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If all variables are positive, is (w/x) > (y/z)?

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If all variables are positive, is (w/x) > (y/z)? [#permalink] New post 06 Jan 2014, 00:11
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57% (01:00) correct 43% (01:10) wrong based on 7 sessions
If all variables are positive, is (w/x) > (y/z)?

(1) w = y + 50
(2) x = z + 50

[Reveal] Spoiler:
1) and 2) alone are INS.

(y+50)/(z+50) > y/z
yz +50z > yz +50y
50y > 50z
y > z

Both together are INSUFICIENT so E)

My question is, however: The first time that I worked it out, I simply did this...

If w = y + 50 , then w > y
If x = z + 50 , then x > z

Divide equations by each other:

(w > y) ÷ (x > z) = (w/x) > (x/z)

Why isn't this right?
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jan 2014, 01:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Concentration: Entrepreneurship, Finance
Schools: Yale '16, ISB '15, NUS
GMAT Date: 04-30-2014
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Re: Inequality Question [#permalink] New post 06 Jan 2014, 00:20
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boyanshmorhun wrote:
If all variables are positive, is (w/x) > (y/z)?

1) w = y + 50
2) x = z + 50

1) and 2) alone are INS.

(y+50)/(z+50) > y/z
yz +50z > yz +50y
50y > 50z
y > z

Both together are INSUFICIENT so E)

My question is, however: The first time that I worked it out, I simply did this...

If w = y + 50 , then w > y
If x = z + 50 , then x > z

Divide equations by each other:

(w > y) ÷ (x > z) = (w/x) > (x/z)

Why isn't this right?



Hi

You can't divide with relational operators as all the w,x,y,z are variable and +ve. They can be equal in pairs.

consider the simple example :

y= 1
w= 1+50= 51

z = 1
x= 51

So w/x = y/z

So doing a division operation in relational operator with variable is not so good appraoch where you dont know the range of the number.
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AB

+1 Kudos if you like and understand.

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Re: Inequality Question [#permalink] New post 06 Jan 2014, 00:28
boyanshmorhun wrote:
If all variables are positive, is (w/x) > (y/z)?

1) w = y + 50
2) x = z + 50

1) and 2) alone are INS.

(y+50)/(z+50) > y/z
yz +50z > yz +50y
50y > 50z
y > z

Both together are INSUFICIENT so E)

My question is, however: The first time that I worked it out, I simply did this...

If w = y + 50 , then w > y
If x = z + 50 , then x > z

Divide equations by each other:

(w > y) ÷ (x > z) = (w/x) > (y/z)

Why isn't this right?


Hi,

Let us take a simple example:
3 > 2 (As w > y)
-3 > -4 (As x > z)

Now (w/x) > (y/z) so (3/-3) > (2/-4) i.e. -1 > -0.5 (NOT true)

In inequalities we cannot use algebra with freedom as we do not know the sign of variables.

So if x/y > 1 we cannot say x > y

I hope it helps.
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Re: Inequality Question [#permalink] New post 06 Jan 2014, 00:43
bhatiamanu05 wrote:
boyanshmorhun wrote:
If all variables are positive, is (w/x) > (y/z)?

So doing a division operation in relational operator with variable is not so good appraoch where you dont know the range of the number.


Thanks!

What about multiplying?

We know that w>y and x>z.

(w>y) * (x>z) = wx>yz --> INSUFFICIENT because this doesn't solve whether (w/x) > (y/z).
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Re: Inequality Question [#permalink] New post 06 Jan 2014, 00:44
boyanshmorhun wrote:
bhatiamanu05 wrote:
boyanshmorhun wrote:
If all variables are positive, is (w/x) > (y/z)?

So doing a division operation in relational operator with variable is not so good appraoch where you dont know the range of the number.


Thanks!

What about multiplying?

We know that w>y and x>z.

(w>y) * (x>z) = wx>yz --> INSUFFICIENT because this doesn't solve whether (w/x) > (y/z).


You can add or subtract a variable however cannot divide or multiply a variable.
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Re: Inequality Question   [#permalink] 06 Jan 2014, 00:44
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