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If y is odd, is z odd? [#permalink]
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26 May 2013, 10:22
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If y is odd, is z odd? (1) xyz = odd (2) yz = x
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Re: If y is odd, is z odd? xyz = odd yz = x [#permalink]
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26 May 2013, 12:12
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1st Statement says XYZ = odd : say y = 3 now 1st eqn becomes 3(XZ) = Odd. Hence XZ has to be odd in order for the equality to hold. Since its not given that x y z are integers we don't know if Z is an integer Z can be 1/2 and X can be 6 and still XYZ be odd. Hence this is not sufficient
2nd Statement is not clear about X,Y,Z Not Sufficient
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Re: If y is odd, is z odd? [#permalink]
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26 May 2013, 12:57



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Re: If y is odd, is z odd? [#permalink]
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08 Jun 2013, 01:55
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atalpanditgmat wrote: If y is odd, is z odd?
(1) xyz = odd (2) yz = x 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
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Re: If y is odd, is z odd? [#permalink]
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08 Jun 2013, 02:10
arpanpatnaik wrote: atalpanditgmat wrote: If y is odd, is z odd?
(1) xyz = odd (2) yz = x 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.
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Re: If y is odd, is z odd? [#permalink]
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08 Jun 2013, 02:18
Bunuel wrote: arpanpatnaik wrote: atalpanditgmat wrote: If y is odd, is z odd?
(1) xyz = odd (2) yz = x 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
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Re: If y is odd, is z odd? [#permalink]
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08 Jun 2013, 02:21
arpanpatnaik wrote: Bunuel wrote: arpanpatnaik wrote: 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.
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Re: If y is odd, is z odd? [#permalink]
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08 Jun 2013, 02:26
Quote: 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
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Re: If y is odd, is z odd? [#permalink]
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11 Jun 2013, 11:41
If we were told x,y, and z are integers would the answer be A?



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Re: If y is odd, is z odd? [#permalink]
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11 Jun 2013, 11:44
bugzor wrote: If we were told x,y, and z are integers would the answer be A? Yes, (1) xyz = oddholds ture only if every term is odd (given that they are integers)
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Re: If y is odd, is z odd? [#permalink]
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21 Mar 2017, 18:59
atalpanditgmat wrote: If y is odd, is z odd?
(1) xyz = odd (2) yz = x 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).
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Re: If y is odd, is z odd? [#permalink]
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22 Mar 2017, 02:31
atalpanditgmat wrote: If y is odd, is z odd?
(1) xyz = odd (2) yz = x 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



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Re: If y is odd, is z odd? [#permalink]
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29 Nov 2017, 12:13
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)




Re: If y is odd, is z odd?
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