Given: x is a factor of y; y is a factor of z

St 1: x.z is even

2 cases: either x is even or z is even

case 1: x is even
X is a factor of Y and X being an even number when multiplied with another number gives an even product.
So Y is even.
Y is a factor of Z. So Z is also even.

case 2: z is even
In both cases Z is even...Suff

St 2: y is even
Y is a factor of Z, so Z is also even...Suff
[spoiler]
Answer is D[/spoiler]

I could not get it. Could you plz tell where I am wrong?

Given: x is a factor of y; y is a factor of z i.e y/x and z/y are integers.

1. x.z is even
The cases possible are
a. x is even z is even
b. x is odd z is even
c. x is even z is odd
If x is even, y is even and hence z should be even.
But what if x is odd, y can be odd or even..eg: 15/3 or 12/3. In that case we can't say that z is even.

2. y is even , hence z should be even.

Hence I go with B.

The positive integers x, y and z are such that x is a factor of y and y is a factor of z. Is z even?

(1) x z is even.

(2) y is even.



Given,

Y = xK
z=yJ=xKJ

(1) XZ is even means either one of them should be even or both even

if x is even then y is even then z is even
if z is even well...

Sufficient

(2) if Y is evn then z has to be even.
Hence D.

Beauty of GMATPrep..simple but tricky..

srn wrote:I could not get it. Could you plz tell where I am wrong?

Given: x is a factor of y; y is a factor of z i.e y/x and z/y are integers.

1. x.z is even
The cases possible are
a. x is even z is even
b. x is odd z is even
c. x is even z is odd
If x is even, y is even and hence z should be even.
But what if x is odd, y can be odd or even..eg: 15/3 or 12/3. In that case we can't say that z is even.

2. y is even , hence z should be even.

Hence I go with B.

With x is a factor of y; y is a factor of z; x is obviously a factor of z. If x z is even, z is even irrespective of what the positive integers x was. Statement 1 is hence, SUFFICIENT.

sanju09 wrote:The positive integers x, y and z are such that x is a factor of y and y is a factor of z. Is z even?

(1) x z is even.

(2) y is even.

One of the most common mistakes people make in attacking DS questions is rushing to the statements too quickly. By taking the time to think about the question, you make working with the statements much easier. Hence:

Step 1 of the Kaplan Method for DS: Analyze the Question Stem

If x is a factor of y and y is a factor of z, then if x is even, y and z will both be even; if y is even, z will be even. So, if any of the 3 are even, we'll get a yes answer. Accordingly, let's rephrase the question:

Is at least one of the integers x, y and z even?

(1) xz is even.

The only way to get an even product from integers is to include an even integer. Accordingly, at least one of x and z is even: sufficient.

(2) y is even.

At least one of x, y and z is even: sufficient.

Each statement is sufficient alone, choose (D).