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Folks - started to solving it like this, but got stuck after a while. Can someone please help?

Basically this question is asking whether y is factor of both w & x?

Considering Statement 1

w/x=1/z+1/x. Since w>x , the quotient w/x must be greater than 1. For example 17/6 = 2 5/6

Since w/x is greater than 1 then 1/z + 1/x must be greater than 1 as well. For 1/z+1/x to be greater than 1 both z and x has to be 1. x cannot be equal to 1 because of the condition w > x > y > z > 0, therefore, z must be equal to 1. We can now re write the equation as w/x = 1+1/x which will give us w = x+1. If w= x+1 then mean x and w are consecutive integers and therefore no number other than 1 can be a factor of both.

I got stuck after this. But from my guess work I think this statement looks sufficient. I know not the right way of doing MATHS, therefore can someone please help?

Considering Statement 2

w^2-wy-2w = 0

If we factor above into, that will give w(w-y-2) = 0. This means that either w = 0 or w-y-2 = 0. w cannot be zero because w > x > y > z > 0, therefore w-y-2 = 0 or w = y+2. Since x falls between y and w we can tell that x, y and w are consecutive integers and therefore no number other than 1 can be a factor of both. Since y >1 it cannot be a factor of both w & x. Therefore this statement is sufficient to answer the question.

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

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 -> \frac{w}{x}=\frac{1}{z}+\frac{1}{x} --> multiply both sides by xz --> wz=x+z --> z(w-1)=x --> since w>x>z then z=1 (if z>1 then x>w which contradicts given condition) --> w=x+1 --> w and x are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus w and x don not share any common factor more than 1 and as y>1 (from y>z>0) then y is not a common factor of w and x. Sufficient.

(2) w^2 – wy – 2w = 0 --> w(w-y-2)=0, since w>0 then: w-y-2=0 --> w-2=y -->s w>x>w-2 (substituting y in the given inequality) --> x=w-1. The same as above. Sufficient.

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]
10 Sep 2012, 22:26

Expert's post

experts Initially I went for B but after carefully examining bunuel's solution, though it was a bit tough one, I arrived on D. Moreover I went a little further and deduced the values of w,x,y,z as 4,3,2,1. Please let me know if these values are correct. Thank you. _________________

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

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 -> \frac{w}{x}=\frac{1}{z}+\frac{1}{x} --> multiply both sides by xz --> wz=x+z --> z(w-1)=x --> since w>x>z then z=1 (if z>1 then x>w which contradicts given condition) --> w=x+1 --> w and x are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus w and x don not share any common factor more than 1 and as y>1 (from y>z>0) then y is not a common factor of w and x. Sufficient.

(2) w^2 – wy – 2w = 0 --> w(w-y-2)=0, since w>0 then: w-y-2=0 --> w-2=y -->s w>x>w-2 (substituting y in the given inequality) --> x=w-1. The same as above. Sufficient.

Answer: D.

Hope it helps.

Bunuel, How to start in this type of question? In both the statement i arrived up to w=x+1 and w-y=2 , but after that I lost my way...

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]
26 May 2013, 20:45

3

This post received KUDOS

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

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]
26 May 2013, 23:21

Expert's post

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

From F.S 1, we know that \frac{w}{x}= z^{-1}+x^{-1} --> w = \frac{x}{z}+1

or w-1 = \frac{x}{z}.

On the number line,we can have this arrangement ---x ----- x/z ---- w ; where \frac{x}{z} and w are consecutive integers.

Note that we can not have ---x/z---x---w as because there cannot be an integer(x) between 2 consecutive integers.

Thus, we have \frac{x}{z}>x --> z<1. However, this is not possible as z is at-least 1. Thus, the only solution is when \frac{x}{z} and x coincide i.e. \frac{x}{z} = x --> z=1.

As w and x are consecutive integers, y(which is not equal to 1) can never be a divisor for both x and w. Sufficient.

From F. S 2, we know that asw\neq{0} w = y+2. Thus, on the number line,

--y ---(y+1)----w. Thus, there is only one integer between y and w and as y<x<w, we have x = y+1. Just as above, y can't be a divisor for both w and x. Sufficient.

Folks - started to solving it like this, but got stuck after a while. Can someone please help?

Basically this question is asking whether y is factor of both w & x?

Considering Statement 1

w/x=1/z+1/x. Since w>x , the quotient w/x must be greater than 1. For example 17/6 = 2 5/6

Since w/x is greater than 1 then 1/z + 1/x must be greater than 1 as well. For 1/z+1/x to be greater than 1 both z and x has to be 1. x cannot be equal to 1 because of the condition w > x > y > z > 0, therefore, z must be equal to 1. We can now re write the equation as w/x = 1+1/x which will give us w = x+1. If w= x+1 then mean x and w are consecutive integers and therefore no number other than 1 can be a factor of both.

I got stuck after this. But from my guess work I think this statement looks sufficient. I know not the right way of doing MATHS, therefore can someone please help?

Considering Statement 2

w^2-wy-2w = 0

If we factor above into, that will give w(w-y-2) = 0. This means that either w = 0 or w-y-2 = 0. w cannot be zero because w > x > y > z > 0, therefore w-y-2 = 0 or w = y+2. Since x falls between y and w we can tell that x, y and w are consecutive integers and therefore no number other than 1 can be a factor of both. Since y >1 it cannot be a factor of both w & x. Therefore this statement is sufficient to answer the question.

Awesome sum, here's my small bit

\frac{w}{x} = \frac{1}{z} +\frac{1}{x} , Given

\frac{w}{x} - \frac{1}{x}= \frac{1}{z}, bringing terms with same denominator to LHS

\frac{w-1}{x}=\frac{1}{z}, simplification

now if z>1 then RHS is a fraction so LHS is a fraction too as

now if LHS is a fraction then x>w-1

lets look at this statement x>w-1, since w and x are positive and x \neq w hence x>w-1 is never possible. Lets take a few positive integers such that w>x and test this. e.g. x = 6 and w =7 , 6 > 6 not possible, x= 6 w=12 , 6>11 not possible , hence x> w-1 not possible if w>x and positive.So LHS cannot be a fraction and hence RHS also cannot be a fraction so Z cannot be > 1

if Z>1 not possible then Z=1 and as 0 < z \not> 1 ( z is greater than 0 but not greater than 1 and z is an integer so z can be 1 only)

so if z = 1 then from \frac{w-1}{x}=\frac{1}{z} we have w-1 = x , which tells us that x is one less than w or x and w are consecutive integers. Hence they are co prime . Therefore they cannot have a common factor other than 1, but we know z is 1 so y cannot be 1, because y>z , Hence y is not a common factor for w and x as y is not 1 but >1. Sufficient. _________________

- Stne

Last edited by stne on 31 Jul 2014, 02:20, edited 1 time in total.

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]
16 May 2014, 13:51

Another easy approach is to introduce y in the equation by inserting...w=py and x=qy The equation formed after putting in the values in I is not valid & hence w & x do not have a common factor x ...as the equation cannot be untrue _________________

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]
16 May 2014, 23:49

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. _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]
01 Aug 2014, 15:10

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/z + 1 -> x = kz and w = k + 1 ( with k is an positive integer) since w > x -> k+1> kz -> k(z-1) < 1 because of k and z are positive integers, z must equal 1 -> x = k , and w = k + 1 -> y is not a common divisor of w and x.

(2) w = y + 2 -> x = y + 1 ..... -> y is not a common divisor of w and x.

answer D _________________

......................................................................... +1 Kudos please, if you like my post

gmatclubot

Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common
[#permalink]
01 Aug 2014, 15:10