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by GMATGuruNY » Fri Sep 05, 2014 6:22 am
j_shreyans wrote:If v≠0, is |w|<|v|?
(1) w/v<1

(2) w^2/v^2<1
Since an absolute value must be NONNEGATIVE, we can safely rephrase the question stem by squaring both sides:
(|w|)² < (|v|)²
w² < v².

Question stem, rephrased:
Is w² < v²?

Statement 1: w/v < 1
If w=0 and v=1, then w² < v².
If w=-1 and v=1. then w² = v².
INSUFFICIENT.

Statement 2: w²/v² < 1
Since v≠0, we know that v²>0, implying that we can safely multiply each side by v²:
(w²/v²) * v² < 1 * v²
w² < v².
SUFFICIENT.

The correct answer is B.
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by [email protected] » Fri Sep 05, 2014 10:38 am
Hi j_shreyans,

This question is perfect for TESTing Values (and a bit of Number Properties).

We're told that V≠0 and we're asked if |W|<|V|. This is a YES/NO question.

Fact 1: W/V < 1

Let's TEST VALUES:
If W = 1, V = 2, then the answer to the question is YES.
If W = -3, V = 2, then the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: W^2/V^2 < 1

This becomes W^2 < V^2.

The Number Property here is that "squaring" a term has the same "effect" as the absolute value signs in the question: any negative signs are removed. This ultimately means that the absolute value of V will ALWAYS be greater than the absolute value of W. The answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT.

Final Answer: B

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by GMATinsight » Fri Sep 05, 2014 10:46 am
j_shreyans wrote:If v≠0, is |w|<|v|?
(1) w/v<1

(2) w^2/v^2<1

OAB
Question :Is |w|<|v|?

Given : v≠0

Statement 1) w/v<1

Less than 1 could be positive or negative
for w and v to be positive w will be lesser than v
and for w/v to be positive |w| can be greater than |v| (e.g. w = 2 and v = -1 )
NOT SUFFICIENT

Statement 2) w^2/v^2<1
Cross multiplying (as both w^2 and v^2 have to positive due to being squares)
w^2 < v^2

For this to be true |w| will certainly be lesser than |v| (Substitute values for certainty)
SUFFICIENT

Answer: Option B
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by j_shreyans » Fri Sep 05, 2014 8:53 pm
Guys ,

Thanks for your reply but i am still confused in statement 2 i.e. w^2/v^2<1

If we multiply v^2 both side we will get w^2<v^2 right?

If i put the value w=-1 and v=1 so this is not satisfying can you pls help me more here.

Thanks in advance.

Shreyans
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by GMATinsight » Fri Sep 05, 2014 9:34 pm
j_shreyans wrote:Guys ,

Thanks for your reply but i am still confused in statement 2 i.e. w^2/v^2<1

If we multiply v^2 both side we will get w^2<v^2 right?

If i put the value w=-1 and v=1 so this is not satisfying can you pls help me more here.

Thanks in advance.

Shreyans
True that w=-1 and v=1 are not satisfying therefore they are not acceptable values for this constraint.

If you take any any set of values e.g.w = 1 and v = 2 which satisfies the given constraint w^2<v^2 then you would see that |w| will always be less than |v| therefore Consistent Answer.
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by j_shreyans » Fri Sep 05, 2014 10:16 pm
Hi ,

But how do we come to know that we should not consider True that w=-1 and v=1 are not satisfying therefore they are not acceptable values for this constraint and why we should test the value .

I am getting confused here and not getting it :(
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by GMATinsight » Fri Sep 05, 2014 10:25 pm
j_shreyans wrote:Hi ,

But how do we come to know that we should not consider True that w=-1 and v=1 are not satisfying therefore they are not acceptable values for this constraint and why we should test the value .

I am getting confused here and not getting it :(
Testing values is a good method therefore it's STRONGLY suggested that one must learn using it however in this particular question if you wish to avoid testing values then you must understand that

1) w^2 < v^2 only suggest that signs of w and v are insignificant and for answering the question as well signs of w an v are insignificant as we are talking about the absolute values of w and v

2) w^2 will be less than v^2 only when absolute value of w is lesser than v as the power of both w and v are same on both sides and therefore this observation answers the question
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by Matt@VeritasPrep » Sat Sep 06, 2014 9:23 am
j_shreyans wrote:Guys ,

Thanks for your reply but i am still confused in statement 2 i.e. w^2/v^2<1

If we multiply v^2 both side we will get w^2<v^2 right?

If i put the value w=-1 and v=1 so this is not satisfying can you pls help me more here.
An important note here: you can't use w = 1 and v = 1 because they DON'T SATISFY a statement you know to be true. In other words, you can ONLY pick values of w and v that satisfy the statement w²/v² < 1, and it's NOT TRUE that (-1)²/1² < 1.

To give a real world example of this, suppose I told you that I'd give you any amount of money less than $1. You can't say, "Hey, give me $1.50!" because that doesn't fit the terms of the offer. You have to satisfy the conditions given.
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