Three pieces of cake of weight 3\(\frac{1}{2}\) lbs, 5\(\frac{1}{3}\) and 6\(\frac{1}{2}\) lbs respectively are to be divided into parts of equal weights.  Further, each part must be as heavy as possible.  If one part is served to each guest, then what is the maximum number of guests that could be entertained?

None of these

Number of guests = \(\frac{Weight\;of\;piece\;of\;cake}{HCF}\) = Other factor

Other factors = 3\(\frac{1}{2}\)  :  5\(\frac{1}{3}\)  :  6\(\frac{1}{2}\)

                    =   \(\frac{7}{2}\)  :  \(\frac{16}{3}\)  :  \(\frac{13}{2}\)

                    =   \(\frac{21}{6}\)  :  \(\frac{32}{6}\)  :  \(\frac{39}{6}\)  =  21  :  32  :  39

Total number of guests = Sum of other factors

Sum of other factors = 21 + 32 + 39 = 92