A merchant has 435 litres,493 litres,and 551 | vedclass

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(A) To find the least number of casks,the capacity of each cask must be the greatest possible value that divides $435$,$493$,and $551$ exactly. This c

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(A) To find the least number of casks,the capacity of each cask must be the greatest possible value that divides $435$,$493$,and $551$ exactly. This capacity is the $H.C.F.$ of $435$,$493$,and $551$.First,find the $H.C.F.$ of $435$ and $493$:$493 = 435 	imes 1 + 58$$435 = 58 	imes 7 + 29$$58 = 29 	imes 2 + 0$So,the $H.C.F.$ of $435$ and $493$ is $29$.Now,check if $551$ is divisible by $29$:$551 \div 29 = 19$.Thus,the $H.C.F.$ of $435, 493,$ and $551$ is $29$.The capacity of each cask is $29$ litres.Total number of casks required $= (435 \div 29) + (493 \div 29) + (551 \div 29)$$= 15 + 17 + 19 = 51$.

$A$ merchant has $435$ litres,$493$ litres,and $551$ litres of three different kinds of milk. Find the least number of casks of equal size required to store all the milk without mixing.

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