KVPY Sample Paper KVPY Stream-SX Model Paper-30

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 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

Correct Answer: D

Solution :

\[HCF\left( \frac{9}{2},\frac{27}{4},\frac{36}{5} \right)=\frac{HCF(9,27,36)}{LCM(2,4,5)}=\frac{9}{20}lbs\]

= weight of each piece.

\[Total\text{ }weight=\frac{9}{2}+\frac{27}{2}+\frac{36}{5}=18.45\,\,;\]

\[Maximum\text{ }number\text{ }of\text{ }guests=\frac{18.45\times 20}{9}=41\]