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Each of the numbers w, x, y, and z is equal to either 0 or [#permalink]
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19 Jul 2008, 18:18
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Each of the numbers w, x, y, and z is equal to either 0 or 1. What is the value of w + x + y + z ? (1) w/2 + x/4 + y/8 + z/16 = 11/16 (2) w/3 + x/9 + y/27 + z/81 = 31/81
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Re: GmatPrep. DS. [#permalink]
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19 Jul 2008, 18:39
1) 8W + 4X + 2Y + Z = 11 => w=y=z=1 and x=0 So answer is A or D.
2) 27W+9X+3Y=Z = 31 => w=y=z=1 and x=0
So answer is D.



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Re: GmatPrep. DS. [#permalink]
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19 Jul 2008, 18:40
i think D.
stat 1 ) * by 16, we get 8w+4x+2y+z=11. so (w=1,x=0,y=1,z=1) stat 2, * by 81, we get 27w+9x+3y+1=31.( w=1,x=0,y=1,z=1) so D. am i right?



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Re: GmatPrep. DS. [#permalink]
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19 Jul 2008, 18:45
sondenso wrote: Each of the numbers w, x, y, and z is equal to either 0 or 1. What is the value of w + x + y + z ?
(1) w/2 + x/4 + y/8 + z/16 = 11/16
(2) w/3 + x/9 + y/27 + z/81 = 31/81 D? 1)w=8; x=0; y=2; z=1 2) w=27; x=0; y=3; x=1



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Re: GmatPrep. DS. [#permalink]
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19 Jul 2008, 20:16
Thanks, I should think twice I quickly choose E because " Each of the numbers w, x, y, and z is equal to either 0 or 1", and I reasoned that do not know when they each are 0 and when they each are 1. Therefore, they will produce more than 1 options.
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Re: Each of the numbers w, x, y, and z is equal to either 0 or [#permalink]
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18 Oct 2013, 06:20
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Any other alternatives for such problems Take LCM, we get => 8W + 4X + 2Y + Z = 11 In both equations if we take w=1 FS 1 4x+2y+z=3 ( 118) if x=1 then the sum can't equal 3 so X must be 0 and y,z=1 Same logic applies to Statement 2... Does anyone have any alternative solution?
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Re: Each of the numbers w, x, y, and z is equal to either 0 or [#permalink]
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13 Feb 2017, 16:14
statement (1) multiply through by 16, to give 8w + 4x + 2y + z = 11 since each of the variables stands for either 0 or 1, you're basically just deciding whether the 8, 4, 2, and/or 1 are present or absent in the final total. if you do a little experimentation, you'll find that the only combination that works is 8 + 2 + 1: i.e., w = y = z = 1 and x = 0. if you need to be more systematic, you can start out with the realization that z must be 1, because that's the only way that you're going to get an odd number out of all that noise. (aside: if you know anything about binary, you'll recognize that this is precisely the setup of binary digits. if you recognize that association, then you'll know at once that this statement is sufficient, because there's only one way to write each integer in binary.)  statement (2) multiply through by 81, to give 27w + 9x + 3y + z = 31 this one is actually easier to figure out, because the numbers are so far apart. w has to be 1, because the other numbers are way too small to give a total as big as 31 (even if they're all 1's). since w = 1, that's 27 already, so x must be 0 (because 27 + 9 is too big). therefore, you need an additional 4, so y = z = 1. therefore w = y = z = 1 and x = 0 (again), so the requisite quantity is 3. sufficient. Hence D
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Re: Each of the numbers w, x, y, and z is equal to either 0 or [#permalink]
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18 Feb 2017, 09:49
anairamitch1804 wrote: statement (1)
multiply through by 16, to give 8w + 4x + 2y + z = 11
since each of the variables stands for either 0 or 1, you're basically just deciding whether the 8, 4, 2, and/or 1 are present or absent in the final total. if you do a little experimentation, you'll find that the only combination that works is 8 + 2 + 1: i.e., w = y = z = 1 and x = 0.
if you need to be more systematic, you can start out with the realization that z must be 1, because that's the only way that you're going to get an odd number out of all that noise.
(aside: if you know anything about binary, you'll recognize that this is precisely the setup of binary digits. if you recognize that association, then you'll know at once that this statement is sufficient, because there's only one way to write each integer in binary.)

statement (2)
multiply through by 81, to give 27w + 9x + 3y + z = 31
this one is actually easier to figure out, because the numbers are so far apart. w has to be 1, because the other numbers are way too small to give a total as big as 31 (even if they're all 1's). since w = 1, that's 27 already, so x must be 0 (because 27 + 9 is too big). therefore, you need an additional 4, so y = z = 1. therefore w = y = z = 1 and x = 0 (again), so the requisite quantity is 3. sufficient.
Hence D I think Both statements alone are sufficient as u can solve both statments alone 1 8w+4x+2y+z=11 w=1 x=0 y=1 and z=1 same for the secodn 1




Re: Each of the numbers w, x, y, and z is equal to either 0 or
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