Bunuel wrote:

If x and y are integers and x + y = 21, is x > 19?

(1) 2y > 19 + y

(2) 3x/y > 51x

Hi

Bunuel,

Are you sure that answer is A? How do you prove that Statement 2 is insufficient?My answer as follows:2) 3x/y > 51x............x/y>17x

If y>0.... then x>17xy.....x-17xy>0....x(1-17y)>0

As y is positive... (1-17y) is negative........x must be negative in order to have x(1-17y)>0

x is NOT greater than 19

If y 51x

x/y > 17x

In the resulting inequality, x can be positive only if y is also positive.

There is ONLY ONE SOLUTION for x+y = 21 such that x>19 and x and y are both positive:

x=20, y=1.

If this solution does not satisfy x/y > 17x, then it is not possible that x>19.

Plugging x=20 and y=1 into x/y > 17x, we get:

20/1 > 17*20

20 > 340.

Doesn't work.

Thus, it is not possible that x>19, implying that the answer to the question stem is NO.

SUFFICIENT.

Thanks

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