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 part 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? (CAT 2001)

• A. 54
• B. 72
• C. 20
• D. None of these

Answer 1

The correct option is D None of these

Total weight of three pieces: $=(\frac{9}{2}+\frac{27}{4}+\frac{36}{5})=\frac{369}{20}=18.45lbs.$

Required weight of a single piece is HCF of $(\frac{9}{2},\frac{27}{4}and\frac{36}{5})=\frac{HCFof(9,27,36)}{LCMof(2,4,5)}=\frac{9}{20}lb.$

Number of guests = $\frac{18.45}{\left(\frac{9}{20}\right)}=41$