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# w, x, y, and z are integers. If w >x>y>z>0, is y a common

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Intern
Joined: 18 Aug 2012
Posts: 10

Kudos [?]: 9 [0], given: 1

GMAT 1: 730 Q50 V39
w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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27 Nov 2015, 01:11
Given to us : w > x > y > z > 0

In such type of questions, we have to start by reducing the given equations to a simple form and then test the simplified equations with a random number, so to quickly come to a required answer.

Using Statement 1

$$w/x=1/z+1/x$$

Simplifying the equation

$$w/x=(x+z)/xz$$
$$wz=x+z$$
$$(w-1)z=x$$ (Simplified Equation)

Now from the given equation $$w>x>y>z>0$$, we definitely know that z minimum will be equal to 1. Substituting z=1 in the simplified equation, we get
$$w=x+1$$.
From this equation, we can thus deduce that w and x are consecutive integers and it also satisfies w>x criteria. So if two numbers are consecutive to each other there can only be one number common to them (because these numbers become co-prime to each other, example 4 & 5) i.e. 1. but we definitely know y cannot be 1. It will definitely be greater than 1 because z=1 and y>z. So, y is not a common divisor of x and w.
Now, in case if z>1, then from the simplified equation, we will have x>w, which will contradict the given condition of (w>x>y>z>0).
Hence, statement 1 is sufficient is sufficient to answer the question that y is not a common divisor of x and w.

Using Statement 2

$$w^2-wy-2w=0$$

Simplifying the equation

$$w(w-y-2)=0$$
So, either $$w=0$$ or $$w-y-2=0$$

w=0 is not possible because $$w>x>y>z>0$$ is given to us.

Considering, $$w-y-2=0$$, we get
$$w=y+2$$.

Using this equation, we can conclude that there will be an integer between w and y, and that is definitely going to be x, because of the equation w>x>y>z>0.
For example, if w=6, then y becomes 4, then x has to be 5 to satisfy the equation.

So, from this statement we definitely know, that w,x and y are three consecutive integers.
Now three consecutive integers can only be co-prime to each other and can never have any common divisor in between them. There can only be one number common in between these numbers i.e. 1

So from this statement we can conclude that y is not the common divisor in between x and w.
Hence, Statment 2 is also sufficient to answer the Question.

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Intern
Joined: 11 Oct 2012
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GMAT 1: 610 Q42 V32
w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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11 Mar 2016, 09:51
gmacforjyoab wrote:
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

1 ) w/x = 1/z + 1/x
since W is greater than x , w/x should be greater than 1
hence 1/z +1/x should also be greater than 1. since x and z are integers grater than 0 , and z<x , the only way , 1/z +1/x can be >1 is if z=1
hence equation becomes - w/x= 1/1 + 1/x ======> w-1=x ..
hence x and w are consecutive integers and since y is not 1 and is >1 ( since z is 1) , the answer is NO
Sufficient

2) w^2-wy-2w=0
w(w-y-2) =0
since w != 0 w=y+2 .. w, x and y are consecutive integers . Same as statement 1 , the answer is NO.
Suff

-Jyothi

Thanks for the nice explanation !! Only this enabled me to understand the logic...
++ kudos

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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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13 Mar 2016, 23:05
Bunuel wrote:
ronr34 wrote:
PiyushK wrote:
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z"
Bunuel - In your solution, for st. 1, you wrote
Bunuel wrote:
since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)
. How did you reach this conclusion?

w > x > y > z > 0, is y a common divisor of w and x?

We have that z(w-1)=x and w>x>z>0. If z>1, say if z=2, then 2(w-1)=x. Now tell me can in this case x be less than w?

I did not understand one concept here
What if W=-10
w-1=-11
z=100
=> z*(w-1) = -200
so x=-200
also x<w is satisfied..

Regards
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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13 Mar 2016, 23:28
Chiragjordan wrote:
I did not understand one concept here
What if W=-10
w-1=-11
z=100
=> z*(w-1) = -200
so x=-200
also x<w is satisfied..

Regards

The stem says that all the variables are positive: w > x > y > z > 0.
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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15 Mar 2017, 01:44
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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09 Aug 2017, 22:30
What is tested is whether one is able to identify a specific set of consecutive integers.

Statement 1: z has to be 1 because w/x is greater than 1. z is the smallest integer and greater than 0. If z=1, we can see that w=x+1. A number cannot be a common divisor of two consecutive integers unless it is 1. y is not 1. Sufficient.

Statement 2: We can see W=y+2 and so w=x+1. Now the same reasoning as that for statement 1 applies. Sufficient.
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common   [#permalink] 09 Aug 2017, 22:30

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