If xy = - 6, what is the value of xy ( x + y )?
(1) x - y = 5
(2) xy^2= 18
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- Raj
Re: If xy = – 6, what is the value of xy ( x + y )?
From (1), x = y+5, substitute in xy = -6 and you get a quadratic from which,
y could be -3 or -2 and hence x could be 2 or 3. The value of xy(x+y) is different in each case.
Hence INSUFFICIENT.
From (2), use xy = -6 in this equation to get y = -3, x= 2. You get a unique value for xy(x+y). SUFFICIENT.
Answer B.
Is this right?
-Raj.
y could be -3 or -2 and hence x could be 2 or 3. The value of xy(x+y) is different in each case.
Hence INSUFFICIENT.
From (2), use xy = -6 in this equation to get y = -3, x= 2. You get a unique value for xy(x+y). SUFFICIENT.
Answer B.
Is this right?
-Raj.
guest 2 Wrote:If xy = - 6, what is the value of xy ( x + y )?
(1) x - y = 5
(2) xy^2= 18
- COOPER2248817
XY
1. x-y=5since we are given xy=-6 either x is (pos/neg) or y is (pos/neg)
x* y
1* -6=-6
2* -3=-6
3* -2=-6
6* -7=-6
the only way that statment one can be true is if:
2* -3=-6
3* -2=-6
When we put this into our equation of xy(x+y) we will yield either a positive or a negative answer ; so BCE
2. xy^2=18
solve for x
x=18/y^2
plug this into xy=-6
18/(y^2)*y=-6
when you solve for Y you will yield y=-3 and then x=2
so the correct answer will be B
x* y
1* -6=-6
2* -3=-6
3* -2=-6
6* -7=-6
the only way that statment one can be true is if:
2* -3=-6
3* -2=-6
When we put this into our equation of xy(x+y) we will yield either a positive or a negative answer ; so BCE
2. xy^2=18
solve for x
x=18/y^2
plug this into xy=-6
18/(y^2)*y=-6
when you solve for Y you will yield y=-3 and then x=2
so the correct answer will be B
- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
these solutions are good.
to readers of this thread: if you don't understand the derivation of any of the solutions - particularly the work (omitted by the posters) inherent in solving the quadratics - then post back to this thread and i'll give the details.
for statement (2), there are also other solutions besides the substitutions put forth here. for instance, you can also DIVIDE each side of the equation xy^2 = 18 by the corresponding side of the equation xy = -6, giving immediately that y = -3. that's pretty nice.
then, you can just substitute this back into either of the equations (although it's probably easier to use xy = -6), giving x = 2. alternatively, once you get a unique value for y, you can just REALIZE that you're going to get a unique value for x, and therefore a unique value for the quantity that's actually the object of the problem; therefore, you don't actually have to bother carrying out the rest of the solution, since a single value is guaranteed to be sufficient.
to readers of this thread: if you don't understand the derivation of any of the solutions - particularly the work (omitted by the posters) inherent in solving the quadratics - then post back to this thread and i'll give the details.
for statement (2), there are also other solutions besides the substitutions put forth here. for instance, you can also DIVIDE each side of the equation xy^2 = 18 by the corresponding side of the equation xy = -6, giving immediately that y = -3. that's pretty nice.
then, you can just substitute this back into either of the equations (although it's probably easier to use xy = -6), giving x = 2. alternatively, once you get a unique value for y, you can just REALIZE that you're going to get a unique value for x, and therefore a unique value for the quantity that's actually the object of the problem; therefore, you don't actually have to bother carrying out the rest of the solution, since a single value is guaranteed to be sufficient.
4 posts Page 1 of 1