Absolute Value: MGMAT QBank Question

[Guest]

Hi All-

Source: MGMAT QBANK EIV "Absolute Range"

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0


Answer is C. However, what I don't understand is the part of the explanation of why one must consider 3 scenarios for statement 1. Someone help.

[StaceyKoprince]

I'll assume you understand how / why we rephrased the question, since you didn't ask about that. Also, this is a very difficult question, so one of the things you shoudl think about here is how you'd make an educated guess and move on - so that you know what to do if something throws you on the actual test.

So the rephrased question is: "Is -1 < x < 1?"

When we evaluate statement 1, we're of course trying to figure out whether we can answer the above question with a "yes (always)" or "no (always)."

Statement 1 is more complicated than statement 2 because we have absolute value signs on both sides of the equation. So we have to consider each absolute value sign in determining the possible ways to solve. Because there are two absolute value signs, there are three possibilities: the contents within both signs are negative, the contents within both are positive, or the contents within one are positive and the contents within the other are negative.

|x + 1| = 2|x -1|
Scenario 1: Both are negative. Ask yourself what it would take to make each one negative. For |x + 1|, x would have to be less than -1 (going down to negative infinity). For |x -1|, x would have to be less than 1 (going down to negative inifinity). If I combine those two, they overlap at -1 going down forever. So, if x < -1, then the contents of both absolute value signs will be negative.

Scenario 2: Both are positive. Follow the same process. For |x + 1|, x would have to be more than -1 (going up to infinity). For |x -1|, x would have to be more than 1 (going up to inifinity). If I combine those two, they overlap at 1 going up forever. So, if x > 1, then the contents of both absolute value signs will be positive.

Scenario 3: One positive, one negative. And now I've already got an idea of the range of numbers I should be testing, because Scenario 1 resulted in x < -1 and Scenario 2 resulted in x > 1. So... let's see what would happen if x is between those two numbers. For |x + 1|, if -1 < x < 1, then the contents will be positive. For |x -1|, if -1 < x < 1, then the contents will be negative.

Hope that clears things up a little!

[Guest]

Stacey- Thank you for the explanation. It definitely helps.

[Guest]

Hi Stacy,

You mention that this is a "very difficult" question. What "level" question is this ? Is it 650-700, 700-750, 750+ ?
Also, how would go about making an educated guess for this question ?

Thanks.

[Guest]

Guys,

I am struggling to understand why the answer here is C.
If we look at statement 2 "|x - 3| > 0", we can see that x > 3. So this enables us to answer the question "is -1 < x < 1 ?" ... the answer is NO. If x > 3, then x is not between -1 and 1 !

Thanks in advance for the explanation.

[brian]

Good question!

Statement 2 tells us that:
|x-3| > 0

In order for this to be true, what do we know about |x|? Based on this statement alone, not much. In fact, there is nothing we know about x, because any value of x will satisfy that statement (i.e. if x=-4, then |-4-3| = 7, if x=7, then |7-3| = 4.

Therefore, statement 2 is not sufficient by itself.

Hopefully that clears things up somewhat.

Best of luck.

-Brian

[StaceyKoprince]

One little point: the only value that will NOT satisfy the second statement is 3 itself - that would make the statement false. This isn't so important in evaluating statement 2 by itself, but it becomes important later on in the problem... :)

[Guest]

"so one of the things you shoudl think about here is how you'd make an educated guess and move on"

Stacy,
In this particular case ... what kinds of things should one look at in order to make an educated guess ?
How would you make an "educated guess" on this one ?

Thanks.

[WannabeatGMAT]

In this question the logic is to solve |x + 1| = 2|x -1| and to solve |x - 3| > 0 and to see which one will give us the answer for Is |x| < 1 ?

So if we solve |x + 1| = 2|x -1| <==> x+1= 2*(x-1) or x+1= -2*(x-1) <==> x=3 or x=1/3 in this step we can’t say that Is |x| < 1 or not because 3 >1

If we solve |x-3| >0 <==> x>3 or x<3 ==> x<>3 this inequality alone don’t also answer the question.

BUT (1)+(2) <==> x=1/3 and now we can say that |x|<1.

Thanks.

[RonPurewal]

In this question the logic is to solve |x + 1| = 2|x -1| and to solve |x - 3| > 0 and to see which one will give us the answer for Is |x| < 1 ?

So if we solve |x + 1| = 2|x -1| <==> x+1= 2*(x-1) or x+1= -2*(x-1) <==> x=3 or x=1/3 in this step we can’t say that Is |x| < 1 or not because 3 >1

If we solve |x-3| >0 <==> x>3 or x<3 ==> x<>3 this inequality alone don’t also answer the question.

BUT (1)+(2) <==> x=1/3 and now we can say that |x|<1.

Thanks.

correct on all counts.

[viksnme]

In this question the logic is to solve |x + 1| = 2|x -1| and to solve |x - 3| > 0 and to see which one will give us the answer for Is |x| < 1 ?

So if we solve |x + 1| = 2|x -1| <==> x+1= 2*(x-1) or x+1= -2*(x-1) <==> x=3 or x=1/3 in this step we can’t say that Is |x| < 1 or not because 3 >1

If we solve |x-3| >0 <==> x>3 or x<3 ==> x<>3 this inequality alone don’t also answer the question.

BUT (1)+(2) <==> x=1/3 and now we can say that |x|<1.

Thanks.

correct on all counts.

Hi Ron, I guess if Scenario 1 (i.e. when x is less than -1) as mentioned by Stacey earlier is also considered then this solution would be correct on all counts. Trust this is ok.

Regards.

[StaceyKoprince]

Yes - though technically, if you continue on with the math, what you discover when you solve for the scenario x<-1 is that x would be equal to 3. That contradicts the parameter of the scenario (that is, x<-1), so it's an invalid solution and we can ignore that scenario. Since this is data sufficiency, it didn't matter - we already knew the statement was insufficient alone and, even with the two statements together, the value of x=3 is the same value you get for the positive scenario, and x=3 is the possibility that gets nixed when you combine the statements. But if it were problem solving, we would care!

[arunpal]

Hi - after reading Stacey's response to prove out statement 1, I am very confused as to how we go about using statement 2. This is a very tricky problem...would you say it is high 700 level?

thx
ARUN

[JonathanSchneider]

In my opinion, Arun, yes, this is definitely a 700+ level problem. I get asked about it in Office Hours quite a bit - a great challenge. I also tend to find that people have a difficult time even with relatively simple absolute value expressions. I highly recommend that whenever you see an absolute value sign or an inequality (they often come together), you draw out a number line to help visualize the problem.

[earthling]

Guys,

I am confused with Stacy's remark "Scenario 3: One positive, one negative. And now I've already got an idea of the range of numbers I should be testing, because Scenario 1 resulted in x < -1 and Scenario 2 resulted in x > 1. So... let's see what would happen if x is between those two numbers. For |x + 1|, if -1 < x < 1, then the contents will be positive. For |x -1|, if -1 < x < 1, then the contents will be negative. "

Why wouldn't you test two more scenarios i.e. Left Side is negative and right sand is positive or Right Side is -ve and left side is positive. This we can establish the condition that x not equal to 1. I guess my confusion is what condition are we testing in Stacy's remark for scenario 3? How does it help in knowing that the left side is positive and right is -ve?

Any help will be appreciated. Thanks