Question 252630
This is a mixture problem. The thing to remember is that water is 0% everything. Here is the table:
liquid . . . . % . . . . Liter . . . . %Liter
acid . . . . . 50 . . . . 10 . . . . . 500
water . . . .  0 . . . . . L . . . . . 0
mixture . . .  80 . . . . 10 + L . . . 800 + 80L
The liters must be added to get the total amount of liquid.
Now,
500 + 0 = 800 + 80L
This is not possible. You can't add water and make it stronger. If it were 80% and you wanted a mixture of 50%, then we are good. Suppose that is what you meant. The new table looks like this:
liquid . . . . % . . . . Liter . . . . %Liter
acid . . . . . 80 . . . . 10 . . . . . 800
water . . . .  0 . . . . . L . . . . . 0
mixture . . .  50 . . . . 10 + L . . . 500 + 50L
Now,
800 + 0 = 500 + 50L
300 = 50L
L = 6. You would need 6 liters of water to dilute the mixture from 80 to 50%.