Three peices of cakes of weights 9/2 lbs, 27/4 ibs and 36/5 lbs respectively are to be divided into parts of equal weights. Futher, each must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained.
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given 3 peices of cakes weights 9/2 lbs, 27/4 ibs and 36/5 lbs respectively.
LCM of the denominators 2, 4 and 5 is 20
Hence (9/2) = (90/20)
(27/4) = (135/20)
(36/5) = (144/20)
These fractions can also be written as shown:
(9/2) = (90/20) = 10 × (9/20)
(27/4) = (135/20) = 15 × (9/20)
(36/5) = (144/20) = 16 × (9/20)
The heaviest part is (9/20).
Maximum number of guests that could be entertained = 10 + 15 + 16 = 41
LCM of the denominators 2, 4 and 5 is 20
Hence (9/2) = (90/20)
(27/4) = (135/20)
(36/5) = (144/20)
These fractions can also be written as shown:
(9/2) = (90/20) = 10 × (9/20)
(27/4) = (135/20) = 15 × (9/20)
(36/5) = (144/20) = 16 × (9/20)
The heaviest part is (9/20).
Maximum number of guests that could be entertained = 10 + 15 + 16 = 41
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