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. |
\[Total\text{ }weight=\frac{9}{2}+\frac{27}{2}+\frac{36}{5}=18.45\,\,;\] |
\[Maximum\text{ }number\text{ }of\text{ }guests=\frac{18.45\times 20}{9}=41\] |
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