6th Grade Olympiad Concentration Problem!! Urgent!! Help!!

The key to solving the problem is that the salt content of the brine remains the same.

Start with 120 70% = 84 (kg) ......After adding water to the original brine, the brine remains unchanged.

The concentration after dividing by saline, that is, the corresponding number divided by the corresponding amount 84 60% = 140 ......Find the weight of the brine after adding water, and then subtract the weight of the original brine from the weight of the later brine, and the extra is the weight of the added water: 140-120=20 (kg).

If x kg of water is added, there is:

70%*120=（120+x）*60%

The weight of the alcohol before and after adding water is the same.

Set the mass of the water to be x kilograms.

120*70%=(120+x)*60%

84-72=60%x

60%x=12

x=20

Set the first pour of alcohol from the nail x liters, the remaining alcohol in the nail 20 x

Pour 6 liters back from B to A, which is equivalent to pouring 6 20 30 of the alcohol in B back into A, the alcohol in A (20 x), and the remaining alcohol in B.

Both are equal. 20 x 20 liters.

Set the first pour of alcohol from a x liter.

20-x+6x/20=x-6x/20

7/5x=20

x=100/7

Solution: Let the mass of a certain concentration of alcohol in the original container be m, containing n cups of alcohol, the mass of a glass of water is m, and the mass of a cup of pure alcohol is the relative density of pure alcohol, then:

The simultaneous equation can be obtained: n = 1, m = so the original concentration = since there is 1 cup of alcohol in the original container, the concentration is.

Add another glass of pure alcohol, the pure alcohol content in the container is 40%, which is equivalent to mixing 1-40% = 60% of a glass of pure alcohol with 25% alcohol, the number of cups with 25% alcohol: 1 (1-40%) 40%-25%) = 4 glasses.

When a glass of water is added, the pure alcohol content in the container is 25%, so the original alcohol of a certain concentration is 4-1 = 3 cups.

Add a glass of water, the total amount of alcohol remains the same, and the original solution contains alcohol 4 25% = 11 3 = 1 3

25%：75%=1:3

It turned out that there was 1 cup of alcohol in the container with a concentration of 1:(1+2) 33%.