Ratio of moles of Fe (II) oxidised by equal volumes of equimolar KMnO4 and K2Cr2O7 solutions in acidic medium will be:

Fe2++MnO4−⟶Fe3++Mn2+nf =7 nf =5Moles of Fe2+ x 1 by KMnO4=moles of MnO4−×5 ...(1)Fe2++Cr2O72−⟶Fe3++Cr3+nf =7 nf =6Moles of Fe2+ x 1 by K2Cr2O7=moles of MnO4−×6 ...(2)(1)+(2)n1n2=56