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The Hog’s Head pub decides to make a mixture that is 50% Butterbeer. How much of an 80% Butterbeer solution should be mixed with 20 gallons of a 20% Butterbeer solution to create this new mixture? Enter a number only as your answer.

1 Answer

6 votes

Answer:

20 gallons or 160 pints or 75.7082 liters

Explanation:

Let v represent the volume (in gallons) of 80% solution that needs to be added. The mixture wants to be 50% Butterbeer, so we have ...

v·80% + 20·20% = (v+20)·50%

80v +400 = 50v +1000 . . . . multiply by 100, eliminate parentheses

30v = 600 . . . . . . . subtract 400+50v

600/30 = v = 20 . . . . . divide by the coefficient of v

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Comment on units

The units for measuring v are the same units used to measure "20". That is, they are gallons. (Our equation does not make sense otherwise.) The problem statement does not specify units, so we can choose any we like, such as pints (suitable for beer), or liters (an international standard). The number entered for an answer could be "1" (20-gallon keg).

For a problem such as this, when you define your variable using a "let ..." statement, you also need to define the units in which it is measured. This can keep you out of trouble (and show you where units conversions are necessary), especially when the problem uses different units for different parts of the problem.

User Brclz
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