Number Game

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zagcollins: Master | Next Rank: 500 Posts; Posts: 167; Joined: Wed Jun 11, 2008 10:58 pmFollowed by: 1 member; Thanked: 2 times

by zagcollins » Mon Jul 28, 2008 9:10 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

OA is D..this is a question from GPREP...

Last edited by zagcollins on Tue Jul 29, 2008 4:34 am, edited 1 time in total.

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Source: Beat The GMAT — Data Sufficiency | Mon Jul 28, 2008 9:10 am

harish: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri May 30, 2008 10:30 pm; Location: India

by harish » Mon Jul 28, 2008 10:34 am

IMO is D
Since w,x,y,z can have either 0 or 1
from (i)

Keep the denominator equal to 16
8(1) + 4(0) + 2(1) + 1(1) = 11

11 Could be acheived only when wxyz = {1011} ==> 8+2+1

So w+x+y+z=3

Similarly with (ii)
Keep the denominator equal to 81
27(1) + 9(0) + 3(1) + 1(1) = 31

So Answer is D

Whats OA???

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ricky: Master | Next Rank: 500 Posts; Posts: 118; Joined: Sun Jun 01, 2008 1:47 pmFollowed by: 1 member; Thanked: 5 times

by ricky » Mon Jul 28, 2008 11:08 am

IMO Ans shud be D

What is OA?

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