clock reaction dilutions for KI Stock solution 0.1M KI solution 100mL

Volume of KI
30
10.5
5.25                                      

Volume of water
0
19.5
14.75 

concentration of KI
0.1
0.05
0.025

How did they calculate these numbers?
 I tried c1 v1 = c2 v2 
For example for 0.05M:
(0.1)(v1) = (0.05)(30/1000)
v1= 0.015L = 15mL, what does this 15mL represent. How did they calculate the volume of KI and water. 

I want to dilute KI and I want the final solution to be 30mL. And the concentrations that I need are shown above (0.1, 0.05, 0.025, 0.0125).

Please help
The formula 

$$c_1\cdot V_1=c_2\cdot V_2$$

is correct.

Since your stock solution has a concentration of $c=0.1\ \mathrm{mol\ l^{-1}}$, the value for the initial concentration $c_1$ is always

$$c_1=0.1\ \mathrm{mol\ l^{-1}}$$

Since you want the final volume of each solution to be $V=30\ \mathrm{ml}$, the value for the final volume $V_2$ is always

$$V_2=30\ \mathrm{ml}$$

You want four individual solutions (labelled $\mathrm a$, $\mathrm b$, $\mathrm c$, and $\mathrm d$) with four different final concentrations ($0.1\ \mathrm{mol\ l^{-1}}$, $0.05\ \mathrm{mol\ l^{-1}}$, $0.025\ \mathrm{mol\ l^{-1}}$, $0.0125\ \mathrm{mol\ l^{-1}}$. These are the individual values for the final concentration $c_2$:

$$\begin{align}
c_{2,\mathrm{a}}&amp;=0.1\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{b}}&amp;=0.05\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{c}}&amp;=0.025\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{d}}&amp;=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$

Now you can rearrange the above-mentioned equation to solve for the initial volume of the stock solution $V_1$:

$$\begin{alignat}{2}
V_1&amp;=\frac{c_2\cdot V_2}{c_1}\\[6pt]
V_{1,\mathrm a}&amp;=\frac{0.1\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&amp;&amp;=30\ \mathrm{ml}\\[6pt]
V_{1,\mathrm b}&amp;=\frac{0.05\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&amp;&amp;=15\ \mathrm{ml}\\[6pt]
V_{1,\mathrm c}&amp;=\frac{0.025\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&amp;&amp;=7.5\ \mathrm{ml}\\[6pt]
V_{1,\mathrm d}&amp;=\frac{0.0125\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&amp;&amp;=3.75\ \mathrm{ml}
\end{alignat}$$

You measure these volumes of the stock solution and transfer them into four separate vessels.

Finally, you have to fill up each solution with pure water to the final volume of $V_2=30\ \mathrm{ml}$. The required additional volume of water $V_{\ce{H2O}}$ is approximately

$$\begin{alignat}{2}
V_{\ce{H2O}}&amp;\approx V_2-V_1\\[6pt]
V_{\ce{H2O},\mathrm a}&amp;\approx30\ \mathrm{ml}-30\ \mathrm{ml}&amp;&amp;=0\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm b}&amp;\approx30\ \mathrm{ml}-15\ \mathrm{ml}&amp;&amp;=15\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm c}&amp;\approx30\ \mathrm{ml}-7.5\ \mathrm{ml}&amp;&amp;=22.5\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm d}&amp;\approx30\ \mathrm{ml}-3.75\ \mathrm{ml}&amp;&amp;=26.25\ \mathrm{ml}
\end{alignat}$$