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 Post subject: GMAT AlgebraPosted: Sat Apr 07, 2012 6:47 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
For any a and b that satisfy |ab| = ba and a > 0, then
|a + 3| + |-b| + |ba| + |ab| =

A. -ab + 3
B. ab + 3
C. -ab + 2b + 3
D. ab + 2b – 2a – 3
E. ab + 2b + 3

(E) We know that |ab| = ba, so ba > 0, or b > a. We also know that a > 0, so b > 0.

Knowing that both a and b are positive we can easily simplify:
|a + 3| + |-b| + |ba| + |ab| =
a + 3 + b + ba + ab =
ab + 2b + 3

The right answer is choice (E).
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How would the equation be exactly transformed given that either a or b or both are negative?

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 Post subject: Re: GMAT AlgebraPosted: Sat Apr 07, 2012 7:07 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
How would the equation be exactly transformed given that either a or b or both are negative?
I'd like to stress that this is a hypothetical situation, because in the question we show that b > a > 0.

So let's try the proposed conditions.
|a + 3| + |-b| + |ba| + |ab| = ?

1) b < 0
In this case we can simplify |-b| = -b, because -b > 0. But we can NOT simplify any other absolute value, because we know nothing about a.

2) a < 0
In this case we can NOT simplify any absolute value, because

|a + 3| can be a + 3 (if a is from -3 to 0)
OR
|a + 3| can be -a – 3 (if a is less than -3)

Any other absolute value contains b and we know nothing about it.

3) a < 0 and b < 0
In this case we can simplify:
|-b| = -b, because -b > 0;
|ab| = ab, because ab > 0.

We can NOT simplify:
|a + 3| for the same reasons as in 2)
|ba| can be ba (if b > a)
|ba| can be -b + a (if b < a).

You may try to plug some numbers in the absolute values we could NOT simplify:
3) a < 0 and b < 0
a = -1 < 0 , b = -4 < 0
|a + 3| = |-1 + 3| = -1 + 3 = 2
in this case we used |a + 3| = a + 3, because (a = -1 is between -3 and 0)
|ba| = |-4 – (-1)| = -(-4) + (-1) = 3
in this case we used |ba| = -b + a, because (b = -4 < -1 = a)

a = -4 < 0 , b = -1 < 0
|a + 3| = |-4 + 3| = -(-4) – 3 = 1
in this case we used |a + 3| = -a – 3, because (a = -1 is less than -3)
|ba| = |-1 – (-4)| = -1 – (-4) = 3
in this case we used |ba| = ba, because (b = -1 > -4 = a)

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