MGMAT Challenege Question from 02/26/07

[JRankoth]

Can you pls. explain the following? I am getting E, while OA is A. Thanks.

8xy^3 + 8x^3y = (2 * x^2 * y^2)/2^-3

What is xy?

(1) y > x

(2) x < 0

Simplifying the stem we get:

8xy^3 + 8x^3y = (2 * x^2 * y^2) * 8

Dividing by 8xy throughout we get:

y^2 + x^2 = 2xy

x^2 + y^2 - 2xy = 0

(x - y)^2 = 0

The questions asks for the value of xy.

stat 1: y>x

y=5, x=-5; (x - y)^2 = 0? Yes; xy = -25
y= 3, x=-3; (x - y)^2 = 0? Yes; xy = -9

Insufficient

stat 2: x < 0
Doesn't tell us anything; insuff.

Answer E.

However the OA is A. Can you pls. explain.

[StaceyKoprince]

When you are factoring something and you care about the factors, NEVER divide out the variables - you lose possible factors that way. Dividing by 8 is fine but DON'T divide by xy. Instead, factor that xy out front to get:
xy(x^2 + y^2 - 2xy) = 0
xy(x - y)^2 = 0

This means that EITHER xy=0 OR (x-y) = 0, which simplifies to x=y. One of the two equations in the prior sentence must be true in order to make the product of these two equal to zero.

Statement 1 says that y>x. If y>x, then y can't equal x, so the x=y equation isn't true. The other solution, xy=0, therefore must be true.

[Guest]

When you are factoring something and you care about the factors, NEVER divide out the variables - you lose possible factors that way. Dividing by 8 is fine but DON'T divide by xy. Instead, factor that xy out front to get:
xy(x^2 + y^2 - 2xy) = 0
xy(x - y)^2 = 0

This means that EITHER xy=0 OR (x-y) = 0, which simplifies to x=y. One of the two equations in the prior sentence must be true in order to make the product of these two equal to zero.

Statement 1 says that y>x. If y>x, then y can't equal x, so the x=y equation isn't true. The other solution, xy=0, therefore must be true.

Excellent! Thanks a bunch.

[Guest]

8xy^3 + 8x^3y = (2 * x^2 * y^2)/2^-3

can be rewritten as: xy^3 + x^3y = 2*x^2*y^2 (by dividing both sides by 8)

but dont see how we get to "xy(x^2 + y^2 - 2xy) = 0" as mentioned by Stacy.

The x^3y is giving me problems !

Stacy, where am I going wrong here ? Thanks.

[RonPurewal]

8xy^3 + 8x^3y = (2 * x^2 * y^2)/2^-3

can be rewritten as: xy^3 + x^3y = 2*x^2*y^2 (by dividing both sides by 8)

but dont see how we get to "xy(x^2 + y^2 - 2xy) = 0" as mentioned by Stacy.

The x^3y is giving me problems !

Stacy, where am I going wrong here ? Thanks.

i think the problem may originate in the rather cumbersome notation required by forums such as this one (because they don't allow proper mathematical notation): "x^3y" here needs to be read as (x cubed) times (y).

once you make that realization, you should see that, if you move 2(x squared)(y squared) over to the other side of the equation, the three terms have a common factor xy. if you pull out that common factor, you get the expression arrived at by stacey.

hth

[hkparikh09]

When you are factoring something and you care about the factors, NEVER divide out the variables - you lose possible factors that way. Dividing by 8 is fine but DON'T divide by xy. Instead, factor that xy out front to get:
xy(x^2 + y^2 - 2xy) = 0
xy(x - y)^2 = 0

This means that EITHER xy=0 OR (x-y) = 0, which simplifies to x=y. One of the two equations in the prior sentence must be true in order to make the product of these two equal to zero.

Statement 1 says that y>x. If y>x, then y can't equal x, so the x=y equation isn't true. The other solution, xy=0, therefore must be true.

Glad I found this. I approached the problem in the same way as the original poster. Thanks for the help!!! I didn't realized that dividing by XY is the wrong approach.

[RonPurewal]

Yep.

As usual, GMAT problems"”regardless of difficulty"”are based on basic, feet-on-the-ground math ideas. Not "advanced" ideas.

Here, that ground-level idea is "Can't divide by zero." (If you divide by something that could be zero, you MUST investigate that possibility separately.)

You don't need to know much math"”but, with the math you do need to know, you must be very much on point.