Question

A solution containing 20% salt is to be mixed with

a solution of 60% salt to make 200L of a solution
that is 50% salt. How much of each solution is to
be used?
*mixture problems

Answer 1

50 liters of 20% solution and 150 liters of 60% solution should be used.

Explanation:

Given that a solution containing 20% salt is to be mixed with a solution of 60% salt to make 200L of a solution that is 50% salt, to determine how much of each solution is to be used the following calculation must be performed:

1 x 0.2 + 0 x 0.6 = 0.2

0.5 x 0.2 + 0.5 x 0.6 = 0.4

0.3 x 0.2 + 0.7 x 0.6 = 0.48

0.2 x 0.2 + 0.8 x 0.6 = 0.52

0.25 x 0.2 + 0.75 x 0.6 = 0.5

200 x 0.25 = 50

200 x 0.75 = 150

Thus, 50 liters of 20% solution and 150 liters of 60% solution should be used.