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]
value of xy?
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- outreach
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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
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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- thephoenix
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- thephoenix
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i think we need to take a value which satisfies the eqn 6xy=yx^2+9youtreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1
insuff
solving the above eqn we get (x-3)^2=0--->x=3
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)...thephoenix wrote:imo A
s1------> x=3 and y=6
suff
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
thephoenix wrote:i think we need to take a value which satisfies the eqn 6xy=yx^2+9youtreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1
insuff
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
- thephoenix
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but then the q has to from gmat .pc80 wrote: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)...thephoenix wrote:imo A
s1------> x=3 and y=6
suff
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
wats the source of this question.
it does not luk like GMAT prep
got it from another BTG-like forum...thephoenix wrote:but then the q has to from gmat .pc80 wrote: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)...thephoenix wrote:imo A
s1------> x=3 and y=6
suff
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
wats the source of this question.
it does not luk like GMAT prep
- eaakbari
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You can only do this if y=not equal to 0, but you dont know thatthephoenix wrote:i think we need to take a value which satisfies the eqn 6xy=yx^2+9youtreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1
insuff
solving the above eqn we get (x-3)^2=0--->x=3
Whether you think you can or can't, you're right.
- Henry Ford
- Henry Ford
- eaakbari
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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
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
Whether you think you can or can't, you're right.
- Henry Ford
- Henry Ford
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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.
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.
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
=> 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
- harshavardhanc
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second E on this one. IMO B.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
Regards,
Harsha
Harsha
- harshavardhanc
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just wondering, why didn't you consider y=0 as the other possibility.thephoenix wrote:i think we need to take a value which satisfies the eqn 6xy=yx^2+9youtreach wrote:1)
y and x can take
y x
0 -3
1 -2
2 -1
3 0
4 1
insuff
solving the above eqn we get (x-3)^2=0--->x=3
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
Harsha