document.write( "Question 988862: Our dining hall needs to have 140 liters of a 30% chocolate milk. They have a 25% chocolate solution and a 60% chocolate solution. How many liters of each should they use to obtain the mix? \n" ); document.write( "
Algebra.Com's Answer #609392 by josgarithmetic(39628) ![]() You can put this solution on YOUR website! The description and question are not adequate. \"Solution\" must mean \"syrup\" or an ingredient material containing chocolate other than the milk. You have both 25% and 60% chocolate stock liquid of some kind, and using the 60% would be more efficient.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "m, how much MILK \n" ); document.write( "c, how much CHOCOLATE SOLUTION-or-SYRUP \n" ); document.write( "Wanted: 140 LITERS of 30% chocolate milk \n" ); document.write( "Instead of m for how much milk, use 140-c, because the dining hall should mix 140-c liters of milk with c liters of the 60% chocolate.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Seven liters of the 60% chocolate \"solution\" or syrup, and 133 liters of milk.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |