Confusing Inequalities + absolute value

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yregister
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Topic: Confusing Inequalities + absolute value
PostFri Nov 06, 2009 8:06 pm Reply with quote
Hi, writing my Gmat on 9th but still really confused about these inequalities. Can someone help and explain ?

if y >= 0, What is the value of x?

1) |x-3| >= y

2) |x-3| <= -y

OA:B
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Harbinder
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PostFri Nov 06, 2009 8:28 pm Reply with quote
Stmt 1 doesn't give a unique value but Stmt 2 can only be true if the both sides of the inequality are zero so x should be 3
so IMO the answer should be B
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PostFri Nov 06, 2009 8:30 pm Reply with quote
Well, there is no algorithmic approach; or such an algorithmic approach would be so inefficient that you end up wasting time.

Here, you are given that y is non-negative.

Look at (2) |x-3| <= -y

-y is either 0 or negative. However, |x-3| is either 0 or positive. So, |x-3| = -y = 0

x = 3. Sufficient on its own.


(1) |x-3| >= y >= 0
x value changes based on y. Insufficient.


Concept: absolute values are always non-negative. Geometrically, any distance is non-negative.

|x-3| >=0 as well.
depending on y, x value changes. Insufficient.
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PostSat Nov 07, 2009 3:20 am Reply with quote
notice that both statements are giving you the same information since the only the sign of the y variable differs on the right hand side, the absolute value remains unchanged.

from either of the statement we get x<=-y+3 or x<=y+3.
subtracting the eqns gives us -2y=0->y=0 so x<=3 which still doesnt help determine an exact x-value.

hope that helps
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PostSat Nov 07, 2009 8:17 pm Reply with quote
yregister wrote:
Hi, writing my Gmat on 9th but still really confused about these inequalities. Can someone help and explain ?

if y >= 0, What is the value of x?

1) |x-3| >= y

2) |x-3| <= -y

OA:B
Hi yregister,

Absolute value measures how far away from zero on the number line something is.

For example:
|x| = 5 would tell us that x is five units either to the right or else to the left of zero. Therefore, x either equals 5 or -5.

|x| < 5 would tell us that x is fewer than five units away from zero. So either x < 5 or else x >-5 (Remember, a "large" negative number is very small; visualize or draw a number line on these questions!)

Because absolute value is a measure of distance, it is always positive (or else zero).

Looking at this question:

The stem tells us that y is positive or zero. To have sufficiency, we need a single value for x.

Statement one:

1) |x-3| >= y

The equation is telling us that x-3 is at least y units away from zero on the number line. But remember that we don't know y's value. It can be zero or any positive number. x's value would clearly change if y was 10 versus 1,000.

Insufficient.

Statement two:

2) |x-3| <= -y

Okay, so we see the absolute value bars on the left. That means the left hand side is most definitely positive. Therefore, the right hand side must also be positive. (The two sides of the equation must equal each other). But, the equation tells us it is less than or equal to the right side...where we see a negative sign.

We have to fix this; we have to find a way to make the right hand side positive or zero. (We have to figure out how to satisfy the information in the statement--that info is always true!)

Well, y must be either positive or zero (from the stem).

But, if y were positive we would have:

|x-3| <= (-)(+)

And because the left hand side must be positive or zero, we would have:

pos or zero <= some negative number

That's clearly imposstible.

Thefefore, y can't be positive and must be zero.

Se we have:

|x-3| <= 0

But x-3 cannot be fewer than zero units away from zero (because distance is positive), so we have:

|x-3| = 0

x-3 = 0

x= 3

Statement one is not sufficient by itself; statement 2 is; choose B
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