How to solve this

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How to solve this

by gmattester » Thu Aug 21, 2008 6:17 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) 2

After getting ab=a+b, looks like all can be values of b
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Re: How to solve this

by Sunny22uk » Thu Aug 21, 2008 6:31 pm
gmattester wrote: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) 2

After getting ab=a+b, looks like all can be values of b
if you substitute b as 1, you get a=a+1 which cannot be true.
The OA should be be B
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Re: How to solve this

by gmattester » Thu Aug 21, 2008 6:57 pm
Sunny22uk wrote:
gmattester wrote: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) 2

After getting ab=a+b, looks like all can be values of b
if you substitute b as 1, you get a=a+1 which cannot be true.
The OA should be be B
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Re: How to solve this

by 4meonly » Fri Aug 22, 2008 5:45 am
Sunny22uk wrote:
gmattester wrote: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) 2

After getting ab=a+b, looks like all can be values of b
if you substitute b as 1, you get a=a+1 which cannot be true.
The OA should be be B
So, what is the answer?

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by leugene » Fri Aug 22, 2008 1:49 pm
Algebraically, if you think about it this way:

a+b-ab=0

a-ab=-b (just reworking the problem to isolate a)

a(1-b)=-b (still working)

a=[(-b)/(1-b)]

And since you can't have 0 be in the denominator, the only way that would occur is if b=1. Therefore, b cannot equal 1, so answer is B.