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statements must be true for all integers a, b and c?

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statements must be true for all integers a, b and c?

by Nina1987 » Sun Dec 27, 2015 8:30 am

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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)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

I know how to solve it using algebra or by number picking. But Im looking for a more perceptive approach here. Can someone provide an intuitive approach to III w/o resorting to number picking and algebra? maybe something in terms of commutativity and associativity? Thanks
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by Matt@VeritasPrep » Sun Dec 27, 2015 5:41 pm
Nina, you'd just do it algebraically. For instance

I::

a @ b = a + b - ab

b @ a = b + a - ba

Since a + b - ab = b + a - ba for all integers b, a, I is always true.

You'd repeat the process for the other two.
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by gnakul » Mon Dec 28, 2015 12:18 am
Answer : c
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by sanju09 » Thu Dec 31, 2015 10:15 pm
Nina1987 wrote: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)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III

I know how to solve it using algebra or by number picking. But Im looking for a more perceptive approach here. Can someone provide an intuitive approach to III w/o resorting to number picking and algebra? maybe something in terms of commutativity and associativity? Thanks
Commutative and Associative Laws are not good for minuses and divisions in general, but in the present case in particular, Commutative Law by chance works. Hence I & II can be taken in safely, leaving life-size doubts upon the legitimacy of Associative Law. Hence, [spoiler](C) I and II only[/spoiler] can be a smart answer.
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by alanforde800Maximus » Sun May 20, 2018 8:43 am
Is there any better approach to solve this question?
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by GMATGuruNY » Sun May 20, 2018 9:12 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)

Answers:
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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