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For me it is B.
w-y-2=0
w=y+2
hence
x=y+1
so: y, x and w are consecutives.
And they are not 1,2 and 3 as there is still z which is greater than 0.
y is not a common divisor of w and x

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

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

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

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

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

Therefore, D?

I too thought on the same lines. It has to be D . Can we have the OA please ?

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

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

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

For both statements, you can prove that w and x are consecutive integers so that there can be no common divisor that is an integer.

If you look at statement 1:

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

Since you know that w > x, then w/x must be >1. The last term, (1/x), has to be a fraction because x is >1, therefore in order to make this equation true, z must equal 1.

So now, we have:

w/x = 1 + (1/x)

Multiply both sides by x and we get:

w = x + 1. Therefore, w and x are consecutive integers.

Statement 2:

w^2 - wy - 2w. Factor out a w and you end up with w-y-2=0.

Like Caas explained above, this is the same as w=y+2. thus, y,x, and w are consecutive integers and they are not 1, 2, and 3 because z needs to be a positive integer as well.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...