Solution

$$MCO_3\rightarrow MO+CO_2\uparrow$$
$$M'CO_3\rightarrow M'O+CO_2\uparrow$$
Equivalent of $$CO_2 =$$ Equivalent of carbonates of metals
$$1$$ mol of $$CO_2 \equiv 1$$ mol of $$CO_3^{2-}$$
$$44$$ g of $$CO_2 \equiv $$60$$ g of $$CO_3^{2-}$$
$$1.10$$ g of $$CO_2 \equiv \dfrac {60}{44}\times 1.10=1.5$$ g of $$CO_3^{2-}$$
% of $$CO_3^{2-}=\dfrac {1.5\times 100}{3}=50$$%
% of one metal $$ = $$ $$15$$ %
% of another metal $$ = 100-50-15=35$$%