document.write( "Question 159806: One grade of oil has 0.50% of an additive, and a higher grade has 0.75% of the additive. How many liters of each must be used to have 1000 L of a mixture with 0.65% of the additive. \n" ); document.write( "
Algebra.Com's Answer #117866 by KnightOwlTutor(293)\"\" \"About 
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Convert % in decimal format\r
\n" ); document.write( "\n" ); document.write( ".5%=.5/100=0.005
\n" ); document.write( ".75%=.75/100=.0075
\n" ); document.write( ".65%=.65/100=.0065\r
\n" ); document.write( "\n" ); document.write( "X= the higher grade of oil that has .75% of additive\r
\n" ); document.write( "\n" ); document.write( "1000-X= volume of grade of oil that has .5% additive\r
\n" ); document.write( "\n" ); document.write( "The total volume on both sides of the equation =1000L\r
\n" ); document.write( "\n" ); document.write( ".0075X+.005(1000-X)=.0065(1000)\r
\n" ); document.write( "\n" ); document.write( "Use the distributive property\r
\n" ); document.write( "\n" ); document.write( ".0075X+5-.005X=6.5\r
\n" ); document.write( "\n" ); document.write( ".0025X+5=6.5\r
\n" ); document.write( "\n" ); document.write( "subtract 5 from both sides\r
\n" ); document.write( "\n" ); document.write( ".0025X=1.5\r
\n" ); document.write( "\n" ); document.write( "divide both sides by .0025\r
\n" ); document.write( "\n" ); document.write( "X=600L of .75% additive oil\r
\n" ); document.write( "\n" ); document.write( "1000-600=400L of .50% additive\r
\n" ); document.write( "\n" ); document.write( "Plug in to check answer\r
\n" ); document.write( "\n" ); document.write( "600(.0075) + 400(.005)=.0065(1000)\r
\n" ); document.write( "\n" ); document.write( "4.5 + 2=6.5\r
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