Question 495151
A certain mixture consists of acid and water.
 If 5 gallons of acid is added to the mixture, it will produce a mixture that is one-half acid and one-half water.
 If, on the other hand, 5 gallons of water is added, the resulting mixture is one-third acid.
 How many gallons of acid does the original mixture contain?
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Let x = the decimal value of of the percent acid in the original mixture
Let y = the amt (in gallons of the original mixture)
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write an equation for each statement, based on the amt of acid:
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"If 5 gallons of acid is added to the mixture, it will produce a mixture that is one-half acid and one-half water."
xy + 5 = .5(y+5)
xy + 5 = .5y + 2.5
xy = .5y + 2.5 - 5
xy = .5y - 2.5
x = {{{((.5y-2.5))/y}}}
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"If, on the other hand, 5 gallons of water is added, the resulting mixture is one-third acid."
xy = {{{1/3}}}(y+5)
3xy = y + 5
replace x with {{{((.5y-2.5))/y}}}; find y
3y({{{((.5y-2.5))/y}}}) = y + 5
y cancels (fortunately) so we have
3(.5y - 2.5) = y + 5
1.5y - 7.5 = y + 5
1.5y - y = 5 + 7.5
.5y = 12.5
multiply both side by 2
y = 25 gal is the amt of the original mixture
:
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Find x using x = {{{((.5y-2.5))/y}}}
x = {{{((.5(25)-2.5))/25}}}
x = {{{((12.5-2.5))/25}}}
x = {{{10/25}}}
x = .4, the original mixture was 40% acid
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See if that checks out in the first statement
If 5 gallons of acid is added to the mixture, it will produce a mixture that is one-half acid and one-half water."
.4(25) + 5 = .5(25 + 5)
10 + 5 = .5(30); confirms our solution
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An interesting problem (Most of them aren't)
Did all this make sense to you? Make an effort to understand each step, ask me in the comment section, if you have a question.  C