If y is odd, is z odd?

I was wondering if [C] can be the answer?!

As Bhuwan has already mentioned. Statement 1 is flawed since it doesn't tell if x is an integer or not. Furthermore, the concept of odd and even can not be applicable on fractions (They are neither odd or even). So we have rejected Statement 1. But when we put a closer look on the Statement 2,

\(yz = x\), now y is considered odd, hence can be assumed as an integer, and z is required and to be ascertained as odd or even, so also can be assumed as integer. Hence x, the product has to be an integer. The other way to look at it would be to simply replace the values of x = yz into statement 1, and conclude that \(x^2\) = odd. Which would imply that x is odd and an integer and in turn prove that z must be odd as well.

Hence by combining both the statements, we do arrive at the conclusion that z is odd! Am I doing something wrong?

Regards,
Arpan

Nope, the answer is E. Consider:
x=y=z=1 (y=odd, xyz=odd, yz = x) --> answer YES;
x=1, y=3, z=1/3 (y=odd, xyz=odd, yz = x) --> answer NO.

Hope it's clear.

Umm Bunuel, Please correct me if I am wrong. I understand that no specific restriction is being made on z to be an integer or a fraction. But if we assume z to be a fraction, are we not removing it from the very odd and even scenario? If z is a fraction, clearly its neither odd or even. So the whole question is invalidated. I was under the impression that if we are to determine z to be odd or even, we are to assume it must be an integer first and not a fraction.

Regards,
Arpan

We are not told that z is an integer. The question just asks: is z odd? It could be odd, even, fraction or irrational number.

Got it! Just wanted to clarify Thanks Bunuel!

Regards,
A

If we were told x,y, and z are integers would the answer be A?

Yes,

(1) xyz = odd

holds ture only if every term is odd (given that they are integers)

OFFICIAL SOLUTION

Solution: E

Statement (1) tells us that product xyz is odd, so if both x and z are integers then x and z must both be odd, as any even number times an odd must yield an even result. But x and z could also be fractions – consider what would happen if y = 35, x = 1/5, and z = 1/7; NOT SUFFICIENT. Statement (2) gives us three variables with no mention of the properties of x and z; NOT SUFFICIENT. Together yz must be odd, but z could still be a fraction or an odd integer; (E).

St 1: xyz = odd. INSUFFICIENT as the decision totally depends on x if x = 3/2 and z = 2, therefore z will be even and if all are integerz will be odd.

St 2: yz = x. INSUFFICIENT as the decision totally depends on x

St 1 & St 2: INSUFFICIENT as the decision totally depends on x


Option E

Tricky question:

If y is odd, is z odd?

(1) xyz = odd
(2) yz = x

Statement 1: xyz = odd
May be x= 7, y = 7, z = 7 => xyz is odd, z is odd
or \(x = 2/7\), \(y = 7\), \(z = 7/2\) => xyz is odd, but z is not odd

Statement 2:
\(yz = x => y = x/z\)
May be x = 7, y = 1, z = 7, z is odd
or \(x = 7/2\), \(z = 1/2\) and \(y = 7\), but z is not odd

Not sufficient

Combining (1)+(2),
xyz is odd and x = yz => \(yz * yz\) = odd
=> \(y^2 * z^2\) = odd
since \(y^2\) is odd integer, \(z^2\) also must be odd integer, but \(z^2\) need not be a perfect square in which case, z need not be an integer.

Not Sufficient. Answer (E)

Hello Bunuel. I have a small doubt in case. I am just not able to fit two validity of two statements together.
Statement 1 says : xyz is odd. Sure, z could be a fraction here.
Statement 2 says , x=yz . Again does not have to be z as odd.
Now if i combine and take two statements to be true.

I have ,

xyz= odd
Now if y is odd, can i say , y must be an integer
and if yz=x & xyz = odd, can I say x^2 = odd. Now Can i say for a perfect square to be Odd , x must be an odd integer.

Now if x is an odd integer and y is an odd integer , & xyz= odd , z can be a fraction or an integer agreed. But if you try to put z as fraction, i think this would contradict the second statement i.e. yz = x. Cannot be possible. I tried few combinations but failed. So i think Z must be an odd integer for statement 2 to be vaild,

Hence answer should be C.

If its E , can you show with examples for making both the statements valid , how can it be E.

Will appreciate the help.

Thanks

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