Please help

I have no idea how to answer this question even though I've been revising for 2 hours on this topic:
This question is about the chemistry of elements in the d-block of the Periodic Table.
Under certain conditions, dichromate(VI) ions, Cr2O7can oxidise manganese(II) ions, Mn2+ In this reaction, dichromate(VI) ions are reduced to chromium(III) ions, in acidic conditions, according to the half-equation Cr2O7 2−(aq) + 14H+(aq) + 6e− 2Cr3+(aq) + 7H2O(l)
In an experiment it was found that 20.0 cm3 of 0.100 mol dm−3 potassium dichromate(VI) was required to oxidise 30.0 cm3 of 0.200 mol dm−3 manganese(II) sulfate solution. Use these data to calculate the final oxidation state of the manganese

Start by calculating the moles of Mn^2+ ions and the moles of Cr2O7^2- ions.

Then use the equation and the moles of Cr2O7^2- ions to work out how many moles of electrons are transferred.

I have no idea how to answer this question even though I've been revising for 2 hours on this topic:
This question is about the chemistry of elements in the d-block of the Periodic Table.
Under certain conditions, dichromate(VI) ions, Cr2O7can oxidise manganese(II) ions, Mn2+ In this reaction, dichromate(VI) ions are reduced to chromium(III) ions, in acidic conditions, according to the half-equation Cr2O7 2−(aq) + 14H+(aq) + 6e− 2Cr3+(aq) + 7H2O(l)
In an experiment it was found that 20.0 cm3 of 0.100 mol dm−3 potassium dichromate(VI) was required to oxidise 30.0 cm3 of 0.200 mol dm−3 manganese(II) sulfate solution. Use these data to calculate the final oxidation state of the manganese

did you figure out how to do it?
im stuck at the part where i found the ratio 1:3
but the markscheme says something like 3 mn2+ lose 6 electrons..
how ??? the ratio is 1:3

The mole ratio means that 1 mole of Cr2O72- reacts with 3 moles of Mn2+. Since you know from the equation that 1 mole of Cr2O72- gains 6 electrons. That means 6 electrons had to be lost from the 3 Mn2+. Therefore each Mn2+ had to lose 2 electrons as 6/3 =2. Therefore the final oxidation state of the Mn2+ would be +4 as it has lost 2 electrons. I hope I explained it well.

The mole ratio means that 1 mole of Cr2O72- reacts with 3 moles of Mn2+. Since you know from the equation that 1 mole of Cr2O72- gains 6 electrons. That means 6 electrons had to be lost from the 3 Mn2+. Therefore each Mn2+ had to lose 2 electrons as 6/3 =2. Therefore the final oxidation state of the Mn2+ would be +4 as it has lost 2 electrons. I hope I explained it well.