Statement 1: a = -y = | x |

If a = | x |, then a is positive.

If -y = | x |, then y is negative.

x could be negative or positive.

If x is positive, then ax > y.

If x = -1 then ax = y

If x < -1 then ax < y

Insufficient.

Statement 2: a < 1

Tells us nothing about x or y.

Insufficient.

Statements Combined:

The combined statements indicate the following.

-1 < y < 0

0 < a < 1

-1 < x < 0 or 0 < x < 1

Case 1: a and x are both positive, ax > y

Case 2: a and x are both fractions and x is negative, then ax > y.

Sufficient.

The correct answer is C.

(Did you provide the correct OA? If so, if you give me the question #, I will report the error to BellCurves.)

manik11 wrote:If x ≠ 0, is ax > y?

(1) a = -y = | x |
(2) a < 1

Statement 1:
Since |x| > 0, y must be a NEGATIVE VALUE.

Case 1: y=-1, implying that a=1 and x=±1
If a=1 and x=1, then ax = 1.
Here, ax > y.
If a=1 and x=-1, then ax = -1.
Here, ax = y.
INSUFFICIENT.

Statement 2:
No information about x or y.
INSUFFICIENT.

Statements combined:
To satisfy both statements, y must be a NEGATIVE FRACTION.

Case 2: y=-1/2, implying that a=1/2 and x= ±1/2.
If a=1/2 and x=1/2, then ax = 1/4.
If a=1/2 and x=-1/2, then ax = -1/4.
In each case, ax > y.

Implication of Case 2:
Since y must be a NEGATIVE FRACTION, ax must be either a positive fraction or a negative fraction greater than y.
Thus, ax > y.
SUFFICIENT.

The correct answer is C.

Let's write the question as

Is ax - y > 0 ?

S1::

a = |x|, so a ≥ 0
-y = |x|, so 0 ≥ y

We still don't know the sign of x, so this isn't sufficient.

S2::

1 > a

We don't know anything about x and y, so this isn't sufficient.

S1+S2::

We know that 1 > a ≥ 0, that y = -a, and that 1 > x > -1. So the question becomes

Is ax > -a ?

Let's consider two cases: x > 0 and 0 > x.

If x > 0, then we know ax > 0 and 0 > -a, so yes, ax > -a.

If 0 > x, then we have two possibilities:

0 > ax > -a
0 > -a > ax

But if the second case is true, we have

-a > ax

Dividing by a (which is positive), we have

-1 > x

But this contradicts what we learned about x, so it is impossible. Hence we must have the other case, ax > -a, and in either case (x > 0 or 0 > x), we're left with only ax > -a, or ax > y.