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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]

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10 Sep 2012, 23:26

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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]

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26 May 2013, 21:45

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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]

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27 May 2013, 00:21

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 as\(w\neq{0}\) 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, 03:20, edited 1 time in total.

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

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16 May 2014, 14:51

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 _________________

Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

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

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17 May 2014, 00:49

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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]

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01 Aug 2014, 16:10

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

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

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03 Sep 2014, 14:54

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)

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? _________________

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

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08 Mar 2015, 07:03

Bunuel 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

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.

Can someone pls explain how Z =1 in case 1?

Thanks. _________________

Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.

I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.-Mohammad Ali

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]

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03 Jun 2015, 22:35

Bunuel 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

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.

Hi bunuel : my logic to option "A" is this :

reducing equ we get X/ (w-1) = Z ( integer) --> means w-1 is a factor of X. Now W>X and w-1 to be factor of X then W& X are consecutive integers. then only this is possible else W-1 will never be factor of X.

Is this logic right. I am thinking it's same logic you explained above but the explanation language is different language.

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

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26 Jul 2015, 12:52

Bunuel 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

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.

Hello Bunel, For statement 1 z(w-1)=x => either w-1=x or z=x but z cannot be equal to x as x is greater than z

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