document.write( "Question 241234: Could someone please help me with this mixture problem? I have been pouring over it for two days (total of over 4 hours), and have tried various methods none of them yielding the correct answer.
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document.write( "A mechanic has 363 gallons of gasoline and 17 gallons of oil to make gas/oil mixtures. He wants one mixture to be 6% oil and the other mixture to be 2.5% oil. If he wants to use all of the gas and oil, how many gallons of gas and oil are in each of the resulting mixtures? \n" );
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Algebra.Com's Answer #176699 by edjones(8007)![]() ![]() You can put this solution on YOUR website! Let x=gals of gas required to make a .025 mixture. \n" ); document.write( "Let y=gals of oil required to make a .025 mixture. \n" ); document.write( ". \n" ); document.write( "A) y/(x+y)=.025 \n" ); document.write( "B) (17-y)/(363-x+17-y)=.06 \n" ); document.write( ". \n" ); document.write( "A) \n" ); document.write( "y=.025x+.025y \n" ); document.write( ".975y=.025x \n" ); document.write( "y=.0256x \n" ); document.write( ". \n" ); document.write( "B) \n" ); document.write( "(17-.0256x)/(380-x-.0256x)=.06 \n" ); document.write( "17-.0256x=22.8-.0615x \n" ); document.write( ".0359x=5.8 \n" ); document.write( "x=161.56 gal \n" ); document.write( ". \n" ); document.write( "A) \n" ); document.write( "y/(161.56+y)=.025 \n" ); document.write( "y=4.039+.025y \n" ); document.write( ".975y=4.039 \n" ); document.write( "y=4.142 gal \n" ); document.write( "It wont be exact because of rounding errors. \n" ); document.write( ". \n" ); document.write( "Ed \n" ); document.write( " |