Celebrating Agebra with my friend

Let us assume that the glass that we both were holding can hold $10 \space ml$ .

Let us start with the mixture that Ashish had $\implies$ $9 \space ml :1 \space ml$ , total $=10 \space ml$

Now the mixture that I was holding was $\implies$ $4:1=8 \space ml: 2 \space ml$ , total $=10 \space ml$ .

Now, adding both the mixtures we get $(9+8) \space ml : (1+2) \space ml$ = $17 \space ml : 3 \space ml$

So, $x=17, \space y=3$ .

So $x+y=\color{#3D99F6} {\boxed{20}}$

Now, let the volume of each glass be $x$ , then the amount of alcohol in Abhay's glass is $\dfrac{4x}{5}$ and water is $\dfrac{x}{5}$ .
Now, in Ashish's glass amount of alcohol is $\dfrac{9x}{10}$ and water is $\dfrac{x}{10}$ .
So, total water is $\dfrac{x}{5} + \dfrac{x}{10} = \dfrac{3x}{10}$
Total alcohol = $\dfrac{4x}{5} + \dfrac{9x}{10} = \dfrac{17x}{10}$
So, ratio = $\dfrac{17x}{10} × \dfrac{10}{3x} = \dfrac{17}{3}$
So, $a + b = 17 + 3 = \color{#3D99F6}{\boxed{20}}$

I see that @Ashish Siva , you've been very busy writing wikis, drinking alcohol, lending dictionaries and loaning money.

Anyway, let the volume of water in Ashish's glass be $x$ .

This means the volume of alcohol in Ashish's glass is $9x$ .

The total volume of the glass $=9x + x=10x$

The volume of water in Abhay's glass $=10x \times \dfrac{1}{1+4} = 2x$

The volume of alcohol in Abhay's glass $=10x \times \dfrac{4}{1+4} = 8x$

The contents in both glasses were mixed. This means we have

$x+2x = 3x$ of water and

$9x+8x = 17x$ of alcohol

The ratio of alcohol : water $=17x:3x = 17:3$

$x=17,\;y=3,\;x+y=17+3=\boxed{20}$