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Gmac retired test question

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by msd_2008 » Tue Dec 02, 2008 5:38 pm
Hi,

Need help with the following question:

1. 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

OA is B. Dont know how.

Regards
MSD
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by muzali » Tue Dec 02, 2008 5:56 pm
ab=0 implies a+b-ab=0
or a+b=ab

Putting b=1 gives a+1 =a ----> not possible.

so b =1 is the answer
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by msd_2008 » Tue Dec 02, 2008 7:02 pm
But then same can be true for b=2. What makes b=1 different and unique in this case?

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by hwiya320 » Tue Dec 02, 2008 8:24 pm
msd_2008 wrote:But then same can be true for b=2. What makes b=1 different and unique in this case?

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MSD
when a +b = ab, b=1, that will give you a + 1 = a*1
a+1 cannot be a.

you said same for b = 2.

a + b = 2a, this can't be true?
if a was equal to 2,
2+2 = 2x2, correct, so this CAN be true.
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Re: Gmac retired test question

by hwiya320 » Tue Dec 02, 2008 8:31 pm
msd_2008 wrote:Hi,

Need help with the following question:

1. 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

OA is B. Dont know how.

Regards
MSD
We'll try each value choices for verification.
Since we know that a+b-ab=0, we know a+b = ab
(A) a+2 = a*2, true when a=2
(B) a+1 = a*1, no value can give a+1=a, check
(C) a+0 = a*0, true when a=0
(D) a-1 = a*-1 or -a, true when a=1/2
(E) a-3/2 = a*3/2, true when a = -3
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