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VP  D
Joined: 31 Oct 2013
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Concentration: Accounting, Finance
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x=2y-7+3z, which of the following must be odd?  [#permalink]

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Question Stats: 54% (02:09) correct 46% (02:22) wrong based on 84 sessions

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If x, y , z are integers and x=2y-7+3z, which of the following must be odd?

a) y
b) z
c) xy-1
d) xz-1
e) y + zx

Originally posted by KSBGC on 26 Jan 2018, 15:41.
Last edited by KSBGC on 19 Aug 2018, 06:39, edited 1 time in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 8008
Re: x=2y-7+3z, which of the following must be odd?  [#permalink]

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selim wrote:
If x, y , z are integers and x=2y-7+3z, which of the following must be odd?

a) y
b) z
c) xy-1
d) xz-1
e) none of these

x=2y-7+3z......
y can be anything but 2y-7 will be even-odd=odd..
So x=odd+3z...x-3z=0dd..
Therefore one of x and z is odd and other even...
xz=O*E=even..... and xz-1= even-1=odd..
D
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Joined: 15 Mar 2015
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Re: x=2y-7+3z, which of the following must be odd?  [#permalink]

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Given : x= 2y-7 +3z
With this given equation, we do not know y or z must be odd. It can be even or odd. So, option A and B is not correct.
Next ,
2y - always even , 2y - 7= Even - Odd = Odd
Now, x = Odd + 3z
Here, if z is even , x= Odd + Even = Odd
If z is odd , x= Odd + Odd = Even
In either case, xz = O*E or E* O = Even and xz-1 = Even -1 must always be odd.
Hence , option D)
Manager  B
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Re: x=2y-7+3z, which of the following must be odd?  [#permalink]

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x=2y-7+3z

x-3z = 2y-7

2y is even for sure so, 2y-7 would be odd.

so, x-3z is odd.

x and z can't be both odd or even.
One of them has to be odd and other even otherwise their sum wont be odd.

IF that is the case, then xz is even and xz-1 is odd. Re: x=2y-7+3z, which of the following must be odd?   [#permalink] 26 Sep 2019, 22:21
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x=2y-7+3z, which of the following must be odd?

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