For all numbers a and b, the operation is defined by ab = a^2 - ab.

Question

For all numbers a and b, the operation  is  defined by a@b = a^2 - ab. If xy ≠ 0, then which of the following can be equal to zero?

I. x@y

II. (xy)@y

III. x@(x + y)

A. II only
B. I and II only
C. I and III only
D. II and III only
E. All of the above

KarishmaB · Accepted Answer

a@b = a^2 - ab = a(a-b)

x and y both are not 0.

I. x@y
x@y = x(x - y) 
This will be 0 when either x = 0 (not possible as given), or x - y = 0 i.e. x = y (this is possible)

II. (xy)@y
(xy)@y = xy(xy - y) = xy^2(x - 1)
This will be 0 when either x or y is 0 (not possible) or x = 1 (this is possible)

III. x@(x + y)
x@(x + y) = x(x - (x+y)) = -xy
This will be 0 when at least one of x and y is 0 - (both not possible as given)

Hence only (I) and (II) can be 0. Answer (B)

peachfuzz · Answer

xy ≠ 0

I. x@y
x^2 - 2xy
xy will always result in a non zero number

II. (xy)@y
(x^2)(y^2) - 2xy^2
If x and y equal 2, this could equal 0

III. x@(x + y)
-(x^2 + 2xy) 
If x is 4 and y is -2, this could equal 0

Answer: D

EMPOWERgmatRichC · Answer

Hi peachfuzz,

I think there's a typo in your work that has impacted ALL of your calculations:

The original 'symbol' in the prompt is....

a@b = a^2 - ab

It looks like your work is based on....

a@b = a^2 - 2ab

Your approach is perfect for this prompt though, so if you make the necessary adjustments to your work, you should get the correct answer.

GMAT assassins aren't born, they're made,
Rich

chetan2u · Answer

hi,
lets look at the equation in the question and simplify...
a@b = a^2 - ab=a(a-b).... so a@b will be zero if a=b or a=0.. but a cannot be equal to 0.. as per Q, x and y can take any int value except 0...

now lets look at the choices..
I. x@y
when x=y, it will be 0... so ok...

II. (xy)@y
when we put xy=y, it is possible when x=1 and y any integer... so ok again

III. x@(x + y)
when we put x=x+y.... only possibility when y=0 and it is given x and y cannot be 0....so not possible

only l and ll possible ans B....

peachfuzz · Answer

Whoops, you are correct! Thanks for catching my mistake.

I. x^2 - xy
can be true if x and y is equal to 1

II. (x^2)(y^2) - x(y^2)
can be true if x and y is equal to 1

III. -xy
will always equal a number that is not zero

B. I and II

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION: stranop_exp.png

mill · Answer

Is that really a 700 question? Looks too straight

Posted from my mobile device

Bunuel · Answer

Difficulty level (seen in tags) is calculated automatically based on the timer stats from the users who attempted the question.

Mohammadmo · Answer

I: x^2-xy= x(x-1)
II:(xy)^2-xy^2=xy^2(x-1)
III:x^2- x^2-xy
Only if X=1 the cases I & II can be zero.
Option B

Posted from my mobile device

VasundharaS · Answer

ATQ, xy should not be zero but per the option, x*(x+y) should not be zero so ideally 2 should also fit

For all numbers a and b, the operation is defined by ab = a^2 - ab.

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For all numbers a and b, the operation is defined by ab = a^2 - ab.

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