idontknowwhy94 wrote:

If x>0 on the number line, is x>y?

1) |x-y|>|y|

2) |x-y|=3

So x is positive, hence it lies to the right of zero on the number line. We need to know whether x > y or not (whether the distance between x and 0 is greater than the distance between y and 0 on the number line or not). Now on the number line, if y lies to the right of x, then y > x but if y lies to the left of x, then x > y.

(1) |x-y|>|y|

This states that the distance between x and y on the number line is more than the distance between y and 0. Now we are already given that x lies to the right of 0. If y further lies to the right of x, then distance between x and y can never be more than distance between y and 0. Illustration:

...0.....x....y . as we can see, since y lies to the right of x, distance between y and 0 will automatically become more than distance between x and y. BUT if y lies to the left of x, then distance between x and y CAN be greater than distance between y and 0. Illustration:

...y....0....x OR ....0....y......x. If y is negative, then automatically distance between x and y will always be greater than that between y and 0. And if y is also positive but less than x, its still possible for distance between y and x to be greater than distance between 0 and y.

so if x > 0, then the only way for |x-y|>|y| to be true if when x lies to the right of y, or when x > y. So first statement is sufficient.

(2) |x-y|=3

This states that distance between x and y on the number line is '3'. But this could happen both where x > y (eg, x=6, y=3) or where y > x (eg, x=3, y=6). So we cant say. Insufficient.

Hence

A answer