amanvermagmat wrote:

X, Y, Z are integers. Is 3*X > Z?

(1) X + 2*Y > 3*Z

(2) 2*Z > 2*Y - X

As all we're given is equations, we'll work with equation-related tools such as simplification.

This is a Precise approach.

(1) Multiplying our inequality by 3 gives 3x+6y>9z --> 3x > 9z - 6y.

Without knowing if y is positive or negative we definitely cannot know if 3x > 9z, let alone 3x>z.

Insufficient.

(2) We'll once again try to create "3x is larger than" in our expression: Multiplying by 3 gives

6z>6y-3x --> 3x>6y-6z. Similarly to the before, we need some information about y to solve.

Insufficient.

Combined.

We'll notice that one equation has '6y' and another has '-6y'. Adding them together cancels out the y and gives

3x+3x > (3z-6y)+(6y-9z) --> 6x > -6z --> 3x > -3z.

So, to answer the question we need to know if -3z > z. But is it? If z is positive this inequality is always false and if it is negative this inequality is always true.

Insufficient.

(E) is our answer.

_________________

David

Senior tutor at examPAL

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