B) |3-y|=11
|y-3|=11
if y>=3,y=14
if y<=3, y=-8. 2 solutions. insufficient
A) 3|x^2-4|=y-2
x value is not known. thus insufficient.

A&B) if y=-8, 3* |x^2 - 4| = -10 <0 not possible.
Thus y=14 sufficient
IMO C

Statement 1 is clearly not sufficient because x can be anything. However, it does give us a lower limit on the value of y. The smallest that |x^2-4| can be is zero because it is an absolute value. Hence, the smallest possible value for y is 2.

Statement 2 tells us that y could be 14 or -8.

Statements 1&2 tell us that y must be 14 because -8 is below the lower limit for y established in statement 1.

Ans: C

GmatMathPro wrote:Statement 1 is clearly not sufficient because x can be anything. However, it does give us a lower limit on the value of y. The smallest that |x^2-4| can be is zero because it is an absolute value. Hence, the smallest possible value for y is 2.

Hi Pete,
I am a bit confused. Some times we have to consider two cases(+/-) of an absolute inequality or equality. eg |x-2|=4.In this case we write (x-2)=4 or (x-2)=-4
How is statement above different from the example I gave.
Why cant we say, 3|x^2-4|=y-2 ----> 3(x^2-4)= +/- (y-2)
I am clearly lacking this concept. Please help!

Let's look at some slightly simpler cases to clarify why we do what we do with absolute values.
If you're solving something like |x-2|=4, the thought process would go like this: The absolute value of SOMETHING is 4. The only way SOMETHING can have an absolute value of 4 is if that SOMETHING is either 4 or -4. In this case the SOMETHING is x-2. So this equation would only work if that SOMETHING, x-2, is either 4 or -4, which yields the two equations x-2=4 and x-2=-4.

However, what if we have the equation |x-2|=-4. Using the same thought process: The absolute value of SOMETHING is -4. But now we have a problem. an absolute value can NEVER be negative. So there are no values of x that solve this equation; we're finished before we start. No matter how hard we look there are no x-values that can change the fact that absolute values are always positive. So, we see that the rule "set what's inside the absolute value bars equal to plus and minus the other side of the equation" should not be applied blindly.

The thing to take away here is that when we rewrite |x-2|=4 as x-2=4 or x-2=-4, we are writing a set of equations that is EQUIVALENT to the original equation. That is a key principle in any legitimate mathematical manipulation. If you rewrite something, you'd better be sure it's equivalent to what you started with, or you're changing the nature of the problem. In this case, |x-2|=4 has the exact same solutions as the the two equations x-2=4 and x-2=-4: 6, -2. |x-2|=4 is not IDENTICAL to x-2=4 or x-2=-4, but it is EQUIVALENT in the sense that they give us the same solutions. On the other hand, if you rewrite |x-2|=-4 as x-2=4 or x-2=-4, you've taken something with zero solutions and rewritten it as something with two solutions. This is clearly not equivalent, which is why it is not allowed.

Now, let's look at the case of y=|x-2|. If we try to follow the same thought process it is a bit trickier. If we again blindly apply the rule, we get x-2=y or x-2=-y. Rewriting, we get y=x-2 or y=2-x. Now, the question is, are these two equations taken together equivalent to the original y=|x-2|. And the answer is no. Look at the following graphs:

y=|x-2|


and y=x-2 y=2-x plotted on the same graph


Notice that the first equation produces a graph that is entirely above the x-axis. That is, every point on the graph has a positive y-coordinate, as it should because y is defined to be the absolute value of something. In the second graph, though, we get the same V-shaped graph as in the first graph, but we also get the reflection across the x-axis which includes a bunch of points with negative y-coordinates. So, clearly, y=x-2 and y=2-x is not equivalent to y=|x-2|. Typically if you want to rewrite an absolute value function like this you will have to put restrictions on the x-value. In this case we could say that y=|x-2| is equivalent to y=x-2 for all x>2 and y=2-x for all x<=2. Notice that if we restrict x to these values, y is still always positive.

Now, finally, getting back to the original problem, you COULD rewrite y-2=3|x^2-4| as +-(y-2)=3(x^2-4) IF YOU INCLUDE THE PROPER RESTRICTIONS. However, with no restrictions, it is not equivalent to the original form because, among other things, it would now allow negative y-values, whereas before when you have y=3|x^2-4|+2, there is no way y can be negative because it is equal to 3*(something that is never negative) plus 2. Specifically, there is no way it could be -8.

I hope that helps.

Pete,
Thanks for taking time in explaining some really good stuffs about the absolute value. Specially, the line representation of absolute values and the importance of restrictions while considering the possible sub-equations clarified all my doubts related to this specific case.Now, I feel lot more comfortable on absolute values.

Thanks a lot!
-Smack