A mixture is made by combining 110 mL of 0.15 M $\ce{HCl}$ and 215 mL of 0.055 M $\ce{HI}$. What is the pH of the solution?
$$\ce{HCl -> H+ + Cl-}$$ $$\ce{HI -> H+ + I-}$$ According to what I thought, the molarity of $\ce{H+}$ is the same as $\ce{HCl}$, because it is a strong acid and the mole ratio. So, I added the molarity of both acids: $\pu{0.15M} + \pu{0.055M} = \pu{0.205M}$ for $\ce{H+}$ in solution.
$$\mathrm{pH} = -\log[\ce{H+}]$$ $$\mathrm{pH} = -\log[0.205]$$ $$\mathrm{pH} = 0.689$$
You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. here. Therefore you can assume, that the concentration of protons is equal to the concentration of the acid. \begin{align} c_{\ce{HI}}(\ce{H+}) &= c(\ce{HI}) & c_{\ce{HCl}}(\ce{H+}) &= c(\ce{HCl}) & \end{align}
Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$
You have to add the amount of protons in each solution and determine the combined volume. \begin{align} n_\mathrm{tot}(\ce{H+}) &= n_{\ce{HI}}(\ce{H+}) + n_{\ce{HCl}}(\ce{H+})& V_\mathrm{tot} &= V(\ce{HI}) + V(\ce{HCl}) \end{align}
To calculate the pH of the resulting concentration, just take the negative logarithm. $$\mathrm{pH} = -\log c_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n_\mathrm{tot}(\ce{H+})}{V_\mathrm{tot}}\right]$$
Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.
The approach you chose can only work, if you have equal volumes of the two solutions.
$$\begin{align} M_1V_1+M_2V_2&=M\left(V_1+V_2\right)\\ 0.015\times110+0.055\times215&=M\left(110+215\right)\\ 1.65+11.825&=M\cdot325\\ M&=\frac{13.475}{325}\\ M&=0.04146=4.416\times10^{-2}\\ \mathrm{pH}&=-\log\left(4.416\times10^{-2}\right)\\ \mathrm{pH}&=-\log4.416+2\log10\\ \mathrm{pH}&=-0.617+2\\ \mathrm{pH}&=1.383 \end{align}$$
You can go this way also $$\begin{align} \text{HCl conc}&=0.015\ \mathrm M\\ &=0.015\times\frac{110}{1000}\ \mathrm{mol}\\ &=1.65\times10^{-3}\ \mathrm{mol} \end{align}$$
$$\begin{align} \text{HI conc}&=0.055\ \mathrm M\\ &=0.055\times\frac{215}{1000}\\ &=11.825\times10^{-3}\ \mathrm{mol} \end{align}$$
$$\begin{align} \text{Total}\ [\ce{H+}]&=\left(1.65\times10^{-3}+11.825\times10^{-3}\right)\ \mathrm{mol}\ \text{in}\ 325\ \mathrm{ml}\\ &=13.475\times10^{-3}\ \mathrm{mol}\ \text{in}\ 325\ \mathrm{ml}\\ \text{so in}\ 1000\ \mathrm{ml}&=13.475\times10^{-3}\times\frac{1000}{325}\\ &=4.146\times10^{-2}\\ \text{Thus}\ \mathrm{pH}&=1.383 \end{align}$$
