$5.$ To convert the transportation problem into a maximisation model we have to $($a$)$ write the inverse of the matrix, $($b$)$ Multiply the rim requirements by $1,($c$)$ To multiply the matrix by $1,($d$)$ We cannot convert the transportation problem into a maximisation problem, as it is basically a minimisation problem. $()6.$ In a transportation problem where the demand or requirement is equal to the available resource is known as $($a$)$ Balanced transportation problem, $($b$)$ Regular transportation problem, $($c$)$ Resource allocation transportation problem, $($d$)$ Simple transportation model. $()7.$ The total number of allocation in a basic feasible solution of transportation problem of m $\backslash$times n size is equal to $($a$)$ m $\backslash$times n$,($b$)($m $/$ n $)1,($c$)$ m $+$ n $+1($d$)$ m $+$ n $1.()8.$ When the total allocations in a transportation model of m x n size is not equals to m $+$ n $1$ the situation is known as $($a$)$ Unbalanced situation, $($b$)$ Tie situation, $($c$)$ Degeneracy, $($d$)$ None of the above. $()9.$ The opportunity cost of a row in a transportation problem is obtained by: $($a$)$ Deducting the smallest element in the row from all other elements of the row, $($b$)$ Adding the smallest element in the row to all other elements of the row, $($c$)$ Deducting the smallest element in the row from the next highest element of the row,