When the concentration of a certain alcohol solution is 60% and 100 liters of pure alcohol is added to form an 80% alcohol solution, how many liters of the original alcohol?

When the concentration of a certain alcohol solution is 60% and 100 liters of pure alcohol is added to form an 80% alcohol solution, how many liters of the original alcohol?

100L
60%x+100%×100=80%×（100+x）
x=100

There are several liters of 55% alcohol solution. After adding one liter of 80% alcohol solution, the concentration of alcohol solution becomes 60%. If you want to get 70% alcohol solution, how many liters of 80% alcohol solution do you need to add?

(1 × 80% - 1 × 60%) / (60% - 55%), = 0.2 ／ 0.05, = 4 (L), [(4 + 1) × 70% - (4 + 1) × 60%] / (80% - 70%), = 0.5 ／ 0.1, = 5 (L); a: add another 5 L of 80% alcohol solution

There are a number of 80% alcohol, + 1000 grams of water, the concentration of 60%, the original concentration of 80% alcohol how many grams?

3000g
Set 80% alcohol as X
80%X==（1000+X）60%
X==3000
In principle, the quality of pure alcohol is certain