Posted by **parth** on Tuesday, November 22, 2011 at 9:03pm.

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?

- maths -
**Steve**, Wednesday, November 23, 2011 at 12:08am
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.

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