If xy=-6, what is the value of xy(x+y)?
(1) x-y=-5
(2) xy^2=18
OA:B
[spoiler]
Why won't it be D?
From St. 1:
(x-y)^2=(x+y)^2 - 4xy
25=(x+y)^2 + 24
(x+y)^2=1
x+y= sq.root (1) = 1
What'm I doing wrong?[/spoiler]
Source: OG Quant Review
If xy=-6, what is the value of xy(x+y)?
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[spoiler]Why won't it be D?
From St. 1:
(x-y)^2=(x+y)^2 - 4xy
25=(x+y)^2 + 24
(x+y)^2=1
x+y= sq.root (1) = +/- 1..insufficient
u have to consider the sign[/spoiler]
From St. 1:
(x-y)^2=(x+y)^2 - 4xy
25=(x+y)^2 + 24
(x+y)^2=1
x+y= sq.root (1) = +/- 1..insufficient
u have to consider the sign[/spoiler]
"If you don't know where you are going, any road will get you there."
Lewis Carroll
Lewis Carroll
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its not mentioned that x and y positive.If nothing is mentioned and we have (x+y)^2 while doing the square root we have to consider both +ve as well as -vejube wrote:but I thought for sq. rt. (x) we take ONLY the +ve no.?liferocks wrote:
u have to consider the sign
"If you don't know where you are going, any road will get you there."
Lewis Carroll
Lewis Carroll
If xy = -6, what is the value of xy(x+y)?
Rephrase the question: What are the values of x and y?
From xy = -6 , there are only 8 sets of values for { x , y } for the eqn to be valid. These values are { (1, -6) or (-6,1) or (-1,6) or (6,-1) or ( -3,2) or (2, -3) or (-2, 3) or (3,-2) }
Stmt 1.
(1) x-y=-5
Both the following sets of (x, y) satisfy the eqn . (-3,2) and (-2, 3) , So Insufficient.
Stmt 2.
(2) xy^2=18
xy^2=xy(y) = 18. But xy = -6 , so y must be a -3 to yield a positive 18. Thus from the above sets, there is only one set that has y = -3, which is ( 2, -3). Therefore, B is sufficient.
Rephrase the question: What are the values of x and y?
From xy = -6 , there are only 8 sets of values for { x , y } for the eqn to be valid. These values are { (1, -6) or (-6,1) or (-1,6) or (6,-1) or ( -3,2) or (2, -3) or (-2, 3) or (3,-2) }
Stmt 1.
(1) x-y=-5
Both the following sets of (x, y) satisfy the eqn . (-3,2) and (-2, 3) , So Insufficient.
Stmt 2.
(2) xy^2=18
xy^2=xy(y) = 18. But xy = -6 , so y must be a -3 to yield a positive 18. Thus from the above sets, there is only one set that has y = -3, which is ( 2, -3). Therefore, B is sufficient.
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from (2) xy² = 18 , but xy = -6 so y = 18/-6 = -3jube wrote:If xy=-6, what is the value of xy(x+y)?
(1) x-y=-5
(2) xy^2=18
so x = 2
but (1) x-y = -5
but if x = 2 & y = -3 , x-y = +5 not -5
though the answer is still B.
cos x-y = + or - y , when substituted in xy = -6 will result in a quadratic eq.
like y²+5y+6 = 0 or x²+5x+6 = 0
solving it will result in 2 values for x & y.
@kstv, you're correct in that a solution has to be applicable to both stmts. However, If this is a question from the OG Quant Review book then either the answer MUST satisfy both statements OR there is something missing in the equations.
Please do me a favor and confirm from the book that we are not missing anything on the question.
Thanks!
Please do me a favor and confirm from the book that we are not missing anything on the question.
Thanks!
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given
xy=-6 - (1)
option 1
x-y=-5 - (2)
sub value of y from 1 in 2
x+(6/x)=-5
solving
x=-3,-2
similarly y can have more than values..
hence insuff
option 2
xy^2=18
(x*y)*(y)=18 - (3)
substitue value of xy from 1 in 3
-5 * y = 18
y= (-18/5)
similarly we will get value of x
hence suffi
answer should be B
xy=-6 - (1)
option 1
x-y=-5 - (2)
sub value of y from 1 in 2
x+(6/x)=-5
solving
x=-3,-2
similarly y can have more than values..
hence insuff
option 2
xy^2=18
(x*y)*(y)=18 - (3)
substitue value of xy from 1 in 3
-5 * y = 18
y= (-18/5)
similarly we will get value of x
hence suffi
answer should be B
jube wrote:If xy=-6, what is the value of xy(x+y)?
(1) x-y=-5
(2) xy^2=18
OA:B
[spoiler]
Why won't it be D?
From St. 1:
(x-y)^2=(x+y)^2 - 4xy
25=(x+y)^2 + 24
(x+y)^2=1
x+y= sq.root (1) = 1
What'm I doing wrong?[/spoiler]
Source: OG Quant Review
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