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value of xy?

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value of xy?

by pc80 » Thu Apr 15, 2010 10:29 am
1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y - x = 3
(2) x^3< 0

I dont have the OA...but IMO [spoiler]C

how do I get it? from the given question...and statement 1 -> i get x=3,-3...so INSUFF. Statement 2 says x^3 <0...so C [/spoiler]
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by outreach » Thu Apr 15, 2010 10:39 am
1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1

insuff
2)

x has to be negative
no idea abt y
insuff


combining 1 and 2

x y
0 -3
1 -2
2 -1

insufff

hence E
pc80 wrote:1. If 6*x*y = x^2*y + 9*y, what is the value of xy?
(1) y - x = 3
(2) x^3< 0

I dont have the OA...but IMO [spoiler]C

how do I get it? from the given question...and statement 1 -> i get x=3,-3...so INSUFF. Statement 2 says x^3 <0...so C [/spoiler]
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by thephoenix » Thu Apr 15, 2010 10:40 am
imo A
s1------> x=3 and y=6
suff
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by thephoenix » Thu Apr 15, 2010 10:44 am
outreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1

insuff
i think we need to take a value which satisfies the eqn 6xy=yx^2+9y
solving the above eqn we get (x-3)^2=0--->x=3
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by pc80 » Thu Apr 15, 2010 10:47 am
thephoenix wrote:imo A
s1------> x=3 and y=6
suff
as far as I am aware...in GMAT...both statements need to indicate that the answer is correct (I am not talking about sufficiency here)...

so according to you if you take x=3 from the second statement x^3<0...doesnt seem to be even correct...

am I missing something
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by pc80 » Thu Apr 15, 2010 10:50 am
thephoenix wrote:
outreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1

insuff
i think we need to take a value which satisfies the eqn 6xy=yx^2+9y
solving the above eqn we get (x-3)^2=0--->x=3

hmmm...can we just cancel out y from the original equation wouldnt y = (x+3)...from statement 1? and so if you substitute x+3 in the equation...wouldnt that become the root of the equation as well? then from there you'll get y=3 (from (x-3)^2) and y=-3
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by thephoenix » Thu Apr 15, 2010 10:52 am
pc80 wrote:
thephoenix wrote:imo A
s1------> x=3 and y=6
suff
as far as I am aware...in GMAT...both statements need to indicate that the answer is correct (I am not talking about sufficiency here)...

so according to you if you take x=3 from the second statement x^3<0...doesnt seem to be even correct...

am I missing something
but then the q has to from gmat .
wats the source of this question.
it does not luk like GMAT prep
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by pc80 » Thu Apr 15, 2010 10:53 am
thephoenix wrote:
pc80 wrote:
thephoenix wrote:imo A
s1------> x=3 and y=6
suff
as far as I am aware...in GMAT...both statements need to indicate that the answer is correct (I am not talking about sufficiency here)...

so according to you if you take x=3 from the second statement x^3<0...doesnt seem to be even correct...

am I missing something
but then the q has to from gmat .
wats the source of this question.
it does not luk like GMAT prep
got it from another BTG-like forum...
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by eaakbari » Thu Apr 15, 2010 10:59 am
thephoenix wrote:
outreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1

insuff
i think we need to take a value which satisfies the eqn 6xy=yx^2+9y
solving the above eqn we get (x-3)^2=0--->x=3
You can only do this if y=not equal to 0, but you dont know that
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by eaakbari » Thu Apr 15, 2010 11:04 am
Stem
yx^2 -6xy +9y = 0

If y=0 , then we cannot determine x solely from this
If y is not 0 , then x^2 - 6x +9 =0
which gives x = 3

Statement one

y-x = 3

Does not say whether y is zero or non zero.
For eg
If y = 0 x = -3
0-(-3) = 3
Also y = 6 x = 3
6-3 = 3
Hence Insuff


Statement two
x^3<0
Tells us x is negative. That means x is not equal to 3.
Hence Case 1 from stem gets satisfied where y = 0

Hence xy= 0

Suff

Answer B
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by Fiver » Fri Apr 16, 2010 5:27 am
Agree with B.

Simplifying the question:
x^2y + 9y - 6xy = 0
y(x^2 - 6x + 9) = 0
y(x-3)^2 = 0
Therefore either y = 0 for any value of x OR x = 3 for any value of y.
Find x*y = ?

st1] y - x = 3
This means y = 3+x
Therefore if y=0 then x*y = 0 and if x=3 then y=6 and x*y = 18
Insuff.

St2] x^3< 0
This means x<0, and according to our simplification if x is any value other than 3, then y must be 0.
Hence product xy will always be 0.
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by eaakbari » Fri Apr 16, 2010 11:03 am
Since we do not have OA, will experts pleasde verify?
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by ronsom » Fri Apr 16, 2010 12:17 pm
6xy = x^2y + 9y
=> 6xy = y(x^2 + 9)
=> 6x = x^2 + 9
=> x^2 -6x + 9 = 0
=> (x-3)^2 = 0
=> x = 3

from Statement 1)
y -x = 3
substituting value of x from above
y-3 = 3
=> y = 6
Therefore statement 1 does the job.
xy = 3*6 = 18

from Statement 2)
x^3 < 0
this simply implies x < 0
Given this, we cannot calculate the value of xy

Therefore, A
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by harshavardhanc » Fri Apr 16, 2010 10:38 pm
eaakbari wrote:Stem
yx^2 -6xy +9y = 0

If y=0 , then we cannot determine x solely from this
If y is not 0 , then x^2 - 6x +9 =0
which gives x = 3

Statement one

y-x = 3

Does not say whether y is zero or non zero.
For eg
If y = 0 x = -3
0-(-3) = 3
Also y = 6 x = 3
6-3 = 3
Hence Insuff


Statement two
x^3<0
Tells us x is negative. That means x is not equal to 3.
Hence Case 1 from stem gets satisfied where y = 0

Hence xy= 0

Suff

Answer B
second E on this one. IMO B.
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by harshavardhanc » Fri Apr 16, 2010 10:45 pm
thephoenix wrote:
outreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1

insuff
i think we need to take a value which satisfies the eqn 6xy=yx^2+9y
solving the above eqn we get (x-3)^2=0--->x=3
just wondering, why didn't you consider y=0 as the other possibility.

IMO, stem says : Y can be 0 OR X can be 3. We then have to start, taking this information with us.

Statement 1 doesn't tell us definitely that whether X=3 or Y= 0. Hence, value of XY won't be unique. Therefore, this will be insufficient.

Statement 2 clearly tells us that X is not 3 , hence the other possibility , Y=0, is definite. Therefore, XY will always be 0 whatever X may be.

This statement is sufficient.
Regards,
Harsha
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