Question

Three pieces of cake 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 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?

• A. $54$
• B. $72$
• C. $20$
• D. $41$

Answer 1

The correct option is D $41$

The pieces of cakes can be simplified as $4\frac{1}{2}=\frac{9}{2}$, $6\frac{3}{4}=\frac{27}{4}$ and $7\frac{1}{5}=\frac{36}{5}$

To get the heaviest piece that each of the cakes can be divided into, we will take their HCF.

The HCF of 2 fractions is HCF of numerators divided by the LCM of denominators $=\frac{9}{20}$.

Pieces given by first cake $=\frac{9}{2}÷\frac{9}{20}=10$

Pieces given by 2nd cake $=\frac{27}{4}÷\frac{9}{20}=15$

Pieces given by 3rd cake $=\frac{36}{5}÷\frac{9}{20}=16$

Total guests that can be served $=10+15+16=41$.