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a^2 - b^2 = b^2 - c^2 . Is a = |b| ? 1. b = |c| 2. b = |a| *

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VP
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a^2 - b^2 = b^2 - c^2 . Is a = |b| ? 1. b = |c| 2. b = |a| * [#permalink]

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New post 09 May 2009, 11:29
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D
E

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\(a^2 - b^2 = b^2 - c^2\) . Is \(a = |b|\) ?

1. \(b = |c|\)
2. \(b = |a|\)

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

What is the difference between a=|b|, |a|=b and |a|=|b|? I can see that a is +ve in one case and b is +ve in other, but if we square to get rid of the modulus operator all of them are the same right?

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Re: a=|b|?? [#permalink]

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New post 09 May 2009, 15:58
No It is not C try again.

When ever I see modulus, I tend to do two things

(1) get rid of the modulus by squaring on both sides

(2) consider all cases.

When to do (1) and when to do (2)? Any ideas


Here is what I did and got it wrong

a^2 - b^2 = b^2 - c^2 . Is a = |b| ?

1. b = |c|
2. b = |a|

Q is asking Is a = |b| ? ie a^2=b^2 (Do you guys agree?)

(0) a^2 - b^2 = b^2 - c^2

(1) b = |c| ie b ^2 = c^2

Substitute 1 in (0) , (0) becomes a^2=b^2

In 2 it is given that b=|a|, if i square it off it is b^2=a^2

Apparently I arrived at D and it is not the answer.

So I am wondering when to square and when to consider the +ve and _ve of the absolute value.

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Re: a=|b|?? [#permalink]

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New post 10 May 2009, 01:23
i also go along with E.

Q is asking Is a = |b| ? ie a^2=b^2 (Do you guys agree?). No.
You know clearly that a^2=b^2 a can be +b or -b.

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Re: a=|b|?? [#permalink]

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New post 10 May 2009, 07:08
a^2 - b^2 = b^2 - c^2 .

1. b = |c|
2. b = |a|

Look at these expressions carefully. They are not sensitive to substitution a for -a. And if a = |b|, a can be also a=-|b|
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Re: a=|b|?? [#permalink]

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New post 10 May 2009, 09:31
icandy wrote:
What is the difference between a=|b|, |a|=b and |a|=|b|? I can see that a is +ve in one case and b is +ve in other, but if we square to get rid of the modulus operator all of them are the same right?


1. If a = |b|, a is +ve but b could be +ve or -ve.
2. If b = |a|, b is +ve but a could be +ve or -ve.
3. If |b| = |b|, a and b could be +ve or -ve.

icandy wrote:
\(a^2 - b^2 = b^2 - c^2\) . Is \(a = |b|\) ?

1. \(b = |c|\)

2. \(b = |a|\)


\(a^2 - b^2 = b^2 - c^2\)
\(a^2 +c^2 = 2b^2\)

The basic question: is "a" +ve?

1. \(b = |c|\)
\(b^2 = c^2\)

\(a^2 + c^2 = 2b^2\)
\(a^2 + b^2 = 2b^2\)
\(a^2 = b^2\)
\(|a|=|b|\)
Since b is +ve, \(b = |a|\) .........nsf...

2. \(b = |a|\)
Its the same as avove in 1. So nsf again ....

Togather: They both are same. So nsf...
E.
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Re: a=|b|?? [#permalink]

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New post 10 May 2009, 09:39
walker wrote:
And if a = |b|, a can be also a=-|b|


How is that possible?
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Re: a=|b|?? [#permalink]

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New post 10 May 2009, 10:51
GMAT TIGER wrote:
walker wrote:
And if a = |b|, a can be also a=-|b|


How is that possible?


a^2 - b^2 = b^2 - c^2 .

1. b = |c|
2. b = |a|

values of a^2 and |a| is not sensitive to sign of a. So, if positive a is a solution, -a is also solution. In our case if a=|b| is a solution, a=-|b| is also solution and answer is E
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Re: a=|b|?? [#permalink]

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New post 17 May 2009, 23:51
First of all, if a^2 = b^2, then this means |a| = |b| and not a = |b| or b = |a|.

With this, stmt1 tells that b^2 = c^2. Hence, from the original equation, a^2 = b^2 and hence |a| = |b|.

Now, if I combine the second stmt, then,
b = |b|. That means, b is positive. But, this still does not tell whether a is +ve or -ve.

Hence, answer should be E.

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Re: a=|b|??   [#permalink] 17 May 2009, 23:51
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a^2 - b^2 = b^2 - c^2 . Is a = |b| ? 1. b = |c| 2. b = |a| *

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