same idea) is that if |a| = b, then a = b or a = -b. So, if we had a DS question that asked "what is the value of x?", then a statement that said "|x| = 3" would not be sufficient, because x could be 3 or -3.

It's tempting to think that any time you're presented with an absolute value DS question, therefore, a single statement would have to be insufficient. However, there are a few exceptions...

1. An absolute value equation might have only one solution (rather than two) if one of the solutions, when plugged back into the equation, doesn't work. This will only happen when you have the same variable both inside and outside the absolute value sign. You can't assume - you have to test them.

2. When you solve both the positive and negative case, it's possible to end up with 2 solutions that are in fact the same value. This is similar to solving quadratics: usually we get 2 solutions, but with perfect squares we get 2 of the same solution.

3. If your absolute value is set equal to zero, as in |y - 4| = 0. You can't have "positive 0" and "negative 0," so you'll just get a single solution.

(1) |6 - 3x| = x - 2
When we solve, we get 2 solutions:

positive case:
6 - 3x = x - 2
8 = 4x
x = 2
Plug it back in to test:
|6 - 3(2)| = 2 - 2
|0| = 0 --> this works!

negative case:
6 - 3x = -(x - 2)
6 - 3x = -x + 2
4 = 2x
x = 2

Our 2 solutions were the same value, so this statement is actually sufficient. (And we don't need to test again for statement 2).

(2) |5x + 3| = 2x + 9

positive case:
5x + 3 = 2x + 9
3x = 6
x = 2
Plug back in:
|5(2) + 3| = 2(2) + 9
|13| = 13 --> this works

negative case:
5x + 3 = -(2x + 9)
5x + 3 = -2x - 9
7x = -12
x = -12/7
Plug back in:
|5(-12/7) + 3| = 2(-12/7) + 9
|-60/7 + 21/7| = -24/7 + 63/7
|-39/7| = 39/7 --> this works!

Because we got 2 different solutions that worked, this statement is insufficient.

The answer is A.

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Last edited by ceilidh.erickson on Fri Oct 21, 2016 4:23 pm, edited 1 time in total.
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by ceilidh.erickson » Thu Oct 20, 2016 7:27 pm
Here's an example of exception #1 that I listed above:
https://www.beatthegmat.com/simple-yet-a ... tml#714086
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by crackverbal » Thu Oct 20, 2016 9:25 pm
Hi Mo2men,

Let me try to explain these types of questions in GMAT little more elaborately.

There are basically you could expect four types of modulus questions (particularly in DS).

1. LHS is a modulus of some equation and the RHS is just a equation and the question is a YES-NO DS question
2. LHS is modulus of some equation and the RHS is just a equation and the question is a Value of DS question
3. Modulus and inequalities.
4. Both side modulus.

This question is of the second type here. I will explain at the last how we can solve when asked of the first type.

General rule of a modulus (absolute value),

|x| = x , whenever x > = 0

|x| = (-x), whenever x < 0

Here in this question,

We have to find the value of x,

I. |6-3x| = x-2

First case,

6-3x = x-2 , whenever (6-3x) > =0, that is x < = 2

4x = 8,

So x= 2 which matches condition x < = 2

Second case,

6-3x = -(x-2) , whenever (6-3x) < 0 , that is x > 2

Again this gives the same value.

So there is only value possible is 2.

So statement I is sufficient.

Eliminate the answer choices B, C and E. Keep the answer choices A and D.

II. |5x + 3| = 2x + 9

First case,

5x + 3 = 2x + 9 , whenever 5x >=3 , that is x >= (3/5)

3x = 6

x = 2 which matches the condition x >= (3/5)

Second case,

5x + 3 = -( 2x + 9) , whenever 5x <3 , that is x < (3/5)

7x = -12

x = -12/7, which matches the condition x < (3/5)

So we get here 2 values of "x".

So not sufficient.

So the answer is A.

Now, if this question of the first type which I mentioned above, lets see how to do it.

That is YES-NO DS question.

Let say the question is like this,

Is x > 0 ?

I. |6 - 3x| = x - 2

II. |5x + 3| = 2x + 9

Now for this question, we no need to do all the process we need above,

We know that LHS in the |6 - 3x| = x - 2 cannot be negative.

So we write,

x -2 > = 0

that says x >= 2

which is positive. sufficient.

Now the second statement,

|5x + 3| = 2x + 9, again LHS cannot be negative.

So 2x+9 >= 0,

So , x > = -9/2

So it could be negative or positive or zero.

So not sufficient.

So NET-NET, it depends on what question type of modulus GMAT asks. So you could smartly reduce time you spend
on questions when you know how to do the above types clearly.

Hope this is clear.
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by Mo2men » Fri Oct 21, 2016 4:18 am
Dear Ceilidh,

Thanks for your explanation.

Dear Crackverbal,

You exactly hit the point I wanted to discuss. I thought that we can solve type I using the same way of type II. I solved my question above as follows:

What is the value of X?

I.|6-3x| = x-2


We know that LHS in the |6 - 3x| = x - 2 cannot be negative.

So we write,

x -2 > = 0

that says x >= 2

So we can conclude that we have ONE solution. So Sufficient

II.|5x + 3| = 2x + 9

So 2x+9 >= 0,

So , x > = -9/2

So it could be negative or positive or zero.

So we can conclude that we have two solutions solution. so Insufficient

What do you think?
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by GMATGuruNY » Fri Oct 21, 2016 4:23 am
|6x-3|
We need to consider two cases if the expression inside the absolute value can represent both a positive and a negative value.
Case 1: If 6-3x ≥ 0, then |6x-3| = 6-3x.
Case 2: If 6-3x < 0, then |6x-3| = -6+3x.

An example of Case 1: x=0
Here, 6-3x ≥ 0.
As a result, |6-3x| = 6-3x:
|6 - 3*0| = 6 - 3*0
6 = 6.

An example of Case 2: x=3
Here, 6-3x < 0.
As a result, |6-3x| = -6+3x:
|6 - 3*3| = -6 + 3*3
3 = 3.
Statement 1: |6 - 3x| = x - 2
In this equation, since the absolute value on the left side cannot be equal to a negative value, the right side must be NONNEGATIVE:
x-2≥0, implying that x≥2.

Given this constraint, only Case 2 is possible.
Implication:
|6 - 3x| = x - 2 has only ONE solution.
Since the value of this solution can be determined, Statement 1 is SUFFICIENT.

Solution:
In Case 2, |6-3x| = -6+3x.
Substituting |6-3x| = -6+3x into |6-3x| = x-2, we get:
-6+3x = x-2
2x = 4
x = 2.
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