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Vicky
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On his birthday, Praet brings two pieces of cakes weighing 22 Sep 2003, 04:13
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On his birthday, Praet brings two pieces of cakes weighing 3/5 lbs & 7/20lbs. He wants to cut them into pieces of equal weights such that weight of each piece is maximum possible. Find the number of maximum guests praet could invite to his house. (assume: Praet will give atleast one cake to every guest).



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On his birthday, Praet brings two pieces of cakes weighing 22 Sep 2003, 04:47
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At the first glance: 3/5 and 7/20 are not terminating fractions
3/5=12/20
12/20 and 7/20 can be equally divided among 19 people.

19?
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Vicky
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On his birthday, Praet brings two pieces of cakes weighing 22 Sep 2003, 05:03
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thats correct.
Now if two pieces are 6/5lbs & 9/10 lbs. What would be your answer?
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MartinMag
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On his birthday, Praet brings two pieces of cakes weighing 22 Sep 2003, 17:38
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Maybe 7? Each piece weighting 3/10 lbs
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Vicky
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On his birthday, Praet brings two pieces of cakes weighing 22 Sep 2003, 23:45
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there u go.. yes it would be 7 each weighing 3/10 lbs.
we have to divide weights such that each has maximum possible weight.
21 will give each cake of weight 1/10lbs.
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On his birthday, Praet brings two pieces of cakes weighing 21 May 2004, 12:52
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can anybody show the workings better?
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On his birthday, Praet brings two pieces of cakes weighing 21 May 2004, 14:35
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Assume that Praetorian divides 3/5 pound cake into n pieces and 7/20 pound cake into m pieces. m and n pieces have equal weight
Then
weight of each cake piece = (3/5)/n = 3/5n Since each cake piece has same weight
(3/5)/n = (7/20)/m
3/n = 7/4m therefore m = n * 7/12

Now we want minimum pieces because minimum pieces yeild maximum size and we know that m and n are integers
if n = 12 then m = 7
there fore praetorian will invite 7+12 = 19 guests.
I hope I am included :lol:



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