VJesus12 wrote:If x, y and z are integers, is x odd?
(1) yz = x
(2) x - y = z
The OA is the option C.
How can I get an answer here? I've couldn't solve this DS question. <i class="em em-sob"></i>
Hello vjesus12.
We have to answer if x odd.
(1) yz = x
This statement alone doesn't help us, because if y=1 and z=1 then x=1 ODD, but if y=2 and z=1 then x=2 EVEN.
Hence, this statement alone is
NOT SUFFICIENT.
(2) x - y = z
Again, if y=0 and z=2 then x=2 EVEN but if y=0 and z=3 then x=3 ODD. Hence, this statement alone is
NOT SUFFICIENT.
(1) yz = x +
(2) x - y = z
Replacing the first equation in the second one we get: $$yz-y=z\ \ \Rightarrow\ \ y\left(z-1\right)=z\ \ \Rightarrow\ \ y=\frac{z}{z-1}.$$ Now, since y is an integer we get that the options for z are
- z=0, which implies that y=0 and therefore x=0. EVEN.
- z=2, which implies that y=1 and therefore x=2. EVEN.
For any other integer z, y won't be an integer.
Since in both cases we got that x is EVEN we can conlcude that x is NOT ODD.
Hence, both statements together are
SUFFICIENT.
Therefore, the correct answer is the option
C.
