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If |a+1|=|b-1|, what is the value of a-b? (1) ab>0

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If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post Updated on: 18 Feb 2017, 06:21
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Challenging Question!



If \(|a+1|=|b-1|\), what is the value of \(a-b\) ?

(1) \(ab>0\)

(2) \(\frac{a}{b}≠ -1\)

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Originally posted by hazelnut on 18 Feb 2017, 05:15.
Last edited by hazelnut on 18 Feb 2017, 06:21, edited 2 times in total.
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Feb 2017, 10:00
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ziyuenlau wrote:

Challenging Question!



If \(|a+1|=|b-1|\), what is the value of \(a-b\) ?

(1) \(ab>0\)

(2) \(\frac{a}{b}≠ -1\)



Hi...

Let us solve \(|a+1|=|b-1|\)...
Square both sides.. \(a^2+2a+1=b^2-2b+1\)..
\(a^2-b^2+2a+2b=0.....(a-b)(a+b)+2(a+b)=0.....(a-b+2)(a+b)=0\)..
So either (a-b)=-2 OR a=-b or both..

Let's see the statements..
(1) \(ab>0\)
So both a and b are of same sign..
If they are of same sign, a-b=-2..
Sufficient

(2) \(\frac{a}{b}≠ -1\)[/quote]
So a ≠ -b..
As per conditions if a is not equal to b, a-b=-2..
Sufficient

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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Feb 2017, 08:09
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Feb 2017, 09:28
Either of the statements is sufficient.
If |a+1|=|b-1| then there can be two cases
1. a and b both are of same sign, in this case the difference a-b will be -2 always
2. a and b are of different sign, in this case absolute value of a and b will be same

Now let's look at statements:
1. ab>0: so a and b are of same sign (both positive or negative)
- sufficient
2. Frac{a}{b}!= -1: so a and b are of same sign
- sufficient

So either statement is sufficient


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If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 25 Mar 2017, 15:52
Hello Chetan,

Unable to understand how S-1 and S-2 are sufficient. From question I infer a-b = -2 or a+b =0. But not sure how to relate this to S-1, S-2. It looks I am skipping a simple point. Request your help.
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 25 Mar 2017, 20:01
coolkl wrote:
Hello Chetan,

Unable to understand how S-1 and S-2 are sufficient. From question I infer a-b = -2 or a+b =0. But not sure how to relate this to S-1, S-2. It looks I am skipping a simple point. Request your help.


Hi

There are two possiblities as you have also mentioned..
1) a-b=-2
2) a=-b

Now SI tells us that ab>0, so both a and b are of same sign that is either BOTH are NEGATIVE or both are POSITIVE..
So a=-b is not TRUE...
Only possibility left is a-b=-2, so we can say a-b is -2 & this is what we have to find this sufficient

SII tells us a/b \(\neq{-1}\) or \(a \neq{-b}\)
So only second case possible is a-b=-2, again we can tell what a-b is..

Hope it helps
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If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Sep 2019, 07:05
chetan2u wrote:
coolkl wrote:
Hello Chetan,

Unable to understand how S-1 and S-2 are sufficient. From question I infer a-b = -2 or a+b =0. But not sure how to relate this to S-1, S-2. It looks I am skipping a simple point. Request your help.


Hi

There are two possiblities as you have also mentioned..
1) a-b=-2
2) a=-b

Now SI tells us that ab>0, so both a and b are of same sign that is either BOTH are NEGATIVE or both are POSITIVE..
So a=-b is not TRUE...
Only possibility left is a-b=-2, so we can say a-b is -2 & this is what we have to find this sufficient

SII tells us a/b \(\neq{-1}\) or \(a \neq{-b}\)
So only second case possible is a-b=-2, again we can tell what a-b is..

Hope it helps



Hi chetan2u,

This looks wrong. I believe the correct answer is A (ie. only statement (1) is sufficient alone).

The Q states: \(|a+1|=|b-1|\)

If \(\frac{a}{b}≠−1\), then there are two solutions, either \(a-b=-2\) or \(a-b=0\)

Example,\(a=3\) and\(b=5\), then both the original condition \(|3+1|=|5-1|\) and\(\frac{3}{5}≠−1\)are satisfied.

But also consider \(a=b=0\): \(|0+1|=|0-1|\) is satisfied. \(\frac{0}{0}≠−1\) is also satisfied, as \(\frac{0}{0}\) is indeterminate.

Hence statement 2 is not sufficient independently, as you do not have a unique solution for (\(a-b\)).
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Sep 2019, 13:23
chetan2u - Can we multiply a variable in an inequality?.. i suppose not
and then a/b!= -1 could be lot of things

Statement A should only be sufficient

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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Sep 2019, 18:42
dushyanta wrote:
chetan2u - Can we multiply a variable in an inequality?.. i suppose not
and then a/b!= -1 could be lot of things

Statement A should only be sufficient

Posted from my mobile device


Hi,

Yes, we cannot multiply inequality --< or >--, but we can always do it for = or \(\neq\)

For example \(\frac{a}{b}\neq{-1}\)
when is \(\frac{a}{b}=-1\)... ONLY when a=-b.
Now, opposite of above when will a/b NOT be -1 --- when a is NOT equal to -b
SO, if a =2, then for \(\frac{a}{b}\neq{-1}\), b should not be -(2) or -(a) or -a
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 18 Sep 2019, 18:44
ngmat12 wrote:
chetan2u wrote:
coolkl wrote:
Hello Chetan,

Unable to understand how S-1 and S-2 are sufficient. From question I infer a-b = -2 or a+b =0. But not sure how to relate this to S-1, S-2. It looks I am skipping a simple point. Request your help.


Hi

There are two possiblities as you have also mentioned..
1) a-b=-2
2) a=-b

Now SI tells us that ab>0, so both a and b are of same sign that is either BOTH are NEGATIVE or both are POSITIVE..
So a=-b is not TRUE...
Only possibility left is a-b=-2, so we can say a-b is -2 & this is what we have to find this sufficient

SII tells us a/b \(\neq{-1}\) or \(a \neq{-b}\)
So only second case possible is a-b=-2, again we can tell what a-b is..

Hope it helps



Hi chetan2u,

This looks wrong. I believe the correct answer is A (ie. only statement (1) is sufficient alone).

The Q states: \(|a+1|=|b-1|\)

If \(\frac{a}{b}≠−1\), then there are two solutions, either \(a-b=-2\) or \(a-b=0\)

Example,\(a=3\) and\(b=5\), then both the original condition \(|3+1|=|5-1|\) and\(\frac{3}{5}≠−1\)are satisfied.

But also consider \(a=b=0\): \(|0+1|=|0-1|\) is satisfied. \(\frac{0}{0}≠−1\) is also satisfied, as \(\frac{0}{0}\) is indeterminate.

Hence statement 2 is not sufficient independently, as you do not have a unique solution for (\(a-b\)).


In GMAT, we deal with ONLY real numbers...
That is why you will never see a case in GMAT when the denominator is 0... Although, the statement II could have mentioned \(b\neq{0}\)
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 19 Sep 2019, 05:11
Hi! Can anyone explain this with plotting the number line approach?
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0  [#permalink]

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New post 23 Sep 2019, 01:37
hazelnut wrote:

Challenging Question!



If \(|a+1|=|b-1|\), what is the value of \(a-b\) ?

(1) \(ab>0\)

(2) \(\frac{a}{b}≠ -1\)


We can also solve by making use of the below concept:

if |a|=|b| , then a=b OR a=-b

Applying the above concept we get:

|a+1|=|b-1|

a+1=b-1 OR a+1=1-b
=> a-b=-2----1) OR a=-b------2)

now, first says: ab>0, thus eqn 2 cannot satisfy as both are of opposite signs and hence eqn 1 suffices. -- Sufficient
second says : a/b not equal to -1, thus eqn 2 again cannot satisfy and hence eqn 1 suffices here too-- Sufficient.
D it is.
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Re: If |a+1|=|b-1|, what is the value of a-b? (1) ab>0   [#permalink] 23 Sep 2019, 01:37
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