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?
Answer & explanation
Correct answer: option 4
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