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

1 Answer

Shwetank Mauria

Apr 15, 2017

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

Explanation:

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

$200 \times 80 % = 200 \times \frac{80}{100} = 2 \times 80 = 160$ $200×80%=200×80/100=2×80=160$ gallons of alcohol.

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

$160 \times \frac{100}{75} = 160 \times \frac{4}{3} = \frac{640}{3} = 213 \frac{1}{3}$ $160×100/75=160×4/3=640/3=213 1/3$ gallons.

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

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