Titration with Ca(OH)2 In the laboratory, $20.0$ milliliters of an aqueous solution of calcium hydroxide, $\ce{Ca(OH)2}$, was used in a titration. A drop of phenolphthalein was added to it to indicate the end point. The solution turned colorless after $20.0$ milliliters of a standard solution of $\pu{0.050 M}$ $\ce{HCl}$ solution was added. What was the molarity of the $\ce{Ca(OH)2}$?
  (A) $\pu{0.010 M}$
  (B) $\pu{0.025 M}$
  (C) $\pu{0.50 M}$
  (D) $\pu{0.75 M}$
  (E) $\pu{1.0 M}$  


I did not know how to answer such kind of question using formulas, that is why I applied logic. At first, I wrote such a reaction: 

$$\ce{\underset{\pu{20 mL}}{Ca(OH)2} + \underset{\pu{0.05 M}}{\underset{\pu{20 mL}}{HCl}} = end point}$$

Then, I understood that if the volume of both reactants is equal, the molarity of $\ce{HCl}$ must be higher than that of $\ce{Ca(OH)2}$, as $\ce{HCl}$ changes the nature of $\ce{Ca(OH)2}$.

As I knew that the molarity of $\ce{Ca(OH)2}$ is just a bit less than that of $\ce{HCl}$, I chose B, and this variant turned out to be the right one.
In the book, I have such explanation:  


  In the titration, the reaction is:
  $$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$
  The acid to base ratio is $2 : 1$, or (moles acid used) = 2(moles base used), so $M_\mathrm{a}V_\mathrm{a} = 2M_\mathrm{b}V_\mathrm{b}$, where $M$ is the molarity and $V$ is the volume expressed in liters. Then
  $$M_\mathrm{b} = \frac{M_\mathrm{a}V_\mathrm{a}}{2V_\mathrm{b}}$$ 
  $$M_\mathrm{b} = \frac{\pu{0.05 M} \times \pu{0.02 L}}{2}$$
  $$\pu{0.02 L} = \pu{0.025 M}$$  


I do not understand the explanation. Although it comes up with the same variant as I did, if I had such formula I would use it this way $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ as we have 2 moles of acid. 

Then, the answer would be $0.1$. The same formula is used on this forum. Can you help me understand what is wrong here please?
(1) Meaning of the textbook solution

$$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$

This reaction clearly tells us that 

$$\frac{moles_{acid}}{moles_{base}}=2$$

moles=molarity* volume(in litres)

Therefore, $$\frac{moles_{acid}}{moles_{base}}=\frac{M_{a}V_{a}}{M_{b}V_{b}}$$

(2) What is wrong with your interpretation?

You wrote $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$.

$2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ actually implies that more number of moles of base need to be taken. For neutralisation, for every one mole of base, we require two moles of acid since each calcium hydroxide will release 2 hydroxide ions (for which two H+ are required).