If A, B, C and D are positive integers such that 4A=9B, 17C=11D, and 5C=12A, then the arrangement of the four numbers from greatest to least is
CDAB
BACD
DCAB
DCBA
BDAC
Numbers Order
This topic has expert replies
-
- Legendary Member
- Posts: 869
- Joined: Wed Aug 26, 2009 3:49 pm
- Location: California
- Thanked: 13 times
- Followed by:3 members
- papgust
- Community Manager
- Posts: 1537
- Joined: Mon Aug 10, 2009 6:10 pm
- Thanked: 653 times
- Followed by:252 members
IMO C
4A = 9B => B = 4/9 A
5C = 12A => C = 12/5 A
17C = 11D => D = 17/11 C = 17/11 (12/5 A) = 204/55 A
So the order from greatest to least is,
204/55 A, 12/5 A, A, 4/9 A
Nothing but, DCAB
4A = 9B => B = 4/9 A
5C = 12A => C = 12/5 A
17C = 11D => D = 17/11 C = 17/11 (12/5 A) = 204/55 A
So the order from greatest to least is,
204/55 A, 12/5 A, A, 4/9 A
Nothing but, DCAB
-
- Senior | Next Rank: 100 Posts
- Posts: 49
- Joined: Tue Nov 11, 2008 8:38 pm
- Thanked: 3 times
- GMAT Score:740
5C = 12A and 4A = 9B; So 5C = 12A = 27B
17C = 11D, so 5C ~ 3D (approximation is fine here, since D seems far enough from A and B)
The master equation is thus set: 5C = 12A = 27B ~ 3D. So D must be the largest, B the smallest and C>A.
Ans: DCAB
17C = 11D, so 5C ~ 3D (approximation is fine here, since D seems far enough from A and B)
The master equation is thus set: 5C = 12A = 27B ~ 3D. So D must be the largest, B the smallest and C>A.
Ans: DCAB
- thephoenix
- Legendary Member
- Posts: 1560
- Joined: Tue Nov 17, 2009 2:38 am
- Thanked: 137 times
- Followed by:5 members