Three pieces of cake weighing \(3\frac{1}{2}\) lb, \(5\frac{1}{3}\) lb and \(6\frac{1}{2}\) lb are to be cut into parts of equal weight, each part being as heavy as possible. If one part is served to each guest, what is the maximum number of guests that can be served?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → None of these
Every piece must be cut into a whole number of equal parts, so the weight of one part must divide all three weights. The heaviest such part is the HCF of the three weights. Writing the weights with a common denominator turns this into an HCF of whole numbers.
\(3\frac{1}{2}=\frac{7}{2}=\frac{21}{6}\) lb, \(5\frac{1}{3}=\frac{16}{3}=\frac{32}{6}\) lb, \(6\frac{1}{2}=\frac{13}{2}=\frac{39}{6}\) lb
HCF of 21, 32 and 39 is 1, since 32 shares no factor with 21 or 39
Heaviest part \(=\frac{1}{6}\) lb
Number of parts \(=21+32+39=92\)
The cake gives 92 parts, so 92 guests can be served. 92 is not among 54, 72 and 20, so the answer is None of these.