How much pure antifreeze must be added to 12 L of a 40% solution to obtain a 60% solution?

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Answer 1 · 2017-06-29T10:14:13

The answer is 6 L.

First we need to determine the amount of antifreeze in the 12 L solution.

#0.4 = x/12# #rArr x = 12*0.4 = 4.8 L#

Then using the formula below:

#0.6# = #(4.8 + x)/(12+x)#

#0.6 (12+x)# = #(4.8 + x)#

#7.2 + 0.6 x# = #4.8 + x#

#7.2 - 4.8# = #x - 0.6 x#

#2.4 = 0.4 x#

#x = 6 L#