BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Need Help with Manhattan Gmat question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

Re: Need Help with Manhattan Gmat question

by camitava » Sat Mar 15, 2008 6:32 pm
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.
Correct me If I am wrong


Regards,

Amitava
Top
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Sat Dec 29, 2007 10:24 pm

by ofolami » Sat Mar 15, 2008 6:47 pm
Thanks for the attempt. I choose E too, but the answer is actually C. I need to figure that out.
Top
Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Aug 23, 2007 5:11 am
Thanked: 2 times

by gmatguy16 » Sun Mar 16, 2008 4:27 pm
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
Top
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 am
Location: India
Thanked: 51 times
Followed by:1 members

Re: Need Help with Manhattan Gmat question

by gabriel » Mon Mar 17, 2008 12:49 am
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
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Mon Dec 10, 2007 1:35 pm
Location: London, UK
Thanked: 19 times
GMAT Score:680

by II » Mon Mar 17, 2008 12:54 am
By looking at the first statement, how can you see that y has to be positive ?

Thanks.
Top
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

by camitava » Mon Mar 17, 2008 1:50 am
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 ... :cry:
Correct me If I am wrong


Regards,

Amitava
Top
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Wed Dec 12, 2007 8:29 am

explanation

by anandsv » Mon Mar 17, 2008 2:00 am
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
Top
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

Re: explanation

by camitava » Mon Mar 17, 2008 3:06 am
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 ... :wink: :roll: :cry: :lol:
Correct me If I am wrong


Regards,

Amitava
Top
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 am
Location: India
Thanked: 51 times
Followed by:1 members

Re: explanation

by gabriel » Mon Mar 17, 2008 4:17 am
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 ... :wink: :roll: :cry: :lol:
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
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Mon Mar 17, 2008 5:58 pm
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).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

by camitava » Mon Mar 17, 2008 8:43 pm
Thanks Stuart and Gabriel, got it now! I initially failed to understand the issue I faced... :cry:
Correct me If I am wrong


Regards,

Amitava
Top
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Mon Jan 19, 2009 12:11 am
Thanked: 3 times
GMAT Score:640

great catch on the '3 x absolute value can't be neg. thing'

by Edthesock » Tue Jan 20, 2009 6:28 pm
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!!!
Top
Post new topic Post Reply
• Page 1 of 1