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Number properties

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by abhasjha » Fri Dec 06, 2013 7:58 am
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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Source: — Data Sufficiency |
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by theCodeToGMAT » Fri Dec 06, 2013 9:49 am
To find: w+x+y+z

Statement 1:
w/2+x/4+y/8+z/16=11/16
(8w + 4x + 2y + z) = 11
This is possible when:
w = 1, x = 0, y = 1, z = 1
SUFFICIENT

Statement 2:
w/3+x/9+y/27+z/81=31/81
(27w + 9x + 3y + z ) = 31
This is possible when:
w = 1, x = 0, y = 1, z = 1
SUFFICIENT

Answer [spoiler]{D}[/spoiler]?
R A H U L
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by [email protected] » Fri Dec 06, 2013 2:49 pm
Hi abhasjha,

Rahul has correctly answered this question. I'm going to add a few more details to the explanation.

We're told that each of the variables is either 0 or 1, which really limits the possibilities. We're asked for the value of W+X+Y+Z.

Fact 1: W/2 + X/4 + Y/8 + Z/16 = 11/16

A little algebra gives us:

8W/16 + 4X/16 + 2Y/16 + Z/16 = 11/16

Then....

8W + 4X + 2Y + Z = 11

With the limitation that each variable is either 0 or 1, we'd have these possibilities:

8W = 0 or 8
4X = 0 or 4
2Y = 0 or 2
Z = 0 or 1

With these options, the ONLY way to get to 11 is 8+0+2+1

So W = 1, X = 0, Y = 1 and Z = 1

With just one possibility, Fact 1 is SUFFICIENT (the answer to the prompt is 1+0+1+1 = 3, but you don't need to do THAT math).

Fact 2: Involves math that is similar to Fact 1, but the fractions and common-denominator are different.

We'd end up with:

27W + 9X + 3Y = Z = 31

27W = 0 or 27
9X = 0 or 9
3Y = 0 or 3
Z = 0 or 1

Here, the ONLY way to get to 31 is 27+0+3+1.

Fact 2 is also SUFFICIENT.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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