document.write( "Question 99077: Joseph mixed a certain amount of 10% isopropyl mixture with 60 gallons of a 20% isopropyl solution, and the resulting mixture is 16% isopropyl. How much of the 10% isopropyl did he add? \n" ); document.write( "

Algebra.Com's Answer #72092 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of 10% mixture that he used
\n" ); document.write( "(1) Now the amount of pure isopropyl in this mixture is 0.10x\r
\n" ); document.write( "\n" ); document.write( "(2) Amount of pure isopropyl in the 20% mixture is 0.20(60)\r
\n" ); document.write( "\n" ); document.write( "(3) Amount of pure isopropyl in the final mixture is 0.16(60+x)\r
\n" ); document.write( "\n" ); document.write( "Now we know that 1+2=3. So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.10x+0.20*60=0.16(60+x) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "0.10x+12=9.6+0.16x subtract 9.6 and also 0.10x from both sides\r
\n" ); document.write( "\n" ); document.write( "0.10x-0.10x+12-9.6=9.6-9.6+0.16x-0.10x collect like terms\r
\n" ); document.write( "\n" ); document.write( "2.4=0.06x divide both sides by 0.06\r
\n" ); document.write( "\n" ); document.write( "x=40 gal--------------------amount of 10% solution used\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "0.10*40+0.20*60=0.16*100
\n" ); document.write( "4+12=16\r
\n" ); document.write( "\n" ); document.write( "16=16\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

\n" );