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DS - Absolute value

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by karthikpandian19 » Tue Jun 12, 2012 7:03 pm
If z>y>x>w>0, is y-w = z-x?

1. y-w = |x-z|

2. ((w^2 - z^2)/(z-w)) = ((x^2 - y^2)/(y-x))

IMO B, but OA is D
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Source: — Data Sufficiency |
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by sam2304 » Tue Jun 12, 2012 9:42 pm
karthikpandian19 wrote:If z>y>x>w>0, is y-w = z-x?

1. y-w = |x-z|
Split into two equations

y-w = x-z --- 1
y-w = z-x --- 2
With the condition z>y>x>w>0, y-w is +ve and x-z is -ve from 1. z-x is again +ve and is equal to y-w from 2. z-x and x-z will result in same absolute value only the sign changes, so y-w = |x-z|
2. ((w^2 - z^2)/(z-w)) = ((x^2 - y^2)/(y-x))
You already know this. Simply factorize it and then cancel the like terms.

IMO D.
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by karthikpandian19 » Tue Jun 12, 2012 11:17 pm
Sorry abt tht......i couldnt get it
sam2304 wrote:
karthikpandian19 wrote:If z>y>x>w>0, is y-w = z-x?

1. y-w = |x-z|
Split into two equations

y-w = x-z --- 1
y-w = z-x --- 2
With the condition z>y>x>w>0, y-w is +ve and x-z is -ve from 1. z-x is again +ve and is equal to y-w from 2. z-x and x-z will result in same absolute value only the sign changes, so y-w = |x-z|
2. ((w^2 - z^2)/(z-w)) = ((x^2 - y^2)/(y-x))
You already know this. Simply factorize it and then cancel the like terms.

IMO D.
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by bobdylan » Wed Jun 13, 2012 2:30 am
I got B! I thought that if you multiply by -1, you need to multiply the whole equation, and not just one side. So statement 1 could be or could not be.
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by sam2304 » Wed Jun 13, 2012 2:40 am
karthikpandian19 wrote:Sorry abt tht......i couldnt get it
We are asked to find whether y-w = z-x ? This simplifies to one thing "does both y - w and z - x result in same value".

We need some more data to prove that y-w = z-x say some values or some eqns stating that they are equal. We know z > y > x > w > 0. Everything is positive.

1.y-w = |x-z|

Let w,x,y,z be 1,2,3,4.
y-w = |x-z|
=> 3-1 = |2-4|
=> 2 = |-2|
=> 2 = 2

x-z and z-x will have same absolute value and since everything is positive and z > x will result in a positive value. With 1 we get a -ve value and modulo of that will give us z-x.

Does that clarify your doubt ? If you cannot solve it algebraically use numbers.
Last edited by sam2304 on Wed Jun 13, 2012 5:30 am, edited 1 time in total.
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by karthikpandian19 » Wed Jun 13, 2012 3:59 am
@sam in your explanation thread please edit this typo error
Let w,x,y,z be 1,2,3,4.
y-z = |x-z|
=> 3-1 = |2-4|
=> 2 = |-2|
=> 2 = 2
sam2304 wrote:
karthikpandian19 wrote:Sorry abt tht......i couldnt get it
We are asked to find whether y-w = z-x ? This simplifies to one thing "does both y - w and z - x result in same value".

We need some more data to prove that y-w = z-x say some values or some eqns stating that they are equal. We know z > y > x > w > 0. Everything is positive.

1.y-w = |x-z|

Let w,x,y,z be 1,2,3,4.
y-z = |x-z|
=> 3-1 = |2-4|
=> 2 = |-2|
=> 2 = 2

x-z and z-x will have same absolute value and since everything is positive and z > x will result in a positive value. With 1 we get a -ve value and modulo of that will give us z-x.

Does that clarify your doubt ? If you cannot solve it algebraically use numbers.
Regards,
Karthik
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by karthikpandian19 » Wed Jun 13, 2012 4:03 am
For me the doubt is cleared.....thanks for your explanation
sam2304 wrote:
karthikpandian19 wrote:Sorry abt tht......i couldnt get it
We are asked to find whether y-w = z-x ? This simplifies to one thing "does both y - w and z - x result in same value".

We need some more data to prove that y-w = z-x say some values or some eqns stating that they are equal. We know z > y > x > w > 0. Everything is positive.

1.y-w = |x-z|

Let w,x,y,z be 1,2,3,4.
y-z = |x-z|
=> 3-1 = |2-4|
=> 2 = |-2|
=> 2 = 2

x-z and z-x will have same absolute value and since everything is positive and z > x will result in a positive value. With 1 we get a -ve value and modulo of that will give us z-x.

Does that clarify your doubt ? If you cannot solve it algebraically use numbers.
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

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by sam2304 » Wed Jun 13, 2012 5:34 am
karthikpandian19 wrote:For me the doubt is cleared.....thanks for your explanation
sam2304 wrote:
karthikpandian19 wrote:Sorry abt tht......i couldnt get it
We are asked to find whether y-w = z-x ? This simplifies to one thing "does both y - w and z - x result in same value".

We need some more data to prove that y-w = z-x say some values or some eqns stating that they are equal. We know z > y > x > w > 0. Everything is positive.

1.y-w = |x-z|

Let w,x,y,z be 1,2,3,4.
y-z = |x-z|
=> 3-1 = |2-4|
=> 2 = |-2|
=> 2 = 2

x-z and z-x will have same absolute value and since everything is positive and z > x will result in a positive value. With 1 we get a -ve value and modulo of that will give us z-x.

Does that clarify your doubt ? If you cannot solve it algebraically use numbers.
Done with the edit. Hope its clear. I know I did a pathetic job in explaining the problem, because I couldn't write down my thoughts clearly maybe you can wait for someone else as well to pour in their thoughts to get a better understanding.
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by GMATGuruNY » Thu Jun 14, 2012 3:03 am
karthikpandian19 wrote:If z>y>x>w>0, is y-w = z-x?

1. y-w = |x-z|

2. ((w^2 - z^2)/(z-w)) = ((x^2 - y^2)/(y-x))
Statement 1: y-w = |x-z|
|x-z| is the DISTANCE between x and z.
The distance between two values must be NON-NEGATIVE.
Since x<z, the distance between x and z = z-x.
In other words, |x-z| = z-x.
Substituting |x-z| = z-x into y-w = |x-z|, we get:
y-w = z-x.
SUFFICIENT.

Statement 2: (w² - z²)/(z-w) = (x² - y²)/(y-x)
(w+z)(w-z) / (z-w) = (x+y)(x-y) / (y-x)
(w+z)(w-z) / (w-z) = (x+y)(x-y) / (x-y)
w+z = x+y
z-x = y-w.
SUFFICIENT.

The correct answer is D.
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by hey_thr67 » Thu Jun 14, 2012 3:52 am
What is the source of this problem ?
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by karthikpandian19 » Thu Jun 14, 2012 5:06 pm
Source: Knewton GMAT
hey_thr67 wrote:What is the source of this problem ?
Regards,
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by karthikpandian19 » Thu Jun 14, 2012 5:07 pm
Thanks for the briefing MITCH
GMATGuruNY wrote:
karthikpandian19 wrote:If z>y>x>w>0, is y-w = z-x?

1. y-w = |x-z|

2. ((w^2 - z^2)/(z-w)) = ((x^2 - y^2)/(y-x))
Statement 1: y-w = |x-z|
|x-z| is the DISTANCE between x and z.
The distance between two values must be NON-NEGATIVE.
Since x<z, the distance between x and z = z-x.
In other words, |x-z| = z-x.
Substituting |x-z| = z-x into y-w = |x-z|, we get:
y-w = z-x.
SUFFICIENT.

Statement 2: (w² - z²)/(z-w) = (x² - y²)/(y-x)
(w+z)(w-z) / (z-w) = (x+y)(x-y) / (y-x)
(w+z)(w-z) / (w-z) = (x+y)(x-y) / (x-y)
w+z = x+y
z-x = y-w.
SUFFICIENT.

The correct answer is D.
Regards,
Karthik
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