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Thank's in advance for helping to solve the problem, the OA should be ( C ) , but I'm not sure 100% about it; a friend gave to me several GMAT exercises for training.

Re: If a ≠ b and |a-b| = b-a, which of the following statements [#permalink]

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27 Mar 2013, 05:06

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If a ≠ b and |a-b| = b-a, which of the following statements must be true ?

I. a < 0 II. a + b < 0 III. a < b

(A) None (B) I only (C) III only (D) I and II (E) II and III

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

Thus, according to the above, since \(|a-b| = b-a=-(a-b)\), then \(a-b\leq{0}\) --> \(a\leq{b}\). Since we also know that \(a\neq{b}\), then we have that \(a<b\). So, III is always true.

As for the other options: I. a < 0 --> not necessarily true, consider a=1 and b=2. II. a + b < 0 --> not necessarily true, consider a=-2 and b=-1.

Re: If a ≠ b and |a-b| = b-a, which of the following statements [#permalink]

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26 Mar 2013, 09:32

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I think the best way here is using real numbers

I) a < 0 a=1, b=5 |1-5|=5-1 4=4, so I is not always true

II) a + b < 0 a=3,b=4 |3-4|=4-3 1=1, so II is not always true

III) a < b |a-b| = b-a if \(a>b\) then \(|a-b| = a-b\) and doesn't equal b-a if \(b>a\) then \(|a-b| = -(a-b) = -a+b = b-a\) so III is true _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If a b and |a-b| = b-a, which of the following statements [#permalink]

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15 Jun 2013, 23:32

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WholeLottaLove wrote:

So, in other words,

I.) |a-b| = b-a II.) b-a is positive because it is equal to an absolute value III.) b must be greater than a because b-a is positive IV.) a-b must be negative V.) |a-b| = -(a-b) VI.) a-b ≤ 0 VII.) a ≤ b

Yes, perfect. Just remember that we are told that \(a\neq{b}\) so

VII)\(a<b\) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If a ≠ b and |a-b| = b-a, which of the following statements [#permalink]

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27 Mar 2013, 00:26

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The given conditions are : i) a is not equal to b ,i,e a-b is non zero. ii) |a -b | = b-a ,i,e -(a-b). So ,considering the above conditions, a - b < 0 => a < b. _________________

Re: If a b and |a-b| = b-a, which of the following statements [#permalink]

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15 Jun 2013, 17:58

So, in other words,

I.) |a-b| = b-a II.) b-a is positive because it is equal to an absolute value III.) b must be greater than a because b-a is positive IV.) a-b must be negative V.) |a-b| = -(a-b) VI.) a-b ≤ 0 VII.) a ≤ b

Zarrolou wrote:

WholeLottaLove wrote:

Because |a-b| = b-a, could we say that b-a is positive (because it is equal to an abs. val.) and therefore, b must be greater than a?

Also, I first tired to solve this problems by:

|a-b| = b-a so:

a-b = b-a 2a = 2b a=b (which isn't true as the stem tells us it isn't)

OR

-a+b=b-a 0=0

But I'm not sure how to interpret that result. Is that a valid way to solve the problem?

The second result tells you that whatever value \(a\) and \(b\) have, that equation will always be true: \(0=0\) always.

0=0 means that that case will always hold, hence that case (b>a) will always be "true"

Re: If a b and |a-b| = b-a, which of the following statements [#permalink]

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16 Jun 2013, 07:58

Ahh - I forgot about that. Thanks!

Zarrolou wrote:

WholeLottaLove wrote:

So, in other words,

I.) |a-b| = b-a II.) b-a is positive because it is equal to an absolute value III.) b must be greater than a because b-a is positive IV.) a-b must be negative V.) |a-b| = -(a-b) VI.) a-b ≤ 0 VII.) a ≤ b

Yes, perfect. Just remember that we are told that \(a\neq{b}\) so

Re: If a b and |a-b| = b-a, which of the following statements [#permalink]

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13 May 2016, 04:42

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Thank's in advance for helping to solve the problem, the OA should be ( C ) , but I'm not sure 100% about it; a friend gave to me several GMAT exercises for training.

Lets interpret the LHS first :----> |a-b| anything that comes out of a mod is positive ; therefore |a-b|=postive ..........equation 1 Now lets check the RHS -------> b-a The question stem tells us that |a-b|=b-a .............equation 2 Rearranging equation 2 gives us b-a =|a-b| .............. equation 3

Equation 1 tells that |a-b| is positive Put value of |a-b| from equation 1 into equation 3 b-a = |a-b|=positive b-a =positive

This is only possible when b > a we cannot say anything with mathematical surety except that b>a

Hence the answer is C (III only) _________________

Posting an answer without an explanation is "GOD COMPLEX". The world does't need any more gods. Please explain you answers properly.

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