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anksbhandari
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A
We can solve the equation further and then use the data in the first statement.
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hrishikesh05
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sudhir3127
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Answer A.
6xy = yx^2 + 9y
we can eliminate y from the equation and solve it to get X= -3
using 1 we y = 0
hence A is sufficient
6xy = yx^2 + 9y
we can eliminate y from the equation and solve it to get X= -3
using 1 we y = 0
hence A is sufficient
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anksbhandari
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Even I did the same way and got the A as answer but OA is B
See Explanation
The equation in question can be rephrased as follows:
x^2y – 6xy + 9y = 0
y(x^2 – 6x + 9) = 0
y(x – 3)^2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3
It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased “What is y?”
(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:
If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18.
If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0.
Since there are two possible answers, this statement is not sufficient.
(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.
The correct answer is B.
See Explanation
The equation in question can be rephrased as follows:
x^2y – 6xy + 9y = 0
y(x^2 – 6x + 9) = 0
y(x – 3)^2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3
It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased “What is y?”
(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:
If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18.
If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0.
Since there are two possible answers, this statement is not sufficient.
(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.
The correct answer is B.
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Stuart@KaplanGMAT
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This is a VERY common mistake that people make when solving equations (especially quadratics).sudhir3127 wrote:Answer A.
6xy = yx^2 + 9y
we can eliminate y from the equation and solve it to get X= -3
The only way you can "eliminate" y is to divide both sides by y. However, when you do so you're ignoring the possibility that y=0, in which case you're not allowed to divide by y.
So, the true solution is:
6xy = yx^2 + 9y
yx^2 - 6xy + 9y = 0
y(x^2 - 6x + 9) = 0
y(x-3)(x-3) = 0
So, y = 0 or x = 3
An analagous equation with only one variable would be:
x^2 - 25x = 0
Some people would solve it as:
x^2 = 25x
x = 25
However, again we're ignoring the possibility that x=0. A safer way to solve is to factor out x on the left side:
x(x-25) = 0
x = 0 or x = 25
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