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DS - Easy problem made to look tricky.

Expert replies
Source: — Data Sufficiency |

by diebeatsthegmat » Wed Oct 27, 2010 5:49 pm
tlt2372 wrote:Image

OA: D

How is this possible?
for what you named the topic you post, it seems you could solve this data?
here is what i did.
w,x,y,z =0 or =1 w+x+y+z=?
1/ w/2+x/4+y/8+z/16=11/16
<=> 8w/16+4x/16+2y/16+1z/16=11/16
<=> 8w+4x+2y+z=11
only w,x,x=1 and y=so so the 8*1+4*1+0+1=11
there's no other numbers/ways
so A is sufficicent
we do the same with statement 2
and D is the answer
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by limestone » Wed Oct 27, 2010 9:20 pm
diebeatsthegmat wrote
<=> 8w+4x+2y+z=11
only w,x,x=1 and y=so so the 8*1+4*1+0+1=11
I think you mistyped something here.

It must be 8*1 + 4*0 + 2*1 + 1 = 11

As x,y,z,w are either Zero or 1
and 8w+4x+2y+z=11

Assume that all are 1, then 8w+4x+2y+z = 15. We must try to let either x,y,z,w to be Zero to reduce the sum to 11. There's only 1 way, as 15 - 11 = 4, and ONLY x has "4" before it. Therefore the unique set must be: x = 0; z,y,w = 1.

The same approach for statement 2.

Thus D is the correct choice.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by Rahul@gurome » Wed Oct 27, 2010 9:26 pm
Given: w, x, y and z are either 0 or 1.

Statement 1: w/2 + x/4 + y/8 + z/16 = 11/16 => 8w + 4x + 2y + z = 11
As w, x, y and z are either 0 or 1, only possible combination of values for them is w = y = z = 1 and x =0.
Thus, w + x + y + z = 1 + 0 +1 +1 = 3

Sufficient.

b]Statement 2:[/b] w/3 + x/9 + y/27 + z/81 = 31/81 => 27w + 9x + 3y + z = 31
As w, x, y and z are either 0 or 1, only possible combination of values for them is w = y = z = 1 and x =0.
Thus, w + x + y + z = 1 + 0 +1 +1 = 3

Sufficient.

The correct answer is D.
Rahul Lakhani
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by diebeatsthegmat » Wed Oct 27, 2010 11:26 pm
limestone wrote:
diebeatsthegmat wrote
<=> 8w+4x+2y+z=11
only w,x,x=1 and y=so so the 8*1+4*1+0+1=11
I think you mistyped something here.

It must be 8*1 + 4*0 + 2*1 + 1 = 11

As x,y,z,w are either Zero or 1
and 8w+4x+2y+z=11

Assume that all are 1, then 8w+4x+2y+z = 15. We must try to let either x,y,z,w to be Zero to reduce the sum to 11. There's only 1 way, as 15 - 11 = 4, and ONLY x has "4" before it. Therefore the unique set must be: x = 0; z,y,w = 1.

The same approach for statement 2.

Thus D is the correct choice.
yep, i mistyped something but the solution was ok, wasnt it? :)
thank you for finding me a mistake... hope wont get mistakes in test day :)
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