Question

Three pieces of cakes of weights $4\frac{1}{2}$lbs,$6\frac{3}{4}$lbs and $7\frac{1}{5}$lbs respectively are to be divided into parts of equal weights. Further 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?

Answer 1

$4\frac{1}{2}$lbs $,6\frac{3}{4}$lbs $,7\frac{1}{5}$lbs
$\frac{9}{2}$lbs $,\frac{27}{4}$ lbs$,\frac{36}{5}$lbs
$\frac{4.5}{0.9}$ lbs$,\frac{6.75}{0.9}$ lbs$,\frac{7.2}{0.9}$lbs
$5+7.5+8$
$=20.5$