document.write( "Question 980892: A tank contain water and oil in the ratio 5 :3. If 16 gallen mixture is taken out from this mixture and 16 gallen oil is added to it then ratio of water and oil becomes 3 :5. How many gallen of mixture are there in the original mixture.
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Algebra.Com's Answer #601968 by ankor@dixie-net.com(22740)\"\" \"About 
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A tank contain water and oil in the ratio 5:3.
\n" ); document.write( " If gallons of the mixture is taken out from this mixture and 16 gallons of oil is added to it, then ratio of water and oil becomes 3:5.
\n" ); document.write( " How many gallons of mixture are there in the original mixture.
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\n" ); document.write( "let x = amt of the mixture
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\n" ); document.write( "a mixture of 5:3 would have a fractional amt of oil of \"3%2F8\"x
\n" ); document.write( "a mixture of 3:5 would have a fractional amt of oil of \"5%2F8\"x
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\n" ); document.write( "An equation based on the amt of oil
\n" ); document.write( "\"3%2F8\"(x-16) + 16 = \"5%2F8\"x
\n" ); document.write( "multiply by 8 to get rid of the denominators, resulting in:
\n" ); document.write( "3(x-16) + 8(16) = 5x
\n" ); document.write( "3x - 48 + 128 = 5x
\n" ); document.write( "3x + 80 = 5x
\n" ); document.write( "80 = 5x - 3x
\n" ); document.write( "80 = 2x
\n" ); document.write( "x = 80/2
\n" ); document.write( "x = 40 gallons in the mixture:
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\n" ); document.write( "Check
\n" ); document.write( "\"3%2F8\"(40-16) + 16 = \"5%2F8\"(40)
\n" ); document.write( "\"3%2F8\"(24) + 16 = \"5%2F8\"(40)
\n" ); document.write( "do the math
\n" ); document.write( "9 + 16 = 25 \n" ); document.write( "
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