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If xy=-6, what is xy(x + y)? 1) x - y = 5 2) x(y^2) = 18

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Director
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If xy=-6, what is xy(x + y)? 1) x - y = 5 2) x(y^2) = 18 [#permalink]

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29 Jan 2008, 18:12
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If xy=-6, what is xy(x + y)?

1) x - y = 5
2) x(y^2) = 18
Senior Manager
Joined: 06 Jul 2006
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Schools: Berkeley Haas
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Re: DS - solve for x and y [#permalink]

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29 Jan 2008, 18:30
B alone is sufficient. So B.
xy = -6
St2: xy.y = 18, Hence y = -3, Hence x = 2.
St1 : x = 5 + y, Hence x = -6 / y = 5 + y, Giving y^2 + 5y +6 = 0. (Can't solve). Not suff.
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Re: DS - solve for x and y [#permalink]

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29 Jan 2008, 18:42
sonibubu wrote:
If xy=-6, what is xy(x + y)?

1) x - y = 5
2) x(y^2) = 18

1) We get x = y + 5. Substituting this in xy=-6, we get
(y+5)(y) = -6
y^2 + 5y + 6 = 0
(y+3)(y+2) = 0
y = -3 or y = -2
if y = -3, x = 2; if y=-2, x = 3
plugging in both we get 6 and -6 --> different results. Hence insufficient.

2) x*(y^2) = 18
also, xy=-6
dividing the 2 equations, we get y = -3
which gives x as 2
Sufficient

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Re: DS - solve for x and y [#permalink]

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29 Jan 2008, 19:59
B

stat 1 , either 6 and -1 or 3 and -2 works, but they give different vals for x+y. insuff

stat 2, only 2 and -3 works. try out the other permuations of 6,-1 , -1,6 , 3-2, -2,3...none works. suff.
Director
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Re: DS - solve for x and y [#permalink]

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30 Jan 2008, 07:35
sonibubu wrote:
If xy=-6, what is xy(x + y)?

1) x - y = 5
2) x(y^2) = 18

1) insufficient because two solutions of quadratic equation
2) sufficient because single solution y = -3, x = 2

B
Director
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Re: DS - solve for x and y [#permalink]

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30 Jan 2008, 07:55
Well done, OA is B.
How long did it take you guys to do? Possible in under 2 min?
Director
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Re: DS - solve for x and y [#permalink]

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30 Jan 2008, 08:21
it took me under 30 sec - the key is not to solve equations but calculate the number of solutions
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Re: DS - solve for x and y [#permalink]

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30 Jan 2008, 08:25
B

Distribute Stem

XY (X + Y) ---> (X^2) (Y) + (X) (Y^2)

Statement 2 tells us that (X) (Y^2) = 18

So in stem we have (X)(Y) = -6

We can square this to give us (X)^2 * (Y^2) = 36

So in order to find

(X^2) (Y) + (X) (Y^2)

we have

(X^2) (Y) + 18 (from statement 2)

and since

(X^2) * (Y^2) = 36

and Y^2 * X = 18

We can divide both statements to give us X = 2

the Y squareds cancel out and X squared divided by X is X....36/18 = 2

we have X=2 Y=-3
Re: DS - solve for x and y   [#permalink] 30 Jan 2008, 08:25
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