Question

Three pieces of cakes of weights $4\frac{1}{2}$ 1bs $6\frac{3}{4}$ 1bs and $7\frac{1}{5}$ 1bs 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?
(lb is for 1 pound)

• A. $54$
• B. $72$
• C. $41$
• D. $45$

Answer 1

The correct option is C $41$

Total weight of the three pieces
= $(\frac{9}{2}+\frac{27}{4}+\frac{36}{5})$ pounds = $\frac{369}{20}$ Pounds
Required weight of a single piece
= H.C.F. of $(\frac{9}{2},\frac{27}{4},\frac{36}{5})$
= $\frac{H.C.F.of(9,27,36)}{L.C.M.of(2,4,5)}=\frac{9}{20}$ pound
$\therefore$ Number of guests
$=\frac{Totalweight}{Weightofasinglepiece}$
$=\frac{369/20}{9/20}=\frac{369}{9}=41$