document.write( "Question 691017: a 20% solution of fertilizer is to be mixed with a 50% solution of fertilizer in order to get 60 gallons of a 40% solution. How many gallons of the 20% solution and 50% solution should be mixed \n" ); document.write( "

Algebra.Com's Answer #426456 by fcabanski(1391) $\"\"$ $\"About$

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Start with two equations. The first is for the gallons and the second for the amount of alcohol. Call the first solution x and the second y.

\n" ); document.write( "Gallons: x+y=60 and thus x=60-y

\n" ); document.write( "Amount of alcohol. .2x + .5y = .4*60 = 24. This makes sense because 60 gallons of a 40% solution has 24 gallons of alcohol in it. 24/60 = .4

\n" ); document.write( "Substitute x in terms of y from the gallons equation into the amount equation:

\n" ); document.write( ".2(60-y) +.5y =24 ---> 12 -.2y +.5y = 24 ---> .3y = 12 ---> y = 40.

\n" ); document.write( "Substitute y=40 into the gallons equation x+40 = 60 ---> x =20

\n" ); document.write( "20 gallons of 20% solution mix with 40 gallons of 50% solution to give 60 gallons of 40% solution.
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