Take 10x litre of each solution, so we are adding 8x & 9x litre of water (not 8x litre each), 2x & x litre of that chemical(not 2x or x litre each).
So, total water = (8x + 9x) litre = 17x litre and total chemical=(2x + x) litre = 3x litre.
So, the resultant ratio is $\frac{17}{3}=5.666...$
Clearly, when we are adding the same amount of the solutions, the same amount of water (or the chemical) are not added.
Trivially, we could take average, only if we had added same amount of water (or the chemical) separately.
Let us assume that the ratio of water and the chemical be a:b in the first solution and c:d in the second.
Let's take (a+b)x litre of the 1st solution and (c+d)y litre of the second.
So, water taken = (ax + cy) litre and the chemical taken = (bx + dy) litre.
If $\frac{ax+cy}{bx+dy}=\frac{\frac{a+c}{2}}{\frac{b+d}{2}}$
$=>\frac{ax+cy}{bx+dy}=\frac{a+c}{b+d}=>(bc-ad)(x-y)=0$
So if bc=ad=>c:d=a:b (i.e., if the ratios of water & the chemical are same), the ratio of water & the chemical will be the ratio of their average for any values of x,y i.e., we can mix the solutions in any ratio.
Here is this question $a=4,b=1,c=9,d=1=>bc-ad=1.9-4.1≠0=>x=y$
So, we need to take (4+1)x and (9+1)x litres of the solutions, respectively to make the ratio of water and the chemical as the ratio of their average.