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

Please explain why the answer is D.

(1) w/2+x/4+y/8+z/16=11/16 ----->8w/16+4x/16+2y/16+z/16 must be equal to 11/16. if you add w+x+y+z=8+4+2+1=15. 15/16 which is not equal to 11/16, so either w,x,y, or z must be equal to 0. if we add 8w/16+2y/16+1z/16=11/16, 11/16 equal to 11/16, then we know that x must be equal to 0, and w,y,z must be 1. statement 1 alone is sufficient.

(2) w/3+x/9+y/27+z/81=31/81---->27w/81+9x/81+3y/81+z/81 must be equal to 31/81. if you add w+x+y+z=27+9+3+1=40. 40/81 which is not equal to 31/81, so either w,x,y, or z must be equal to 0. if we add 27w/81+3y/81+1z/81=31/81, then w,y, z must be 1, and x must be 0. statement 2 alone is sufficient.
thus, the answer is D.

The question says 'Each of the numbers w,x,y and z is equal to either 0 or 1'.!!!

[quote="Capthan"]The question says 'Each of the numbers w,x,y and z is equal to either 0 or 1'.!!![/quote

Correct. This means that all the variables might be 0, or they might all be 1, or some might be 1 and some might be 0. Any combination is acceptable according to the setup in the question as long as each one is either 0 or 1. We thus have to ask what we can deduce from the information in the statements. Let's take statement 1.

(1) w/2+x/4+y/8+z/16=11/16

With so many unknowns it seems that it might be impossible to infer anything. But in fact we can. Let's try both possibilities, x is 1 and x is 0, and see if each is possible.

First, can x be 0? If x is 0, we could make w, y, and z 1. In this case the sum would be 11/16, as stated. So yes, x can be 0.

Second, can x be 1? Let's try. If we also make w equal to 1, then we get a contradiction: 1/2 + 1/4 is 12/16, which is greater than (and hence not equal to) 11/16. OK. What if w is 0 and y and z are 1? Then we get 1/4 + 1/8 + 1/16. This is too small. So it boils down to this: once we commit to making x equal to 1, either we also make w equal to 1 and we are too high, or we make w equal to 0 and we are too low. Like playing blackjack against the dealer!

Therefore, we have found that given statement 1, x MUST equal 1.

Similar logic will work for statement 2. I believe this is the idea suggested by the commenter above - I just fleshed out the details a little more.

Jonathan, i am confused! X must equal 1 or zero???

guys!
people!

kill the fractions!

for the love of all that is good in this world, you should IMMEDIATELY multiply these sorts of equations through, on both sides, by the common denominator. i mean, are you kidding? i don't want to look at those fractions for a second longer than i have to.

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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.)

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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.

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notice the interesting - and unnecessary - symmetry in this problem: not only does the combination w + x + y + z have the same value in both statements, but the same three variables (w, y, z) equal 1 both times.
note that this doesn't have to happen; even if w, x, y were 1's in the first example and w, y, z were 1's in the second example, you'd still have w + x + y + z = 3 both times.
but apparently they want to be nice to you.

hilarious, ron is the man

I agree with Ron: whenever you see fractions, get rid of them!

_________________
Ben Ku
Instructor
ManhattanGMAT

Ron

So in 1

we have

(8w/16) + (4x/16) + (2y/16) + (z/16)
_________________________________ = 11/16
16


why cant I have the options as

8(1) + 2(1) + 1(1) = 11

or

8(0) + 4(2) + 2(1) + 1 (1) = 11

so the options are

1110
0211

so A is insufficent

"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?"

Pay proper attention to the problem statement before posting.

have everyone's questions been answered on this one? if not, let us know so we can help you further..

_________________
Tim Sanders
Manhattan GMAT Instructor