Math problems from the *free* official practice tests and
problems from mba.com
guptakshay
Students
 
Posts: 4
Joined: Fri Sep 09, 2011 7:10 am
 

Is |x| > |y|?

by guptakshay Thu Feb 09, 2012 4:06 pm

Is |x| > |y|?

1. x^2 > y^2
2. x>y

Can someone please explain why the answer is C?
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Is |x| > |y|?

by RonPurewal Thu Feb 09, 2012 8:35 pm

guptakshay Wrote:Is |x| > |y|?

1. x^2 > y^2
2. x>y

Can someone please explain why the answer is C?


any useful explanation needs to work with what you already do and don't know.
therefore, please tell us (1) what you DO understand about the problem, and (2) SPECIFICALLY WHAT you don't understand about it.

thanks.
guptakshay
Students
 
Posts: 4
Joined: Fri Sep 09, 2011 7:10 am
 

Re: Is |x| > |y|?

by guptakshay Fri Feb 10, 2012 3:24 am

[editor's warning: the answer reached in this post is incorrect.]

Is |x| > |y|?

1. x^2 > y^2
2. x>y

Thanks for the guidance Ron.

Please let me know the numbers required to prove statement 1 insufficient. Below are my calculations.

Solving Statement 2. x>y first
By plugging in the following numbers, I can prove that this statement is insufficient.

Option 1. x,y( 2,-3)
When absolute value of (2, -3) results to |x| < |y|
[editor: this is the problem. you can't use x = 2 and y = -3, because the statement x^2 > y^2 is false for these two values.]

Option 2. x,y (4,3)
When absolute value is taken results to (4, 3) therefore |x| > |y|
Therefore statement 2 is insufficient.

Statement 1. x^2 > y^2

Taking square roots results in 4 possibilities:
Base number Square of base Absolute value of base
A. x>y (4,2) (16,4) |x| > |y|
B. x>-y (4,-3) (16,9) |x| > |y|
C. x<-y (-3,-2) (9,4) |x| > |y|
D. x<y (-3,2) (9,4) |x| > |y|



All the possibilities satisfy the statement 1 and lead to |x| > |y|.
I can think of two numbers that can prove this statement insufficient.

However, for C to be true, I do know that I can use the numbers from statement 2 to test statement 1 and get only one solution:

Option 1. x,y( 2,-3)
Square of (2, -3) = (4, 9) does not satisfy statement 1
Option 2. x,y (4,3)
Square of (4, 3) = (16, 9) satisfies statement 1.
guptakshay
Students
 
Posts: 4
Joined: Fri Sep 09, 2011 7:10 am
 

Re: Is |x| > |y|?

by guptakshay Sun Feb 12, 2012 12:28 am

Any thoughts on how I could prove statement 1 insufficient?
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Is |x| > |y|?

by RonPurewal Fri Feb 17, 2012 5:31 am

guptakshay Wrote:@Jamie, Ron: Any thoughts on how I could prove statement 1 insufficient?


you shouldn't be able to prove statement 1 insufficient ... because statement 1 is sufficient!

the inequalities x^2 > y^2 and |x| > |y| convey exactly the same information. there are a couple of different ways you can think about this.

* conceptually: either of these inequalities, from a conceptual standpoint, means "x is bigger in magnitude than y, but we don't know the sign of either number". therefore, they are equivalent.

* algebraically: since both x^2 and y^2 are non-negative quantities, you can take the square root of both sides of the inequality. but the square root of x^2 is |x|, and the square root of y^2 is |y|, so this operation produces an immediate "yes" answer to the question.

* you can also just test a bunch of numbers until you are convinced. no matter what numbers you make up, any numbers satisfying x^2 > y^2 will also satisfy |x| > |y|.

so, this is actually (a), not (c).
shubham_sagijain
Forum Guests
 
Posts: 27
Joined: Wed Jan 18, 2012 6:50 pm
 

Re: Is |x| > |y|?

by shubham_sagijain Thu Mar 01, 2012 8:34 am

Ron Sir,

I approached this problem in the following way. Please let me know whether i am correct.

we need to find whether Mod (X) > Mod (Y).
This is equivalent to asking whether sqrt (X^2) > sqrt (Y^2).

S1 : X^2 > Y^2

So, surely, sqrt (X^2) > sqrt (Y^2).

S2 : Not sufficient

2nd Approach :

Mod (X) > Mod (Y)

From this, we have

Case 1: If x > y > 0
Case 2 : If 0 > y > x

So, S1 is sufficient.

S2 : Just says x > y but do not actually reveal whether they are greater or less than 0. So, insufficient.

Thanks,
Shubh
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Is |x| > |y|?

by RonPurewal Sat Mar 03, 2012 8:44 am

shubham_sagijain, your approaches look good, except in the following part:

Case 1: If x > y > 0
Case 2 : If 0 > y > x


there are also two more cases, y < 0 < x and x < 0 < y, to consider, since we have no information regarding signs.
since you've listed only these two cases, it appears you're assuming that x and y have the same sign (i.e., are either both positive or both negative). there's no reason to make that assumption.
this observation doesn't change the outcome of this problem, but it could very well make a difference on other problems that test similar material.
tanyatomar
Students
 
Posts: 49
Joined: Fri Sep 02, 2011 6:44 am
 

Re: Is |x| > |y|?

by tanyatomar Mon Mar 05, 2012 12:31 pm

Hi Ron,
i have still not understood why C. According to me it should be A.. since x^2 is equivalent to |x|.. then why do we need x>y to be proven... i mean if x=-4, y= -3 => x<y but |x| > |y|... Also if x=4, y = 3 => x > y but still |x| > |y|...
Basically according to me A is sufficient and we donot need B..
But then how the answer is C....
Please do explain....

Tanya
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Is |x| > |y|?

by RonPurewal Wed Mar 07, 2012 8:09 pm

scroll up to my post above, in which i stated that the answer to this problem is (a).

(here's a link to that post:
post62261.html#p62261)