## Between A and B

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### Between A and B

by adi » Fri Mar 14, 2008 7:48 pm
If a and b are nonzero numbers on the number line, is 0 between a and b?

(1) The distance between 0 and a is greater than the distance between 0 and b.
(2) The sum of the distances between 0 and a and between 0 and b is greater than the distance between 0 and the sum of a + b.

Answer should be "Statement (2) alone is sufficient.

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by simplyjat » Fri Mar 14, 2008 10:28 pm

Coming back to the question, when dealing with number line and distances, always think absolute values.
1) |a| > |b|
2) |a| + |b| > |a + b|

Here the second statement is true only when a and b have different sign, because with same signs |a| + |b| = |a + b|.
So we have two numbers with different signs 0 will always be in between...
simplyjat

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by kaf » Mon Apr 13, 2009 3:06 pm
Can someone please elaborate on this solution

thanks

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by vittalgmat » Mon Apr 13, 2009 3:37 pm
kaf wrote:Can someone please elaborate on this solution

thanks
Simplyjat has done excellent job in explaining..
I will try add my 2c.

Try plugging numbers to the absolute equation mentioned above.
But whenever u plug in numbers assume they are numbers on a numberline.
And whenever u see a absolute mentioned, immediately think of as a distance of that number from the origin.

Actually, u can see Stewart/Ian's comments on a previous Absolute!!! problem.

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by gmat740 » Tue Apr 14, 2009 8:28 pm
It is a good idea to draw a number line:

two cases are possible:

a..........b.........0

or a.........0........b

(I) this one talks about a.........0........b

distance between 0 and a> distance between 0 and b

so Insuff

(II)clearly speaks of this:
a.........0........b

the other case will not be satisfied

So (II) Suff

Hence B

Hope this helps

Karan

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by bhumika.k.shah » Sun Apr 04, 2010 12:30 pm
karan,
I did the same thing. though can u explain it with #s?

I'm guessing that would make it quicker and simpler.
gmat740 wrote:It is a good idea to draw a number line:

two cases are possible:

a..........b.........0

or a.........0........b

(I) this one talks about a.........0........b

distance between 0 and a> distance between 0 and b

so Insuff

(II)clearly speaks of this:
a.........0........b

the other case will not be satisfied

So (II) Suff

Hence B

Hope this helps

Karan

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by bhumika.k.shah » Sun Apr 04, 2010 12:31 pm
Also what would be the difficulty level of this Q? and how much time should one spend on this Q?
Can someone break down the second statement, please?

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by blee0810 » Sat Dec 04, 2010 5:39 pm
https://www.beatthegmat.com/a-and-b-on-t ... 37331.html

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by diebeatsthegmat » Mon Dec 06, 2010 11:08 am
adi wrote:If a and b are nonzero numbers on the number line, is 0 between a and b?

(1) The distance between 0 and a is greater than the distance between 0 and b.
(2) The sum of the distances between 0 and a and between 0 and b is greater than the distance between 0 and the sum of a + b.

Answer should be "Statement (2) alone is sufficient.
ohh my answer is B too,
what is the OA? " answer should be statement 2 alone is sufficient" is your answer or OA?

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by GMATGuruNY » Mon Dec 06, 2010 12:30 pm
adi wrote:If a and b are nonzero numbers on the number line, is 0 between a and b?

(1) The distance between 0 and a is greater than the distance between 0 and b.
(2) The sum of the distances between 0 and a and between 0 and b is greater than the distance between 0 and the sum of a + b.

Answer should be "Statement (2) alone is sufficient.
Absolute value means distance from 0.

Statement 1: |a| > |b|
Plug in a=3, b=2.
This works, because |3| > |2|.
Is 0 between 3 and 2? No.

Plug in a= -3, b=2.
This works, because |-3| > |2|.
Is 0 between -3 and 2? Yes.

Since the answer can be both No and Yes, insufficient.

Statement 2: |a| + |b| > |a+b|
Plug in a=3, b=2.
|3| + |2| > |3+2|. Doesn't work, because |3| + |2| = |3+2|.
This shows us that a and b cannot both be positive.

Plug in a = -3, b = -2.
|-3| + |-2| > |-3 + (-2)|. Doesn't work, because |-3| + |-2| = |-3 + (-2)|.
This shows us that a and b cannot both be negative.

Since a and b cannot both be positive -- nor can they both be negative -- 0 must be between them. Sufficient.

For example:
Plug in a=-3, b=2.
This works, because |-3| + |2| > |-3+2|.
Is 0 between -3 and 2? Yes.

Plug in a=3, b=-2.
This works, becuase |3| + |-2| > |3+(-2)|.
Is 0 between 3 and -2? Yes.

Since the answer is consistently Yes, sufficient.
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