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On his birthday, Praet brings two pieces of cakes weighing

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Senior Manager
Joined: 22 Aug 2003
Posts: 257
Location: Bangalore
On his birthday, Praet brings two pieces of cakes weighing [#permalink]

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22 Sep 2003, 04:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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).
SVP
Joined: 03 Feb 2003
Posts: 1604

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22 Sep 2003, 04:47
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?
Senior Manager
Joined: 22 Aug 2003
Posts: 257
Location: Bangalore

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22 Sep 2003, 05:03
thats correct.
Now if two pieces are 6/5lbs & 9/10 lbs. What would be your answer?
Senior Manager
Joined: 22 May 2003
Posts: 329
Location: Uruguay

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22 Sep 2003, 17:38
Maybe 7? Each piece weighting 3/10 lbs
Senior Manager
Joined: 22 Aug 2003
Posts: 257
Location: Bangalore

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22 Sep 2003, 23:45
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.
Senior Manager
Joined: 02 Feb 2004
Posts: 344

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21 May 2004, 12:52
can anybody show the workings better?
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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21 May 2004, 14:35
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
21 May 2004, 14:35
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