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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: \(|5|=5\);

So, for our case, since \(a<1\), then \(a-1<0\) hence according to the above \(|a-1|=-(a-1)=1-a\).

Remember these things regarding absolute values in the GMAT (1) \(\sqrt{x^2}=|x|\) (2) |x| = x ==> if x > 0 (3) |x| = -x ==> if x < 0

Golden rules! Must memorize!

Solution: ***Transform the equation: \(\frac{|1+a| + |a-1|}{|1+a| - |a-1|}\)

***Since we know that a is a positive fraction as given: 0<a<1 ***This means 1+a is always positive. Using property#2 above, |1+a| = 1+a ***For a-1 = (fraction) - 1, we know that a-1 is negative. Using property #3, |a-1| = -(a-1)

Hi Bunuel, I was also thinking along the same lines as you in this question.However, I got stuck deciding which rule to apply to which equation. How did you decide this. Couldn't you have swapped the rules then? |1+a|=-1-a And |a-1|=a-1

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: \(|5|=5\);

So, for our case, since \(a<1\), then \(a-1<0\) hence according to the above \(|a-1|=-(a-1)=1-a\).

Similarly as \(0<a<1\), then \(1+a>0\), hence \(|1+a|=1+a\).

since its mentioned that a is between 0 and 1,we can directly use any of the numbers,say a=1/2.Substitute this in the equation,in the end ul get 2 as the answer. Looking at ans choices,since we started out with a=1/2, 1/a gives us our answer that is 2

So...0<a<1 which means that |a+1| will always be positive. However |a-1| = |some negative value| so, |some negative value| = -(some negative value) therefore, in this case:

|a-1| = -(a-1) ==> =1-a.

Is this correct?

Bunuel wrote:

sharmila79 wrote:

Hi Bunuel, Could you please explain me how a-1 becomes 1-a in the numerator and denominator in the second part of the problem? Thanks!

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

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

So, for our case, since \(a<1\), then \(a-1<0\) hence according to the above \(|a-1|=-(a-1)=1-a\).

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