Consider the following equations:
$$2Fe^{2+} + H_2O_2 \to xA + yB$$ (in basic medium)
$$2MnO_4^- + 6H^+ + 5H_2O_2 \to x'C + y'D + z'E$$ (in acidic medium)
The sum of the stoichiometric coefficients x, y, x', y' and z' for products A, B, C, D and E respectively, is

First we deal with the reaction that occurs in basic medium:

$$2Fe^{2+}+H_2O_2 \rightarrow xA+yB$$

In basic medium, the usual redox couples are

$$Fe^{3+}/Fe^{2+}\;,\qquad H_2O_2/\,OH^-$$

We write the two half-reactions and state the rule that “electrons lost in oxidation must equal electrons gained in reduction.”

Oxidation half-reaction (ferrous to ferric):

$$Fe^{2+}\;\longrightarrow\;Fe^{3+}+e^-$$

Reduction half-reaction (peroxide to hydroxide in basic medium):

$$H_2O_2+2e^-\;\longrightarrow\;2OH^-$$

To equalise electrons, we multiply the oxidation half by 2 so that both halves involve two electrons:

$$2Fe^{2+}\;\longrightarrow\;2Fe^{3+}+2e^-$$

Adding the two balanced halves gives

$$2Fe^{2+}+H_2O_2\;\longrightarrow\;2Fe^{3+}+2OH^-$$

Hence the products are

$$A=Fe^{3+},\qquad B=OH^-$$

with stoichiometric coefficients

$$x=2,\qquad y=2$$

Now we balance the second reaction that takes place in acidic medium:

$$2MnO_4^-+6H^++5H_2O_2 \rightarrow x'C+y'D+z'E$$

In acidic solution the relevant redox couples are

$$MnO_4^-/Mn^{2+}\;,\qquad H_2O_2/O_2$$

Reduction half-reaction (permanganate to manganous):

$$MnO_4^-+8H^++5e^-\;\longrightarrow\;Mn^{2+}+4H_2O$$

Oxidation half-reaction (hydrogen peroxide to oxygen):

$$H_2O_2\;\longrightarrow\;O_2+2H^++2e^-$$

To match the electron count we take the least common multiple of 5 and 2, which is 10. So,

multiply the permanganate half by 2:

$$2MnO_4^-+16H^++10e^-\;\longrightarrow\;2Mn^{2+}+8H_2O$$

multiply the peroxide half by 5:

$$5H_2O_2\;\longrightarrow\;5O_2+10H^++10e^-$$

Adding the two adjusted half-reactions and cancelling the 10 electrons yields

$$2MnO_4^-+16H^++5H_2O_2\;\longrightarrow\;2Mn^{2+}+8H_2O+5O_2+10H^+$$

Subtracting $$10H^+$$ from both sides gives the fully balanced overall equation in acidic medium:

$$2MnO_4^-+6H^++5H_2O_2\;\longrightarrow\;2Mn^{2+}+8H_2O+5O_2$$

Thus the products are

$$C=Mn^{2+},\qquad D=H_2O,\qquad E=O_2$$

with stoichiometric coefficients

$$x'=2,\qquad y'=8,\qquad z'=5$$

Finally we add all the required coefficients:

$$x+y+x'+y'+z'=2+2+2+8+5=19$$

So, the answer is $$19$$.