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a,b,c and d are positive integers such that ab=24, cd=48,

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a,b,c and d are positive integers such that ab=24, cd=48, [#permalink]

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a,b,c and d are positive integers such that ab=24, cd=48, ac=64 and bd=18: Is a=c?


(1) Two of a,b,c and d are equal
(2) a+c > b+d

Last edited by kevincan on 12 Aug 2006, 11:03, edited 1 time in total.
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New post 12 Aug 2006, 10:38
I think the answer is A

From St1 Factors of ab, cd and bd contain a 3 in them so they cannot be split equally. The only possibility is ac = 64 (2^6) hence 8 and 8
SUFF

From St2 a+c<b+d seems to imply that the only possiblity is 8,8 and 18,1 which would imply that the a & b be non integers
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New post 12 Aug 2006, 11:04
Oops, wrong sign! Now corrected
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New post 12 Aug 2006, 11:13
The answer then is D

for bd the possiblities are (1,18) (3,6) (2,9)

if bd are (1,8) or (2,9) then a & b are non integers hence only possiblity is 6,3

if b = 6, d = 3 then a=4 and c = 16 (a+c > b+d)
if b=3, d = 6 then a=8 and c=8 ( a+c < b+d)

Hence ST 2 is also SUFF
  [#permalink] 12 Aug 2006, 11:13
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a,b,c and d are positive integers such that ab=24, cd=48,

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