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Absolute Value Strategy

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Absolute Value Strategy

by michaelfaulkner » Thu Feb 17, 2011 6:27 am
I can't figure out, for an absolute value problem, if and when I should separate the equation into two or more equations when testing cases.

For example, look at this problem:

is x|x| < 2^x?
(1) x < 0
(2) x=-10

So for item (1), if I test negative cases directly into the equation, I get yes on every instance. However, my first instinct for these types of problems is to break the absolute value equation into two equations:

x|x| < 2^x equals...

x^2 < 2^x AND -x(x) > 2^x which equals
-(2^x) < x^2 < 2^x

If I test cases for (1) using that equation, I get yes and no depending on whether I use a negative fraction or negative whole number. The answer to this problem is D, so this method is obviously wrong.

Why doesn't this method work for this problem?
And what is the rule of thumb for when I should leave an absolute value testing cases problem as is versus breaking down into two equations?

I always struggle with this.

Thank you,
Michael
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by DarkKnight » Thu Feb 17, 2011 7:39 am
Michael,

I think you are making it more complicated that it actually is. lets see:

for statement 1: if x<0

that means x*abs(x) is negative. Where as 2^x is a positive value because 2^(-x) = 1/2^x which is positive for any negative x. hence x*abs(x) < 2^x --- sufficient

Statement 2: x=-10 --- this is same as statement 1 (x is negative) but instead giving us definite value. Therefore sufficient.

Answer D.

Hope this helps
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by michaelfaulkner » Thu Feb 17, 2011 9:04 am
^ Right, I understand that.

However, my first instinct for these types of problems is to break it down to 2 equations to make it. For example, is |x| < 1, my first step is always: x<1 and x>-1. Making it: -1 < x < 1

My question is, why won't this approach work for this particular problem?
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by maihuna » Thu Feb 17, 2011 9:52 am
In given scenario it sufice to note whether x<0 as x|x| < 0 for x<0 while 2^x will always be positive.
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by GMATGuruNY » Thu Feb 17, 2011 12:22 pm
michaelfaulkner wrote:I can't figure out, for an absolute value problem, if and when I should separate the equation into two or more equations when testing cases.

For example, look at this problem:

is x|x| < 2^x?
(1) x < 0
(2) x=-10

So for item (1), if I test negative cases directly into the equation, I get yes on every instance. However, my first instinct for these types of problems is to break the absolute value equation into two equations:

x|x| < 2^x equals...

x^2 < 2^x AND -x(x) > 2^x
Your approach illustrates one of the dangers of using algebra when solving a problem with absolute values. The inequality -x(x) > 2^x has no solutions. The left side will always be negative (unless x=0, in which case the left side will equal 0); the right side will always be positive. So the left side can never be greater than the right side.

Here's a much safer approach:

Statement 1: x<0
Plugging in a negative value into x|x| < 2^x, we get:
negative * positive < positive.
negative < positive.
Sufficient.

Statement 2: x= -10
Since in statement 1 we discovered that x<0 is sufficient, knowing that x= -10 also is sufficient.

The correct answer is D.
Last edited by GMATGuruNY on Thu Feb 17, 2011 12:57 pm, edited 2 times in total.
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by michaelfaulkner » Thu Feb 17, 2011 12:47 pm
^Thank you for the great response.

So is it safe to assume that it is usually always a bad idea to use algebra to solve for data sufficiency absolute value questions?
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by GMATGuruNY » Thu Feb 17, 2011 1:55 pm
michaelfaulkner wrote:^Thank you for the great response.

So is it safe to assume that it is usually always a bad idea to use algebra to solve for data sufficiency absolute value questions?
I would tread carefully. The problem above involves a quadratic, making the use of algebra especially tricky.

When there's no quadratic, algebra can be helpful:

Is |2x+5| > |3x-2|?

Let's use algebra to find the points of intersection:
2x+5 = 3x-2
x = 7

2x+5 = -(3x-2)
2x+5 = 2-3x
5x = -3
x = -3/5

The points of intersection (x=7 and x= -3/5) are where the situation will change:

When x>7:
Plug in x=8:
|2x+5| = |2*8 + 5| = 21.
|3x-2| = |3*8 - 2| = 22.

This shows us that when x>7, |2x+5| < |3x-2|

When -3/5 < x < 7:
Plug in x=0:
|2x+5| = |2*0 + 5| = 5.
|3x-2| = |3*0 - 2| = 2.

This shows us that when -3/5 < x < 7, |2x+5| > |3x-2|.

When x<-3/5:
Plug in x= -1
|2x+5| = |2*-1 + 5| = 3.
|3x-2| = |3*-1 - 2| = 5.

This shows us that when x < -3/5, |2x+5| < |3x-2|.

Since |2x+5| > |3x-2| only when -3/5 < x < 7, the question above can be rephrased:

Is -3/5 < x < 7?

Using algebra helps us to rephrase the question, and rephrasing the question makes it much easier to evaluate the two statements. So algebra can certainly be helpful. Just be careful!
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