document.write( "Question 154401This question is from textbook intermediate algebra
\n" ); document.write( ": A mechanic has found that a car with a 16-quart radiator has a 40% antifreeze mixture in the radiator. The mechanic has 70% antifreeze solution on hand. How much of the 40% solution would the mechanic have to replace with the 70% solution to get the solution in the radiator up to 50%. \n" ); document.write( "

Algebra.Com's Answer #113691 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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mechanic has found that a car with a 16-quart radiator has a 40% antifreeze mixture in the radiator. The mechanic has 70% antifreeze solution on hand. How much of the 40% solution would the mechanic have to replace with the 70% solution to get the solution in the radiator up to 50%.
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\n" ); document.write( "Let x = amt of 40% to be removed, and the amt of 70% to be added
\n" ); document.write( ":
\n" ); document.write( ".40(16-x) + .70x = .50(16)
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\n" ); document.write( "6.5 - .4x + .7x = 8
\n" ); document.write( ":
\n" ); document.write( "-.4x + .7x = 8 - 6.4
\n" ); document.write( ":
\n" ); document.write( ".3x = 1.6
\n" ); document.write( "x = $\"1.6%2F.3\"$
\n" ); document.write( "x = 5 $\"1%2F3\"$ quarts of 40% removed
\n" ); document.write( "x = 5 $\"1%2F3\"$ quarts of 70% to be added
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "Check solution on a calc (16 - 5.33 = 10.67):
\n" ); document.write( ".4(10.67) + .7(5.33) = .5(16)
\n" ); document.write( " 4.268 + 3.731 = 7.999 ~ 8, confirms our solution \n" ); document.write( "

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