Kaplan question

[singhsa]

If y >= 0, What is the value of x?

1) |x-3| >= y
2) |x-3| <= -y
[spoiler][/spoiler]

[Testluv]

(1) is clearly insufficient.

(2) is different. We know from the question that y is positive or zero.

Absolute value is always positive or zero. But if y were positive, then "-y" is negative, which means that |x-3| would also be negative (either equal to -y or less than -y--either way, it would be negative).

Thus, y cannot be positive. Thus, y is zero, which means (2) is really saying:

|x-3| = 0, which is solveable for x.

Choose B.

[singhsa]

(1) is clearly insufficient.

(2) is different. We know from the question that y is positive or zero.

Absolute value is always positive or zero. But if y were positive, then "-y" is negative, which means that |x-3| would also be negative (either equal to -y or less than -y--either way, it would be negative).

Thus, y cannot be positive. Thus, y is zero, which means (2) is really saying:

|x-3| = 0, which is solveable for x.

Choose B.

For the first case, we're given |x-3|>=y. If we open up the mod, we get -x+3>=y and x-3>=y.
This ultimately results in 3+y<=x<=3-y. Clearly, this is only possible if y=0 since y cannot be negative.

Does'nt this lead to the same conclusion as statement (2) ?

[Testluv]

When we open up the mod, we have 2 alternative inequalities--not joint inequalities. That is, it is either one inequality OR the other (not AND the other). Thus, we cannot combine them in that manner.

In other words, from (1), we know that

x-3>=y

or

-(x-3)>=y or x-3<=-y

Neither of these two inequalities allows us to compute x (because we don't know y's value).

However, (2) is different because we know that |x-3| = pos or zero. (|x-3| cannot be equal to a negative number for distance cannot be negative.) But if y were a positive number, then we would have |x-3|<= neg. Which is impossible. Thus, y cannot be positive, and since the question stem ruled out the possibility of y being negative, y must be zero.

Thus, |x-3| = 0. |a-b| is always "the distance between "a" and "b" on the number line." So, |x-3| = 0 means that "x is 0 units away from 3 on the number line" or in other words "x must be 3".