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if x and y are positve...

by datonman » Sun Nov 09, 2014 7:36 pm
If 'x' and 'y' are positive, does (x-2)(y-3)=6?

(1) 3x=y(x-2)

(2) (x-8)(y2+5) =0

I got confused with this ds problem because when i first saw this, i initially thought 'oh it's probably not sufficient for both cases but i can't help but look at the second one and wonder. Btw: the y2 is y squared.

I feel like the second one is a good equation. But i still don't know. It looks tricky.
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Source: — Data Sufficiency |

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by [email protected] » Sun Nov 09, 2014 8:05 pm
Hi datonman,

This DS question has a number of restrictions to it, so you're going to have to write everything on the pad and pay careful attention to your work.

We're told that X and Y are POSITIVE. We're asked if (X-2)(Y-3) = 6? This is a YES/NO question.

This prompt can be solved by TESTing VALUES. Here's how:

Fact 1: 3X = Y(X-2)

This Fact interacts in an interesting way with the original prompt. Since X and Y MUST be POSITIVE, X CANNOT = 1 or 2. Here's why:

If X = 1, then the equation in Fact 1 becomes 3 = Y(-1), which makes Y = -3. But that is NOT allowed (Y must be positive)
If X = 2, then the equation in Fact 1 becomes 6 = Y(0), which has NO SOLUTION for Y. That's NOT allowed either (Y must be positive).

This means that X must be GREATER THAN 2.

If X = 3, then 9 = Y(1), so Y = 9 and the answer to the question is (1)(6) = 6 --> the answer is YES.
If X = 4, then 12 = Y(2), so Y = 6 and the answer to the question is (2)(3) = 6 --> the answer is YES.
If X = 2.5, then 7.5 = Y(.5), so Y = 15 and the answer to the question is (.5)(12) = 6 --> the answer is YES.
Fact 1 is SUFFICIENT

Fact 2: (X-8)(Y^2 + 5) = 0

Since this equation is set equal to 0, there's only 1 solution: X MUST = 8. However, we know NOTHING about Y, so it can be anything positive.

If X = 8, Y = 1, then the answer to the question is (6)(-2) = -12 --> the answer is NO
If X = 8, Y = 3, then the answer to the question is (6)(1) = 6 --> the answer is YES
Fact 2 is INSUFFICIENT

Final Answer: A

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by GMATGuruNY » Mon Nov 10, 2014 3:20 am
datonman wrote:If 'x' and 'y' are positive, does (x-2)(y-3)=6?

(1) 3x=y(x-2)

(2) (x-8)(y²+5) =0
FOILING (x-2)(y-3) = 6, we get:
xy - 3x - 2y + 6 = 6
xy - 3x - 2y = 0.

Question stem, rephrased:
Does xy - 3x - 2y = 0?

Statement 1: 3x=y(x-2)
3x = xy - 2y
0 = xy - 3x - 2y.
SUFFICIENT.

Statement 2: (x-8)(y²+5) = 0
It's not possible that y²+5 = 0.
Implication:
The equation above is valid only if x-8 = 0, implying that x=8.

Substituting x=8 into xy - 3x - 2y = 0, we get:
8y - 3*8 - 2y = 0
6y = 24
y = 4.
Question stem, rephrased:
Does y=4?

No way to determine whether y=4.
INSUFFICIENT.

The correct answer is A.
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by Mathsbuddy » Fri Nov 21, 2014 9:35 am
Statement 2 is clearly insufficient as it is higher order to the objective equation.

Let me explain:

(x-2)(y-3)=6 has highest order xy
(if x = y, then it would be quadratic)

(x-8)(y²+5) =0 has highest order xy²
(if x = y, then it would be cubic)

As such, the possible number of solutions to Statement (2) are too many.
OK, maybe not so if the roots not real - but this goes beyond GMAT
OK, maybe not so if the turning point is on the x-axis. However, x and y are positive, so there are no 0 values of x permitted.

Does anyone agree/disagree? Or am I talking nonsense!
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