Q Each of the W,X,Y,Z are 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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Relatively new GMAT PREP QUESTION
This topic has expert replies
-
kanha81
- Master | Next Rank: 500 Posts
- Posts: 431
- Joined: Sat Jan 10, 2009 9:32 am
- Thanked: 16 times
- Followed by:1 members
this is how I would approach:cramya wrote:Q Each of the W,X,Y,Z are 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
1) w=1, x=0, y=1, z=1
1/2 + 0 + 1/8 + 1/16 = (8+2+1) / 16 = 11/16
Suff
2) w=1, x=0, y=1, z=1
1/3 + 0 + 1/27 + 1/81 = (27+3+1) / 81 = 31/81
Suff
[spoiler][D][/spoiler]
I am pretty sure there is more than meets the eye! [/spoiler]
Want to Beat GMAT.
Always do what you're afraid to do. Whoooop GMAT
Always do what you're afraid to do. Whoooop GMAT
-
ajaypatil_am
- Senior | Next Rank: 100 Posts
- Posts: 81
- Joined: Sat Jan 27, 2007 5:18 am
- Location: PUNE
-
Vemuri
- Legendary Member
- Posts: 682
- Joined: Fri Jan 16, 2009 2:40 am
- Thanked: 32 times
- Followed by:1 members
IMO D. This question seems straight forward.cramya wrote:Q Each of the W,X,Y,Z are 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
Stmt1: 8w+4x+2y+z=11. We know that W,X,Y,Z are either 0 or 1. So, the only possible option is when w=1,x=0,y=1,z=1. Sufficient
Stmt1: 27w+9x+3y+z=31. We know that W,X,Y,Z are either 0 or 1. So, the only possible option is when w=1,x=0,y=1,z=1. Sufficient
