The average weight of 4 men A, B, C, and D is | vedclass

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(B) $1$. Initial sum of weights of $A, B, C, D = 4 	imes 67 = 268 \, kg.$$2$. After including $E,$ total men = $5,$ and new average = $67 - 2 = 65 \,

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(B) $1$. Initial sum of weights of $A, B, C, D = 4 	imes 67 = 268 \, kg.$$2$. After including $E,$ total men = $5,$ and new average = $67 - 2 = 65 \, kg.$$3$. Total weight of $5$ men $(A+B+C+D+E) = 5 	imes 65 = 325 \, kg.$$4$. Weight of $E = 325 - 268 = 57 \, kg.$$5$. Weight of $F = E + 4 = 57 + 4 = 61 \, kg.$$6$. After replacing $A$ with $F,$ the new average weight is $64 \, kg.$$7$. Total weight of $5$ men $(F+B+C+D+E) = 5 	imes 64 = 320 \, kg.$$8$. Since $(A+B+C+D+E) = 325,$ we have $(B+C+D+E) = 325 - A.$$9$. Substituting this into the new total: $F + (325 - A) = 320.$$10$. $61 + 325 - A = 320 \implies 386 - A = 320 \implies A = 66 \, kg.$

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