Confusing Inequalities + absolute value

This topic has expert replies
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Wed Oct 28, 2009 1:07 am

Confusing Inequalities + absolute value

by yregister » Fri Nov 06, 2009 8:06 pm
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

Junior | Next Rank: 30 Posts
Posts: 19
Joined: Fri Oct 23, 2009 5:59 pm
Thanked: 2 times

by Harbinder » Fri Nov 06, 2009 8:28 pm
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

Master | Next Rank: 500 Posts
Posts: 199
Joined: Sat Oct 24, 2009 4:43 pm
Thanked: 22 times
GMAT Score:710

by palvarez » Fri Nov 06, 2009 8:30 pm
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.

Master | Next Rank: 500 Posts
Posts: 138
Joined: Mon Mar 02, 2009 12:02 pm
Thanked: 15 times

by life is a test » Sat Nov 07, 2009 3:20 am
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

GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pm
Location: Toronto
Thanked: 539 times
Followed by:164 members
GMAT Score:800

Re: Confusing Inequalities + absolute value

by Testluv » Sat Nov 07, 2009 8:17 pm
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
Kaplan Teacher in Toronto