document.write( "Question 1162590: Tobacco mixture #706 is 10¢ per ounce while mixture #716 is twice that price. How many ounces of the more expensive mixture must be used with 48 ounces of the less expensive mixture to make a new mixture that will sell for 76¢ per 5-ounce tin? \n" ); document.write( "
Algebra.Com's Answer #786419 by ankor@dixie-net.com(22740)\"\" \"About 
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Tobacco mixture #706 is 10¢ per ounce while mixture #716 is twice that price.
\n" ); document.write( " How many ounces of the more expensive mixture must be used with 48 ounces of the less expensive mixture to make a new mixture that will sell for 76¢ per 5-ounce tin?
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\n" ); document.write( "let x = amt of 20 cent mixture
\n" ); document.write( "then
\n" ); document.write( "(x+48) = total amt
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\n" ); document.write( "10(48) + 20x = \"76%2F5\"(x + 48)
\n" ); document.write( "480 + 20x = 15.2(x + 48)
\n" ); document.write( "480 + 20x = 15.2x + 729.6
\n" ); document.write( "20x - 15.2x = 729.6 - 480
\n" ); document.write( "4.8x = 249.6
\n" ); document.write( "x = 52 oz of the 20 cent mixture required
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