Data Sufficiency

Is z an even Number?
1. 3z is even
2. 5z is even

Mayank Choudhary wrote:Is z an even Number?
1. 3z is even
2. 5z is even

1. z can be a fraction or an even number
(z = 2/3 => 3z = 2
z = 2 => 3z = 6)
Not sufficient

2. same reason
(z = 2/5 => 5z = 2
z = 2 => 5z = 10)
Not sufficient

Combination of the 2 statements: 3z and 5z are even
=> 2z is even
Since 2z and 3z are even, z has to be even.

Is it C?

I agree with your solution, but I just don't get it why if 3z and 5z are both even, z should be even?

ganeshrkamath wrote:

Mayank Choudhary wrote:Is z an even Number?
1. 3z is even
2. 5z is even

1. z can be a fraction or an even number
(z = 2/3 => 3z = 2
z = 2 => 3z = 6)
Not sufficient

2. same reason
(z = 2/5 => 5z = 2
z = 2 => 5z = 10)
Not sufficient

Combination of the 2 statements: 3z and 5z are even
=> 2z is even
Since 2z and 3z are even, z has to be even.

Is it C?

thank u guys but i dont get how combination of 2 statements make 2 it can make 8 or 15 to

If z is a positive number, is z an even integer?

(1) 3z is an even integer.

(2) 5z is an even integer.

Statement 1: 3z is even.
It's possible that z = 2, which is an even integer.
It's possible that z = 2/3, which is not an even integer.
INSUFFICIENT.

Statement 2: 5z is even.
It's possible that z = 2, which is an even integer.
It's possible that z = 2/5, which is not an even integer.
INSUFFICIENT.

Every test-taker should know the following:
EVEN - EVEN = EVEN.

Statements 1 and 2 combined:
Here, 5z is even and 3z is even.
Thus:
5z-3z = even - even = even.
Since 5z-3z = 2z, 2z must be even.

Since 3z is even and 2z is even, we get:
3z-2z = even - even = even.
Since 3z-2z = z, z must be even.
SUFFICIENT.

The correct answer is C.