Gmat Prep Absolute Values??

From statement one, it looks like you are being told that x cannot equal to 3. But, it can equal to 4 and -4. When you plug them in, you end up with.

1 = -1 when pluggin in 4, and the answer is no.

7 = 7 when plugging in -4, and the answer is yes.

I would, therefore, say that statement 1 is not sufficient.

It seems that the second statement says that -x multiplied by the absolute value of x is greater than zero. This can only be true if x is a negative number. For example:

-(-2) x |-2| > 0

If x is a positive number then the condition will not hold:

-2 x |2| > 0 is false, since -4 is NOT greater than zero.

So, try two negative numbers, -1 (since 1 is a weird number when it comes to testing) and try -3.

You end up with:

4 =4 when plugging in -1, and the answer is yes.

6 = 6 when plugging in -3, and the answer is yes.

Now, what if you try to plug in -1/2?

7/2 = 7/2

I would say that the second statement is sufficient and the answer is B, if I was taking this CAT.

Hope someone else can improve upon my efforts.

For this question, first we realize that the question stem is true only if x <= 3. For example, If x = 4. the equation is not true. If the x = 2 or -4, the equation is true.

Once we realize that we can look at the 2 alternatives.
A. x <3> 3 or x < 3. So we can not say for sure whether x <= 3. Insufficient.

B. This statement is true only if x < 0. This also means that x < 3. So, B is sufficient.

The first thing I did was square both sides of the equation so it got rid of the square root sign...

For example:

sqrt (x-3)^2 = (3-x)^2

If I squared both sides, then I got:

(x-3)^2 = (3-x)^2

However, when I did that, X could be both negative and positive, and the equation would yield equal results. This gives me the wrong answer, but I'm not sure why.

Can someone explain my mistake? thx.

I think your mistake lies in the following:

Instead of looking back at the original expression that has a sqroot sign, you just look at the expression you ended up with when you simplified. You are correct in saying that any number works for the simplified form - after all, it equates two exact expressions to each other, which means that both will give the same exact output given an exact input. However, you have to test your input numbers in the original expression. This is also true as far as mathematics in general goes. Whenever you get rid of a radical, you must test your input numbers on the original expression, because radicals are tricky that way. So, here goes.

When you square both sides you end up with:

(x - 3)^2 = (3 - x)^2 or (x - 3)(x - 3) = (x - 3)(x - 3), so x = 3.

The first statement tells you that x does not equal to 3, which leaves you with x equaling anything but 3. This does not help you much since x could be positive, in which case the answer is no, or it could be negative, in which case the answer is yes.

The second statement tells you that when you take the absolute value of x and multiply it by a negative of whatever values x is, you end up with a value that is greater than zero. This can only be true if the value of x is negative. For example:

- (-3) x |-3| > 0
3 x 3 > 0 or 9 > 0, which is true. When you plug -3 into the problem, or just about any other negative number, you end up with a yes.