MGMAT CAT 5 DS 700-800

[skprocks]

What is the value of y?

(1) 3|x^2 - 4| = y - 2

(2) |3 - y| = 11

OA is C

I am unable to comprehend that |x^2-4| > 0.And thus y-2 > 0.This is not clear to me.Why can't it be possible that y-2 < 0
as we do not know the value of x.Please explain.The source of the question is a MGMAT CAT.MGMAT staff may reply and ellucidate their explannation.

[Makushr1]

What is the value of y?

(1) 3|x^2 - 4| = y - 2

(2) |3 - y| = 11

OA is C

I am unable to comprehend that |x^2-4| > 0.And thus y-2 > 0.This is not clear to me.Why can't it be possible that y-2 < 0
as we do not know the value of x.Please explain.The source of the question is a MGMAT CAT.MGMAT staff may reply and ellucidate their explannation.

Look at the Y part of 1. y-2. Since 3|x^2-4| has to be positive (the absolute part time 3 is going to be even), you know that Y cannot be less than 2. If it were less than 2, that side would be negative which means the other side needs to be negative as well, and we know it cant since it a positive 3 times an absolute value equation.

With 2, Y can either be -8 or 14. Since, in (1) we know y>2, -8 can't be correct, so it has to be 14, which means both are required to be suff.

[skprocks]

With 2, Y can either be -8 or 14. Since, in (1) we know y>2, -8 can't be correct, so it has to be 14, which means both are required to be suff.

In light of what you have explained about stmt1,stmt 2 shall also be always positive.How can it assume both positive and negative values?(-8/14). Sorry to bother you , but the concept behind treating stmt 1 's absolute value as positive and treating the absolute value in stmt2 as either positive or negative is not clear to me.Thanks!![/quote]

[Makushr1]

With 2, Y can either be -8 or 14. Since, in (1) we know y>2, -8 can't be correct, so it has to be 14, which means both are required to be suff.

In light of what you have explained about stmt1,stmt 2 shall also be always positive.How can it assume both positive and negative values?(-8/14). Sorry to bother you , but the concept behind treating stmt 1 's absolute value as positive and treating the absolute value in stmt2 as either positive or negative is not clear to me.Thanks!!

[/quote]

For statement 2, we are not looking for a solution to the absolute value of 3-y because we already know the answer is positive 11. Rather, for statement 2, we are looking for what Y can be. Y is inside the absolute value, so it can be either positive or negative, and still result in a positive 11.

|3 - y| = 11

Y can either be -8 or 14

|3-14| = |11| = 11

|3- (-8)| = |-11| = 11

So yes, you're correct, the absolute value of 3-y will always be positive, but Y can either be positive or negative.

I got the -8 and 14 by looking at the equation, but if you are unable to simply visualize it, try taking off the absolute value and write 2 equations, one with 11 being positive and one with 11 being negative.

3-Y=11
-Y=11-3
-y=8
y=-8

3-Y=-11
-y=-11-3
-y=-14
y=14

So since statement 2 yields 2 possible answers, we know it's not sufficient.

Now we have 2 answers for what Y can be. In the first statement, Y is NOT the result of an absolute value equation, so it can be either -8 or 14. But, when you plug them in, it will be this:
3|x^2 - 4| = -8-2
3|x^2 - 4| = -10
This is not possible, since a positive 3 times the absolute value of anything will be positive. The equation inside the absolute value doesn't matter, since the result will always be positive. 3 * |anything| = positive

3|x^2 - 4| = 14-2
3|x^2 - 4| = 12



Let me know if you still don't understand. It's hard explaining these things over the computer.

[sk818020]

You can really just think your way through this one.

1) is clearly insufficient to find the value of y because we cannot determine the value of x.

2) Because of the nature of absolute value equations this tells us that that there are two possible values for y. Insufficient.

You know that 2) tells us that there are 2 values for y, you can plug those in to 1) to determine the value of y. But just knowing this allows you to answer C.

Thanks,

Jared