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The operation (*)

Expert replies
by metallicafan » Tue May 07, 2013 11:59 am
If the operation (*) is defined for all integers a and b by a (*) b = a + b - ab, which of the following statements must be true for all integers, a, b, and c?

I. a (*) b = b (*) a
II. a (*) 0 = a
III. (a*b)*c = a*(b*c)

Here was a question that I was wondering if anyone could offer a simpler method for calculating out option III?, a simpler method than using algebra or picking numbers, please?
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Source: — Problem Solving |

by GMATGuruNY » Tue May 07, 2013 12:22 pm
If the operation @ is defined for all integers a and b by a@b = a+b-ab, which of the following statements must be true for all integers a, b, and c?

I. a@b = b@a
II. a@0 = a
III. (a@b) @ c = a @ (b@c)

I only
II only
I and II only
I and III only
I, II, and III
a@b = a + b - ab.
In other words, a@b = SUM - PRODUCT.

Statement I is included in four of the five answer choices.
Thus, it is almost certain that statement I must be true.
Otherwise, a test-taker will be able to eliminate four answer choices simply by evaluating statement I.
To save time, start with statement II.

To make the process easier, plug in values.
Let a=2, b=3, and c=10.

Statement II: a@0 = a
2@0 = 2
2+0 - (2*0) = 2
2 = 2.
On the left side, we simply added and subtracted 0 from the value of a=2.
From this example, we can see that statement II will be true for any integer value of a.
Eliminate A and D, which do not include statement II.

Statement III: (a@b) @ c = a @ (b@c)
First calculate the values INSIDE THE PARENTHESES.

(2@3) @ 10 = 2 @ (3@10)

(2+3 - 2*3) @ 10 = 2 @ (3+10 - 3*10)

-1 @ 10 = 2 @ -17

-1 + 10 - (-1*10) = 2 + (-17) - (2)(-17)

19 = 19.

When a=2, b=3 and c=10, statement III is true.
While not a definitive proof, it seems VERY unlikely that these 3 randomly selected values would prove to be an exceptional case.
Eliminate B and C, which do not include statement III.

The correct answer is E.
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by Blue_Skies » Tue May 07, 2013 12:30 pm
1 and 2 are clearly true. you can solve 3 by plugging numbers or by algebric approach. 3 will also qualify.
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