J.R. S. answered 09/05/18
Tutor
5.0
(145)
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
(x mls)(100%) + (150 mls - x mls)(13%) = (150 mls)(42%)
100 x + 1950 - 13 x = 6300
87x = 4350
x = 50 mls of 100%
150 - 50 = 100 mls of 13%
Done by alligation:
Have.......Want.......Need......
100%.......................29........
................42%.....................
13%.........................58.......
Total parts...............87.......
87parts = 150mls,
1 part = 1.724 mls
1.72 x 29 = 50 mls of 100%
1.72 x 58 = 100 mls of 13%