A mixture of pure k2cr207 and pure kmn04 weighing 0.561g was treated with excess of ki in acidic medium. Iodine liberated required 100ml of 0.15M of sodium thiosulphate solution for exact oxidation. The right answer 43.67 k2cr207 and 56.33 kmn04

My calculation
0.1L x 0.15M = 0.015
x/294g/mol k2cr207 + y/158g/mol kmn04 (0.561g-x)= 0.015

Please ask for Chemistry, not math, help.