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by cramya » Fri Jan 02, 2009 2:50 pm
Lets substitue b's value starting with choice A and see.

b=2

a+b-ab = 0
a+2-2a = 0

a = 2
b=2
a+b-ab=0
Possible

b=1

a+1-a = 0
1 = 0 Not possible

b cannot be 1

Choose B)
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Re: variables

by steveyb » Fri Jan 02, 2009 2:52 pm
[email protected] wrote:16. For any numbers a and b, a • b = a + b – ab.
If a • b = 0, which of the following CANNOT be a value of b?
(A) 2
(B) 1
(C) 0
(D) -1
(E) -3/2

b
0 = a + b - ab
-b = a - ab
-b = a(1-b)
-b/(1-b) = a

thus, b cannot equal 1, or else the denominator will = 0
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by cramya » Fri Jan 02, 2009 2:54 pm
Stevey,
Good algebric approach!

Regards,
CR
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Re: variables

by Brent@GMATPrepNow » Fri Jan 02, 2009 2:54 pm
[email protected] wrote:16. For any numbers a and b, a • b = a + b – ab.
If a • b = 0, which of the following CANNOT be a value of b?
(A) 2
(B) 1
(C) 0
(D) -1
(E) -3/2

b
This is a wonderful question.
set a + b – ab = 0
a + b(1-a) = 0 (factor)
b(1-a)=-a
b=-a/(1-a) or b=a/(a-1) (isolate b)

We can set a/(a-1) equal to each of the answer choices and see whether there is a value for a to let b equal each of those values.

Or we can recognize that a/(a-1) can never equal 1 since a can never equal a-1

So the answer is B
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by ronniecoleman » Sun Jan 04, 2009 10:00 pm
. For any numbers a and b, a • b = a + b – ab.
If a • b = 0, which of the following CANNOT be a value of b?
(A) 2
(B) 1
(C) 0
(D) -1
(E) -3/2

IMO B

if a = 1

1+b -b != 0

for any values of b ...
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