MGMAT DS

A

We can solve the equation further and then use the data in the first statement.

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

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.

sudhir3127 wrote:Answer A.

6xy = yx^2 + 9y

we can eliminate y from the equation and solve it to get X= -3

This is a VERY common mistake that people make when solving equations (especially quadratics).

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