Lets say you have a chemical solution $A$ that has a ratio of $4:1$ of water to some chemical substance, and a solution $B$ that has a ratio of $9:1$ of water to some other chemical substance.
If you combine the same amount of each solution, the resulting solution is not in a ratio of $6.5 : 1$ (the average), but is instead $5.6666 : 1$.
Can someone help me understand why one cant just average the two ratios if the same amount of solution is added together? For some reason its just not clicking.
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.