# How much water must be added to 200 gallons of mixture which is 80 percent alcohol to reduce it to a 75 percent mixture?

Apr 15, 2017

$13 \frac{1}{3}$ gallons of water must be added to reduce it to a $75$ percent mixture.

#### Explanation:

$200$ gallons of mixture which has 80% alcohol, has

200×80%=200×80/100=2×80=160 gallons of alcohol.

If $160$ gallons of alcohol forms 75% of solution, the solutions has to be

160×100/75=160×4/3=640/3=213 1/3 gallons.

Hence, $213 \frac{1}{3} - 200 = 13 \frac{1}{3}$ gallons of water must be added to reduce it to a $75$ percent mixture.