Question 256527
This is a mixture problem. Here is the table based on the given information:
ACID . . . . . . . .% . . . . . . . . . .ML . . . . . . . .%ML
60% . . . . . . . . .60 . . . . . . . . . m . . . . . . . .60m
40%  . . . . . . . . .40 . . . . . . . .200-m . . . .. 8000-40m
MIX . . . . . . . .. . 48 . . . . . . .. 200 . . . . . . .9600
In the 3rd column add the first 2 to get the last number. We get
60m + 8000 - 40m = 9600
20m + 8000 = 9600
20m = 1600
m = 80
200-m = 120.
She needs 120ML of 40% acid.