document.write( "Question 838111: it is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 70 liters of 20% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength? \n" ); document.write( "

Algebra.Com's Answer #504875 by mananth(16946) $\"\"$ $\"About$

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Percent ---------------- quantity
\n" ); document.write( "Antifreeze 100% 100.00% ---------------- x liters
\n" ); document.write( "Antifreeze 20% 20.00% ------ 20 - x liters
\n" ); document.write( "Mixture 70.00% ---------------- 20
\n" ); document.write( "Total 20 liters
\n" ); document.write( "100.00% x + 20.00% ( 20 - x ) = 70.00% * 20
\n" ); document.write( "100 x + 20 ( 20 - x ) = 1400
\n" ); document.write( "100 x + 400 - 20 x = 1400
\n" ); document.write( "100 x - 20 x = 1400 - -400
\n" ); document.write( "80 x = 1000
\n" ); document.write( "/ 80
\n" ); document.write( " x = 12.5 liters 100.00% Antifreeze 100%
\n" ); document.write( " 7.5 liters 20.00% Antifreeze 20%
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca \r
\n" ); document.write( "\n" ); document.write( "The quantity you need to increase the percentage to 70 is the amount to be drained
\n" ); document.write( "
\n" ); document.write( "12.5 liters has to be drained \n" ); document.write( "

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