how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 33 % iodine solution ?
solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 33 / 100 40 x + 700 = 33 x + 1155 7 x = 855 x = 122 lts correct option : e
add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 33), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35)