If 6xy = x2y + 9y, what is the value of xy?
(1) x = -2
(2) x < 0
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
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Manhattan GMAT Data Sufficiency question
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machichi
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Is it 6xy=x^2 * y +9y?
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Target question: What is the value of xy?mparakala wrote:If 6xy = (x^2)y + 9y, what is the value of xy?
(1) x = -2
(2) x < 0
Given: 6xy = (x^2)y + 9y
Set equal to zero: (x^2)y - 6xy + 9y = 0
Factor: y(x^2 - 6x + 9) = 0
Factor more: y(x - 3)(x - 3) = 0
For the above equation to hold true, it must be the case that y=0 or x=3
Statement 1: x = -2
This tells us that x does not equal 3.
If x does not equal 3, then y must equal 0 (see above in bold)
If y=0, then xy must equal 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: x < 0
This is yet another way to tell us that x does not equal 3.
If x does not equal 3, then y must equal 0 (see above in bold)
If y=0, then xy must equal 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer = D
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Last edited by Brent@GMATPrepNow on Wed Oct 12, 2016 4:11 am, edited 1 time in total.
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gmattesttaker2
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Brent@GMATPrepNow wrote:Target question: What is the value of xy?mparakala wrote:If 6xy = (x^2)y + 9y, what is the value of xy?
(1) x = -2
(2) x < 0
Given: 6xy = (x^2)y + 9y
Set equal to zero: (x^2)y - 6xy + 9y = 0
Factor: y(x^2 - 6x + 9) = 0
Factor more: y(x - 3)(x - 3) = 0
For the above equation to hold true, it must be the case that y=0 or x=3
Statement 1: x = -2
This tells us that x does not equal 3.
If x does not equal 3, then y must equal 0 (see above in bold)
If y=0, then xy must equal 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: x < 0
This is yet another way to tell us that x does not equal 3.
If x does not equal 3, then y must equal 0 (see above in bold)
If y=0, then xy must equal 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer = D
Cheers,
Brent
Hello Brent,
I was wondering why the following approach wont be correct here:
6xy = x^2y + 9y
Hence I take y common :
6xy = y(x^2 + 9)
=> 6x = x^2 + 9
=> x^2 + 9 - 6x = 0
=> (x - 3)^2 = 0
=> x = 3
Thanks a lot for your help.
Best Regards,
Sri
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Brent@GMATPrepNow
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Hi Sri,gmattesttaker2 wrote:[
Hello Brent,
I was wondering why the following approach wont be correct here:
6xy = x^2y + 9y
Hence I take y common :
6xy = y(x^2 + 9)
=> 6x = x^2 + 9
=> x^2 + 9 - 6x = 0
=> (x - 3)^2 = 0
=> x = 3
Thanks a lot for your help.
Best Regards,
Sri
The problem is highlighted above in blue.
Once you reached the conclusion that 6xy = y(x^2 + 9), you divided both sides by y to get 6x = x^2 + 9
This is not entirely correct.
When you divided by y, you eliminated the possibility that y = 0
Consider this rudimentary example: 2y = 3y
If we divide both sides by y, we get 2 = 3, which doesn't make any sense.
Notice that, if 2y = 3y, then y must equal zero. So, we can't divide both sides by y.
Likewise, y = 0 is one possible solution 6xy = y(x^2 + 9)
So, we can't arbitrarily divide both sides by y.
In fact, notice that, if y = 0, then x can have ANY value and the equation holds true.
The takeaway here is that we must exercise caution whenever we divide both sides of an equation by a variable, since it's possible that we may be dividing both sides by zero (a big no-no)
I hope that helps.
Cheers,
Brent
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gmattesttaker2
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Hello Brent,Brent@GMATPrepNow wrote:Hi Sri,gmattesttaker2 wrote:[
Hello Brent,
I was wondering why the following approach wont be correct here:
6xy = x^2y + 9y
Hence I take y common :
6xy = y(x^2 + 9)
=> 6x = x^2 + 9
=> x^2 + 9 - 6x = 0
=> (x - 3)^2 = 0
=> x = 3
Thanks a lot for your help.
Best Regards,
Sri
The problem is highlighted above in blue.
Once you reached the conclusion that 6xy = y(x^2 + 9), you divided both sides by y to get 6x = x^2 + 9
This is not entirely correct.
When you divided by y, you eliminated the possibility that y = 0
Consider this rudimentary example: 2y = 3y
If we divide both sides by y, we get 2 = 3, which doesn't make any sense.
Notice that, if 2y = 3y, then y must equal zero. So, we can't divide both sides by y.
Likewise, y = 0 is one possible solution 6xy = y(x^2 + 9)
So, we can't arbitrarily divide both sides by y.
In fact, notice that, if y = 0, then x can have ANY value and the equation holds true.
The takeaway here is that we must exercise caution whenever we divide both sides of an equation by a variable, since it's possible that we may be dividing both sides by zero (a big no-no)
I hope that helps.
Cheers,
Brent
Thanks a lot for the excellent explanation. It is clear now.
Best Regards,
Sri
