If a b and |a-b| = b-a, which of the following statements

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.

Answer: C.

For more check here: math-absolute-value-modulus-86462.html
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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

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.

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"

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)\(a<b\)

Ahh - I forgot about that. Thanks!

Another thing to remember is that, because of the absolute value

|a-b| ≤ b-a

because the absolute value either turns a negative positive or leaves a positive the same..

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)

Since, it is given that
|a-b| = b-a, we can infer following from Absolute Value -
|a-b| = -(a-b) = b-a => a-b must be negative or 0
i.e. a-b < 0 (a=B, so a-b cannot be 0)
=> a<b Hence, only option III satisfies.
Now, let us see other options -

I. a < 0 even without plugging numbers, we can see this cannot be true always. No info ob B, and depending upon b the equation may vary.
II. a + b < 0 B=Not neccessarily tru, since we already know that only a<b is sufficient. It foes not matter whether a is +ve or _ve as long as it is less than b.
III. a < b

Hence, C only option III works.
Tip: Sometimes, I do not tend to plug numbers because they eat away time, and questions like (oops such as ) these can be solve without this strategy.

I am really struggling with the below line

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 \(a<b\)

Why does \(|a-b| = b-a=-(a-b)\)?

I thought you have to check if | a-b| > 0 and < 0?

|a-b| = -(a-b) <---- given

Also, we know that if |x| = -x then x<0 therefore, (a-b) < 0 -----> a<b

|a-b| = |b-a| (GMAT property)

Let x=|b-a|

--> |x|=x
As per the graph of the two, only possible when x>=0

--> |b-a| >= 0 (we can ignore the = as b not equals a)
--> b-a>0 ---> b>a

Hence the answer is C (Only III follows)

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