Need Help with Manhattan Gmat question

ofolami wrote:What is the value of y?

(1) 3|x^2 – 4| = y – 2

(2) |3 – y| = 11

From stmt-1, u can not get the value of y as it is a single equ. with two unknown variables.

From stmt-2, u can get the value of y depending upon the fact whether 3 - y is positive or negative. So little bit uncertainty here also! U can not find the exact value of y. So IMO E.

Thanks for the attempt. I choose E too, but the answer is actually C. I need to figure that out.

a is of course not sufficiet,if you solve for b you have 2 possible values of y 14 and -8..but if you combine both you realize that y has to be positive ..
y = -8 is not possible .hence answer c

camitava wrote:

ofolami wrote:What is the value of y?

(1) 3|x^2 – 4| = y – 2

(2) |3 – y| = 11

From stmt-1, u can not get the value of y as it is a single equ. with two unknown variables.

From stmt-2, u can get the value of y depending upon the fact whether 3 - y is positive or negative. So little bit uncertainty here also! U can not find the exact value of y. So IMO E.

Hey Camitava,

If you take a look at the first statement you will realize that y can take only positive values. So the answer is indeed C

Regards

By looking at the first statement, how can you see that y has to be positive ?

Thanks.

gabriel & Gmatguy18, I am not understanding from stmt-1, how r u getting confirmed that y has to be positive?
Look, if y = -8,
3|x^2 - 4| = -8 - 2
or 3(x^2 - 4) = -10
or 3x^2 = 2 -> x^2 = 2/3
Again,
3(4 - x^2) = -8 - 2
or 12 - 3x^2 = -10
or x^2 = 22/3
So in both cases we are getting some acceptable values for x. So why we are rejecting the negative value of y? Need little more explanation ...

hey Amitava

3|x^2 – 4| three times the absolute value has to be positive always so the value of Y-2 has to be positive...
thus Y has to be positive

hope this helps!

Andy

anandsv wrote:hey Amitava

3|x^2 – 4| three times the absolute value has to be positive always so the value of Y-2 has to be positive...
thus Y has to be positive

hope this helps!

Andy

anandsv, here I am not getting onething why you people are trying to take |x^2 - 4| as positive? Suppose, x = 1 so x^2 - 4 = -3 or x = 3, so x^2 - 4 = 5. So we can not say that |x^2 - 4| means always (x^2 - 4) but in some cases, it can be like (4 - x^2).
Got me, Andy? Now another Qs I am having is if we are trying to take only positive value of |x^2 - 4|, then why did we not take the positive value of
|3 - y|? Why are we trying to consider both the values of (3 - y) and (y - 3)?
Look Andy, I am not trying to confuse u? I am trying to clear my doubt ... :roll:

camitava wrote:

anandsv wrote:hey Amitava

3|x^2 – 4| three times the absolute value has to be positive always so the value of Y-2 has to be positive...
thus Y has to be positive

hope this helps!

Andy

anandsv, here I am not getting onething why you people are trying to take |x^2 - 4| as positive? Suppose, x = 1 so x^2 - 4 = -3 or x = 3, so x^2 - 4 = 5. So we can not say that |x^2 - 4| means always (x^2 - 4) but in some cases, it can be like (4 - x^2).
Got me, Andy? Now another Qs I am having is if we are trying to take only positive value of |x^2 - 4|, then why did we not take the positive value of
|3 - y|? Why are we trying to consider both the values of (3 - y) and (y - 3)?
Look Andy, I am not trying to confuse u? I am trying to clear my doubt ... :roll:

Camitava, |x^2 - 4| is always positive, if x = 1 then x^2-4 = -3 but |x^2 - 4| = 3.

The definition of |x| is,
|x| = x when x>0
or |x| = -x when x < 0

Similarly |x^2 - 4| = x^2-4 when x^2-4 >0
or |x^2 - 4| = 4-x^2 when x^2-4 < 0

As you can see in both cases the value of a modulus is always positive

Regards

Small refinement - absolute value can also be 0, since |0| = 0.

Looking at (1), we can rearrange for y:

3|x^2 – 4| = y – 2

y = 3|x^2 - 4| + 2

Since we know that absolute value is always >= 0, we know that the smallest possible value for y is +2.

So, (1) tells us that y will be >= 2, (2) tells us that y=-8 or y=14; together, we know that y=14: choose (c).

Thanks Stuart and Gabriel, got it now! I initially failed to understand the issue I faced...

I was set to pick E, as I had also gotten -8 and 14 as my y values, and didn't believe that the first statement helped me eliminate either one as I could set up equations with both. Had I taken a closer look, I would have noticed that

3|x2 – 4| = y – 2 is only possible for +14 because 3|x2 – 4| MUST be a positive number, which would make -10, or -8 - 2, an unattainable value. 14 is the only possible y value, so C is the answer...

Well done to those who got this, big brains ftw!!!