document.write( "Question 289107: a 90% antifreeze solution is to be mixed with a 75% solution to make 30 liters of an 80% solution. How many liters of the 90% and 75% solutions should be used? \n" ); document.write( "

Algebra.Com's Answer #209478 by Earlsdon(6294) $\"\"$ $\"About$

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If you think in terms of how much antifreeze you are dealing with, I think that the problem becomes a little easier. For example, 10 liters of a 90% antifreeze solution contains (10*(0.9) = 9) liters of antifreeze, so...let x equal the required number of liters of the 90% antifreeeze solution.
\n" ); document.write( "Since you want to end up with 30 liters, you will also need (30-x) liters of the 75% antifreeze solution. In algebra this can be expressed as:
\n" ); document.write( "0.9x+0.75(30-x) = 0.8(30) Simplify and solve for x.
\n" ); document.write( "0.9x+22.5-0.75x = 24 Combine like-terms.
\n" ); document.write( "0.15x+22.5 = 24 Subtract 22.5 from both sides.
\n" ); document.write( "0.15x = 1.5 Finally, divide both sides by 0.15
\n" ); document.write( "x = 10
\n" ); document.write( "You will need to mix 10 liters of 90% antifreeze solution with (30-10 = 20) liters of 75% antifreeze solution to obtain 30 liters of 80% antifreeze solution.
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