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Re: For any non-zero a and b that satisfy |ab|=ab and |a|=-a [#permalink]

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04 Dec 2012, 02:30

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Given: |ab| = ab and |a| = -a Question: |b-4| + |ab-b| = ?

**** Looking at |ab| = ab tells us that a and b are either both positive or negative **** Looking at |a| = -a tells us that a must be negative **** Combine two observations: a and b are both negative values

Let a=-1 and b=-1 |b-4| + |ab-b| = |-1-4| + |1-(-1)| = 7

Test a) ab-4 = (-1)(-1)-4 = -3 Test b) 2b-ab-4 = (2)(-1) - (1) - 4 = -7 Test c) ab+4 = 1 + 4 = 5 Test d) ab-2b+4 = 1-2(-1)+4=7 BINGO!

Re: For any non-zero a and b that satisfy |ab| = ab and |a| = -a [#permalink]

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04 Jul 2013, 06:11

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gmatfrenzy750 wrote:

For any non-zero a and b that satisfy |ab| = ab and |a| = -a, |b-4| + |ab-b| =

A. ab-4 B. 2b-ab-4 C. ab+4 D. ab-2b+4 E. 4-ab

1.We need to find whether ab and b are -ve or +ve 2. Since |ab|=ab, ab is +ve 3. Since |a|=-a, a is -ve. 4. From (2) and (3), b is -ve. 5.Since b is -ve b-4 is -ve and so |b-4| becomes -(b-4) 6. Since ab is +ve and b is -ve, ab-b is +ve and |ab-b| becomes ab-b 7. (5) + (6) = -b+4+ab-b= ab-2b+4
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Re: For any non-zero a and b that satisfy |ab| = ab and |a| = -a [#permalink]

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09 Jul 2013, 16:49

For any non-zero a and b that satisfy |ab| = ab and |a| = -a, |b-4| + |ab-b| =

|ab| = ab and |a| = -a, |b-4| + |ab-b| =

From |a| = -a we get that a is negative because: |a| = -a and -a must be positive as it is set to an absolute value so: |a| = -(-a) a=a

If a is negative then from |ab| = ab we get that b must be negative as well because: |ab| = (-a)b (-a)b must be positive as it is set to an absolute value |ab| = (-a)(-b) ab=ab

So, both a and b are negative. |b-4| + |ab-b| =

b is negative and ab is positive. Also, because b is negative we know that (ab-b) = (ab-[-b]) = (ab+b)

Re: For any non-zero a and b that satisfy |ab| = ab and |a| = -a [#permalink]

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21 Jul 2014, 07:40

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Re: For any non-zero a and b that satisfy |ab| = ab and |a| = -a [#permalink]

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Re: For any non-zero a and b that satisfy |ab| = ab and |a| = -a [#permalink]

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01 Jan 2017, 12:22

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