GMAT Prep Test - Strange Operator Question - The Beat The GMAT Forum

GMAT Prep Test - Strange Operator Question

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v_schame: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Thu Dec 06, 2012 7:12 amFollowed by: 1 member; Thanked: 1 times

by v_schame » Wed Jan 23, 2013 4:13 pm

Hi,

I was doing Test Prep 1 with the official software and I'm pretty stumpped with one question.

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 (CORRECT ANSWER)

Thanks for any help you can give me!

V.

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Source: Beat The GMAT — Problem Solving | Wed Jan 23, 2013 4:13 pm

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Tommy Wallach: GMAT Instructor; Posts: 451; Joined: Thu Jan 21, 2010 11:58 amFollowed by: 120 members; Location: New York City; Thanked: 188 times; GMAT Score:770

by Tommy Wallach » Wed Jan 23, 2013 4:29 pm

Hey V_schame,

This is a function question, like f(x) = 2x, only they made up a funny symbol to make it more confusing. All it means is that whatever two numbers go on either side of the @ symbol, you need to run them through that little program (add them together, and subtract the product).

I. a@b = a + b - ab
b@a = b + a - ba

Does: a + b - ab = b + a - ba? Indeed it does!

II. a@0= a + 0 - a*0 = a Hey, that one is true, too!

III. This is a complicated one. Let's start with just the left side, and save ourselves time by remembering we already know what a@b is:

(a@b)@c = (a + b - ab)@c

Now, remember that the (a + b - ab) will represent just the "a", in terms of the original function, while the c represents the "b" in the original function:

(a + b - ab)@c = (a + b - ab) + c - (a + b - ab) * c

Phew! That's the left side done! The right will be the same, only with the letters scrambled around:

(b@c) = b + c - bc

So now notice that (b + c - bc) represents the "b" part of the original equation, while the "a" is just the same old a:

a@(b + c - bc) = a + (b + c - bc) - a * (b + c - bc)

Okay, now we have both sides, so let's see if they're equal:

(a + b - ab) + c - (a + b - ab) * c = a + (b + c - bc) - a * (b + c - bc)

a + b - ab + c - ac - bc + abc = a + b + c - bc - ab - ac + abc

If you start doing some adding and subtracting, you'll see that the exact same terms are on both sides. They're equal!

That was a toughie!

-t

Tommy Wallach, Company Expert
ManhattanGMAT

If you found this posting mega-helpful, feel free to thank and/or follow me!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Jan 23, 2013 6:30 pm

v_schame wrote:Hi,

I was doing Test Prep 1 with the official software and I'm pretty stumpped with one question.

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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lazarogb: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Sun Nov 18, 2012 7:09 am

by lazarogb » Sun Mar 03, 2013 7:20 am

Hello Mitch,

Ran into this questions just yesterday on a practice test and got it wrong. I also went for the plug in approach but then I got a little nervous because I thought I might need to test other values like negative integers in order to prove it.

I notice you went for numbers other than 0 or 1 that tend to skew answers when plugging in on certain cases.

Is there any other information that made you think that plugging negative integers was not necessary?

Thanks!

GMATGuruNY wrote:
v_schame wrote:Hi,

I was doing Test Prep 1 with the official software and I'm pretty stumpped with one question.

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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