How do I solve this problem on mass balance from blast furnace iron making? Question

An ironmaking blast furnace produces hot metal of the composition: $\pu{93.6\%}$ $\ce{Fe},$ $\pu{2.1\%}$ $\ce{Si},$ $\pu{3.6\%}$ $\ce{C},$ $\pu{0.7\%}$ $\ce{Mn}.$ The composition of the iron ore is $\pu{78\%}$ $\ce{Fe2O3},$ $\pu{9\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{Al2O3},$ $\pu{1\%}$ $\ce{MnO},$ $\pu{7\%}$ moisture. Coke rate is $\pu{850 kg/thm}.$ Composition of the coke is $\pu{80\%}$ $\ce{C},$ $\pu{10\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{CaO}.$ Enough limestone is charged to make $\pu{45\%}$ $\ce{CaO}$ in slag and composition of limestone is $\pu{100\%}$ $\ce{CaCO3}.$
Calculate (i) Weight of ore used per tonne of hot metal production; (ii) weight of slag produced per tonne of H.M.; (iii) weight of limestone used to produce given slag composition.

Doubt
In such problems all I need to balance is iron, silica, $\ce{MnO}$ etc. However, in case of $\ce{MnO}$ the solution in my book reads $\ce{MnO}$ from ore equals $\ce{Mn}$ in hot metal plus $\ce{MnO}$ in slag.
My question is for balancing $\ce{MnO}$: why do I need to drag in $\ce{Mn}$ content? The $\ce{Mn}$ in iron ore should be equal to $\ce{Mn}$ in slag. In fact, I feel that any content apart from $\ce{MnO}$ should not be dragged into picture. The rest of the contents like alumina, iron etc. quite comply to this argument. I would require a vivid justification to this.
"The Mn in iron ore should be equal to Mn in slag." vs
"the solution in my book reads MnO from ore equals Mn in hot metal plus MnO in slag."

It seems that the calculation desired involves breaking down the MnO in ore to Mn + O, and putting some of the Mn into the metal; the rest of the Mn goes into slag, as MnO. The hot metal would contain only metallic Mn, not MnO. It shouldn't be a huge adjustment in the answers.