Now according to a cup of 200ml water calculation:
A cup of 200ml water and B cup of 200ml alcohol.
The first time: a cup of 20ml alcohol, with a total volume of 220ml；;
The second cup is reduced by 20ml of alcohol, and the total volume is180 ml;
Second time: reduce 20ml of mixed solution in Cup A, and the remaining alcohol volume is:
20-(20 * 20/220)= 20- 1.8 = 18.2ml；
Take 20ml of mixed solution from Cup B, and the volume of alcohol is:
180+(20 * 20/220)= 180+ 1.8 = 18 1.8ml；
The results are as follows: one cup contains 18.2ml alcohol, and the rest is water; (More water)
The second cup contains 18 1.8ml alcohol, and the rest is water. (Too much alcohol)
It is assumed that when alcohol and water are mixed, the volume will not shrink.
(2) Mixed concentration of brine.
First, find out the mass of 30% brine M:
(m*30%+300*20%)/(m+300)=25%
0.3m+60 = 0.25m +75
0.05m = 15
m = 300g
Calculating the mass of 40% brine by cross method;
40% physiological saline: 30- 10=20 portions;
10% physiological saline: 40-30 =10;
A * * * is 300g, so the quality of 40% brine is:
20 * 300g/(20+10) = 200g.