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If, x, y, and z are integers, is x even?
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23 Jun 2015, 02:47
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64% (01:29) correct 36% (01:43) wrong based on 340 sessions
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Re: If, x, y, and z are integers, is x even?
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23 Jun 2015, 05:22
If, x, y, and z are integers, is x even?
(1) 10^x = 4^y*5^z
2^x * 5^x = 2^2y * 5^z equating, we get x=2y (x will be even regardless whether y is even or odd) x=z ( x will be even if z is even and x will be odd if z is odd)
Insufficient
(2) 3^(x + 5) = 27^(y + 1)
3^(x + 5) = 3^3(y + 1) equating, we get
x + 5 = 3(y + 1) x = 3y  2 (x will be even if y is even and x will be odd if y is odd)
Insufficient
Combining statements 1 and 2
we have below equations x=2y (1) x=z (2) x=3y2 (3)
from (1) and (3) 2y = 3y2 y=2 (y is even)
so x will be even
Sufficient.
z will also be even
Ans:C



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Re: If, x, y, and z are integers, is x even?
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23 Jun 2015, 10:36
Bunuel wrote: If, x, y, and z are integers, is x even?
(1) 10^x = 4^y*5^z (2) 3^(x + 5) = 27^(y + 1)
Kudos for a correct solution. Question : Is x Even?Statement 1: 10^x = 4^y*5^zi.e. 2^x * 5^x = 2^2y * 5^z i.e. x = 2y = z Since, y is an Integer and x=2y, therefore x must be a multiple of 2 i.e. an Even Integer Hence SUFFICIENTStatement 2: 3^(x + 5) = 27^(y + 1)i.e. 3^(x + 5) = 3^(3y + 3) i.e. (x+5) = (3y+3) i.e. x = 3y  2 But x will be odd if y is odd and x will be even if y is even Hence NOT SUFFICIENTAnswer: Option
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Re: If, x, y, and z are integers, is x even?
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23 Jun 2015, 10:38
ManojReddy wrote: If, x, y, and z are integers, is x even?
(1) 10^x = 4^y*5^z
2^x * 5^x = 2^2y * 5^z equating, we get x=2y (x will be even regardless whether y is even or odd) x=z ( x will be even if z is even and x will be odd if z is odd)
Insufficient
Hi ManojReddy, Looks like you have made an error in statement 1 (check highlighted part) how can x be even and odd simultaneously x = 2y = z i.e. x MUST be even because y is an Integer and therefore z will also be even Hence, SUFFICIENTI hope it helps!
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Re: If, x, y, and z are integers, is x even?
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23 Jun 2015, 10:46
Bunuel wrote: If, x, y, and z are integers, is x even?
(1) 10^x = 4^y*5^z (2) 3^(x + 5) = 27^(y + 1)
Kudos for a correct solution. 1)10^x=2^x*5^x=2^2y*5^z... equating powers on two sides.. x=2y... suff 2) 3^x*3^5=3^3y*3^3.... so x+2=3y... x can be 4 when y=2 and x will be 7 when y=3... insuff ans A
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Re: If, x, y, and z are integers, is x even?
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07 Aug 2015, 14:58
I saw this question pop up today when I was studying my MGMAT book. In a recent post by mikemcgarry he encouraged me to not dismiss things and to look deeper to really ensure you understand the mechanics of the problem/question rather than just a surface understanding (something about icebergs comes to mind). I see in the explanation here (and in the MGMAT book) that we are able to infer that \(4^y\) = \(2^2^y\). I'm certainly not going to argue that this is true as I have tested the theory with a few different numbers, and the rest of the problem makes sense to me. Rather than just dismiss this I would like to grasp the mechanics of why we are able to do this. I hope that if I understand this better I will be able to recognise this more easily in the future and in what cases I can and more importantly cannot make this inferance. Thanks in advance for the help. p.s. I know this is probably a very simple question for most of you around here but I guess we all get caught up on simple things from time to time
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Re: If, x, y, and z are integers, is x even?
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16 Aug 2015, 11:58
DropBear wrote: I saw this question pop up today when I was studying my MGMAT book. In a recent post by mikemcgarry he encouraged me to not dismiss things and to look deeper to really ensure you understand the mechanics of the problem/question rather than just a surface understanding (something about icebergs comes to mind). I see in the explanation here (and in the MGMAT book) that we are able to infer that \(4^y\) = \(2^2^y\). I'm certainly not going to argue that this is true as I have tested the theory with a few different numbers, and the rest of the problem makes sense to me. Rather than just dismiss this I would like to grasp the mechanics of why we are able to do this. I hope that if I understand this better I will be able to recognise this more easily in the future and in what cases I can and more importantly cannot make this inferance. Thanks in advance for the help. p.s. I know this is probably a very simple question for most of you around here but I guess we all get caught up on simple things from time to time \(4^y=(2^2)^y=2^{2y}\).
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Re: If, x, y, and z are integers, is x even?
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25 Sep 2017, 18:13
Bunuel wrote: If, x, y, and z are integers, is x even?
(1) \(10^x = 4^y*5^z\)
(2) \(3^{(x + 5)} = 27^{(y + 1)}\)
Kudos for a correct solution. Statement 1There is a pattern 10^2= 5^2 *2^2 10^3 =5^2 * 2^ 310^4= 2^4 *5^4 10^5= 2^ 5 *5^5 In order to rewrite 10^3 and 10^5 the 2 would have to be written has 4^(3/2) this cannot be true since it must be an integer so x must be even suff Statement 2Simply rewrite and use algebra 3^ (x+5) = 3^3y + 3 x +5= 3y + 3 x + 2= 3y too many possibilities insuff A



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If x, y, and z are integers, is x even?
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23 Oct 2017, 19:15
if x, y, and z are integers, is x even?
(1) \(10^x\) = \((4^y)(5^z)\) (2) 3^(x+5) = 27^(y+1)
I searched high and low and could not find this question posted. Please redirect and lock thread if already posted. Also, I couldn't figure out how to format the (x+5) and (y+1) correctly. I read through the entire "Writing Mathematical Formulas on the Forum"...



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Re: If x, y, and z are integers, is x even?
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23 Oct 2017, 19:30
SPEEDBOATS wrote: if x, y, and z are integers, is x even?
(1) \(10^x\) = \((4^y)(5^z)\) (2) 3^(x+5) = 27^(y+1)
I searched high and low and could not find this question posted. Please redirect and lock thread if already posted. Also, I couldn't figure out how to format the (x+5) and (y+1) correctly. I read through the entire "Writing Mathematical Formulas on the Forum"... Hi... If you look down below in similar topics, this stands out.. Merging topics
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Re: If x, y, and z are integers, is x even? &nbs
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