Drink 1/4 pure milk first and then add water. At this time, the concentration is1× (1-1/4) =1× (3/4) = 0.75 = 75%.

Drink 1/4 and fill it with water. At this time, the concentration of milk is 75% × (1-1/4) = 56.25%.

the second question

The original sugar content is 900×6%=54 (g).

You can put x grams of sugar, and then it is

54+X= 10%×(900+X)

The solution is x = 40g.

Third question

Raw salt: 8× 15%= 1.2 (L) water: 8- 1.2=6.8 (L).

Similarly, it is assumed that 10% salt water can be obtained by adding x liters of water.

6.8+X=( 1- 10%)×(8+X)

X=4 liters of water.

The fourth question

The title gives the volume and salt content of brine, which can only be calculated according to the mass/volume percentage concentration.

Let the original brine volume be V and the concentration be C%. The volume of water added each time is a.

After adding water for the first time, the concentration: V * C%/(V+A) = 8%;

After adding water for the second time, the concentration: v * c%/(v+2a) = 5%;

According to the above two formulas, 8%*(V+A)=5%*(V+2A).

0.08V+0.08A=0.05V+0. 10A

0.02A=0.03V

a = 0.03v/0.02 = 1.5V；

Original concentration of brine: V*C%/(V+2* 1.5)=5%.

V*C%=4V*5%

c % = 20%；

After adding water for three times, the concentration of brine is:

V * 20%/(v+ 3 * 1.5V)= 0.20V/5.5V≈3.6%。