How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution? Below was question 34 in the USNCO 2017 exam:


  If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?
  
  (A) Roughly 50%
  (B) Roughly 75%
  (C) Roughly 95%
  (D) Over 99%  


My solution is as follows:


Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K_\mathrm{sp}$.
Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$
Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$
Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation: 
$$
(0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$
Find percentage of calcium hydroxide precipitated: 
$$\frac{0.019}{0.02} \times 100\% = 95\%,$$
hence (C)


I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).

Is there a simpler method?
A less analytic aproach:

Initial concentrations: 

$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$

$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.

Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.
The concentrations that remain in solution are:

$\ce{[Ca^{2+}}]=0.0126(1-p)$

$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$

Since the solubility product remains the same:

$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$

Now, since there are only four scenarios, try them out substituting every value of $p$ in  the equation and take the closest one to $8\cdot10^{-6}$:

\begin{array}{|c|c|}\hline
    p&amp;K_\mathrm{sp}\\\hline
    0.50&amp;\pu{7.99E-05}\\
    0.75&amp;\pu{3.56E-05}\\
    0.95&amp;\pu{6.46E-06}\\
    0.99&amp;\pu{1.27E-06}\\\hline
\end{array}

The closest one is for roughly 95% precipitation, without solving any equation.