Mathematics intermediate

Question

Pure acid is to be added to a 10% acid solution to obtain 108 L of a 20% acid solution What amounts of each should be used?

Answer 1

100x +10(108-x) = 20(108)

where x = Liters of pure acide

108-x = Liters of 10%

100x -10x = 20(108) -10(108)

90x = 10(108)

9x = 108

x = 108/9

x = 12 Liters of pure acid and

108-x =108-12 = 96 of 10% acid

mixed to get 108 Liters of 20%

100-20 = 80

20-10 = 10

80/10 = 8 times as much 10% needed as pure acid

96/12 = 8,

extra check on the answers:

12(100) + 96(10)

=1200+960

= 2160

= 20(108)

= 2160