posted by parth .
brand 1 beer contain 8% alcohol and brand2 contain 6.5% of alcohol. how many litres of each alcohol must be added to produce 600 l of a new brand containing 7 % alcohol?
Concentrate on the amount of alcohol. No matter how much water is sloshed around in various concentrations, the amount of alcohol is what you want to keep track of.
You have n liters of 6.5% beer, and thus 600-n liters of 8% beer
Alcohol in n liters of 6.5% = .065n
Alcohol in 600-n liters of 8% = .08(600-n)
At the end, you have 600 liters of 7% beer, holding 600*.07 alcohol.
Add things up:
.065n + .08(600-n) = .07*600
.065n + 48 - .08n = 42
.015n = 6
n = 400
So, mix 400 liters of 6.5% beer with 200 liters of 8% beer
Note that 7% is 1/3 of the way between 6.5% and 8%. Thus, you need twice as much 6.5% beer as 8% beer.