Three whole cakes weigh $$4\frac{1}{2}$$​ lbs, $$6\frac{3}{4}$$lbs and $$7\frac{1}{5}$$​ Ibs, respectively. Each cake has to be divided into pieces of equal weight. Each piece must be as heavy as possible. If one such piece is served to each guest, then what are the maximum number of guests that can be served?

The weights of the three cakes are:

First cake: $$\frac{9}{2}$$​​ lbs

Second cake: $$\frac{27}{4}​$$​ lbs

Third cake: $$\frac{36}{5}​$$​ lbs

HCF of the numerators (9, 27, and 36) = 9

LCM of the denominators (2, 4, and 5) = 20

Therefore, the HCF of the fractions is $$\frac{9}{20}$$​.

Now, the number of pieces for each cake:

For the first cake: $$\frac{9/2}{9/20} = \frac{9 \times 20}{2 \times 9}$$​ = 10 pieces

For the second cake: $$\frac{27/4}{9/20} = \frac{27 \times 20}{4 \times 9}$$ = 15 pieces

For the third cake: $$\frac{36/5}{9/20} = \frac{36 \times 20}{5 \times 9}$$​ = 16 pieces

Thus, Total number of pieces (guests served):10 + 15 + 16 = 41.