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Gmat Prep DS 2

Expert replies
Source: — Data Sufficiency |

by Anurag@Gurome » Fri Jun 22, 2012 10:28 am
bobdylan wrote:Is √(x-3)^2 = 3-x ?

1) x ≠ 3
2) -x | x| >0
√[(x-3)^2] = |x - 3|

Now, for x ≥ 3, |x - 3| = (x - 3)
And, for x < 3, |x - 3| = -(x - 3) = (3 - x)

Hence, the question boils down to whether x is less than 3 or not.

Statement 1: x can have any value other than 3 --> Not sufficient

Statement 2: x|x| < 0 ---> x < 0 < 3 --> Sufficient

The correct answer is B.
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by bobdylan » Fri Jun 22, 2012 11:51 am
Mr Anurag, can you please explain a little more in detail. I'm not grasping the concept behind this question.
Thanks.
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by Anurag@Gurome » Fri Jun 22, 2012 12:05 pm
bobdylan wrote:Mr Anurag, can you please explain a little more in detail. I'm not grasping the concept behind this question.
It would've helped if you had pointed out which steps you do not understand. Anyway I'm guessing that it is "√[(x-3)^2] = |x - 3|"

Every positive number x has two square roots: √x, which is positive, and -√x, which is negative. Together, these two roots are denoted ±√x. The positive square root, √x is known as the principal square root of x.

By definition √x is positive.
Note that the positive or negative sign comes before the √ sign. That's your clue.

Whenever the square root sign '√' is used it is used to mean the principal square root, i.e. the positive square root. Hence, square roots of 4 are ±√4, i.e. -2 and 2. But √4 is always equal to 2 NOT -2.

To remove ambiguities we use the modulus notation.
We write √(x²) = |x|, so that we always get the principal square root, i.e. the positive square root. Now, |x| is always positive. Hence,

For x > 0 --> √(x²) = |x| = x > 0
For x < 0 --> √(x²) = |x| = -x > 0

Thus we always get the positive square root.

Let's take two examples,
1. For x = 2 = √(x²) = √[(2)²] = |2| = 2
2. For x = -2 = √(x²) = √[(-2)²] = |-2| = 2
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by GMATGuruNY » Sat Jun 23, 2012 4:44 am
vittalgmat wrote:Is sqrt [(x-3)^2] = 3-x

1) x is not equal to 3
2) -x*|x| > 0

not sure of the OA B
Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.

Question rephrased: Is |x-3| = 3-x?
In other words:
Is the DISTANCE between x and 3 equal to the DIFFERENCE between 3 and x?

A DIFFERENCE can be negative, 0, or positive.
A DISTANCE must be greater than or equal to 0.
For the DIFFERENCE between two values to be equal to the DISTANCE between the two values, the DIFFERENCE -- like the DISTANCE -- must be greater than or equal to 0:
3-x≥0
x≤3.

Question rephrased: Is x≤3?

Statement 1: x is not equal to 3.
It is possible that x<3 or that x>3.
INSUFFICIENT.

Statement 2: -x*|x| > 0 .
Thus, the left-hand side must be (+)(+) or (-)(-).
Since |x| cannot be negative, both factors on the left-hand side must be positive.
Thus:
-x>0
x<0.
Since x<0, we know that x≤3.
SUFFICIENT.

The correct answer is B.
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by gmattesttaker2 » Sun Jun 24, 2012 5:00 pm
GMATGuruNY wrote:
vittalgmat wrote:Is sqrt [(x-3)^2] = 3-x

1) x is not equal to 3
2) -x*|x| > 0

not sure of the OA B
Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.

Question rephrased: Is |x-3| = 3-x?
In other words:
Is the DISTANCE between x and 3 equal to the DIFFERENCE between 3 and x?

A DIFFERENCE can be negative, 0, or positive.
A DISTANCE must be greater than or equal to 0.
For the DIFFERENCE between two values to be equal to the DISTANCE between the two values, the DIFFERENCE -- like the DISTANCE -- must be greater than or equal to 0:
3-x≥0
x≤3.

Question rephrased: Is x≤3?

Statement 1: x is not equal to 3.
It is possible that x<3 or that x>3.
INSUFFICIENT.

Statement 2: -x*|x| > 0 .
Thus, the left-hand side must be (+)(+) or (-)(-).
Since |x| cannot be negative, both factors on the left-hand side must be positive.
Thus:
-x>0
x<0.
Since x<0, we know that x≤3.
SUFFICIENT.

The correct answer is B.
Hi Mitch,

Thanks for the explanation. In statement 2 explained above though I was not clear about how we are getting x≤3. x<3 is clear but not the x=3 part.

Since x<0, we know that x≤3.

I was wondering if you can please explain? Thanks for your valuable time and help.

Best Regards,
Sri
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