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by heshamelaziry » Sat Nov 28, 2009 10:53 am
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
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by papgust » Sat Nov 28, 2009 7:50 pm
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
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by Mozartain » Sun Nov 29, 2009 2:28 am
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
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by thephoenix » Sun Nov 29, 2009 4:47 am
4A=9B,----->A>B
17C=11D--->D>C
5C=12A---->C>A

COMBINING D>C>A>B

HENCE
DCAB
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