Question 585098
I'll pretend I'm not a chemist.
Assuming volumes are additive (or close enough for our purpose),
(256 x 32+125 x 20+265 x 0) / (256+125+265) = 16.6 (rounding)
(I put the parentheses where they are needed according to algebraic notation when you cannot draw horizontal fraction bar lines. If I use those horizontal lines, I would write
{{{(256 x 32+125 x 20+265 x 0) / (256+125+265) = 16.6}}}
I would actually calculate it as
{{{(256*0.32+125*0.20+256*0.00)/(256+125+265)=0.166}}}
where all the percentages are expressed as fractions/decimals, as in {{{0.32=32/100}}}
If there was a thing that is considered 100% HCl and you wanted what the same people consider 55% HCl, you could mix that 100% with water, or an HCl solution of any concentration to get the desired 55% HCl.
If you started with the original mix, and added x mL of 100% HCl, the final concentration (as a fraction/decimal) would be
{{{(256*0.32+125*0.20+256*0.00+x*1.00)/(256+125+265+x)=0.55}}} --> {{{(256*0.32+125*0.20+256*0.00+x*1.00)=0.55(256+125+265+x)}}} --> {{{106.92+x=0.55(646+x)}}} -- {{{106.92+x=355.3+0.55x)}}} -- {{{x-0.55x=355.3-106.92}}} --> {{{0.45x=248.38}}} --> {{{x=248.38/0.45}}} --> {{{highlight(x=552.0)}}} (rounding).
NOTE:
As a chemist, I really hate this kind of math problem, because the chemistry part is all wrong, but this one is the "wrongest" one I have seen.
1) HCl as pure HCl is a gas in a compressed gas cylinder that the chemists call hydrogen chloride, and sometimes use in chemical synthesis.
A concentrated solution of HCl in water is the convenient form I have been using in labs for the past 40 years, and that is what we usually call hydrochloric acid. 
2) Solution concentrations, when expressed as %, are specified as  "% w/w" (percent weight in weight) or "% w/v" (weight in volume), or even "% v/v" (volume in volume). A commercial concentrated HCl solution probably has analysis results printed in the label, listing assay as 37% or so. That is in terms of % w/w. The list would hopefully include density of the solution.
3) Volumes are not additive. When you mix different solutions, the resulting volume is not exactly the sum of the volumes, and in some cases the difference in volume could be substantial.