A chemist has one solution that is 20% salt and a second solution that is 45% salt. How many liters of each should be used in order to get 20 L of a solution that is 30% salt

8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt

Explanation:

Let x represent the liters of the first salt solution

Let y represent the liters of the second salt solution

x + y= 20equation 1

x= 20-y

The first solution is 20% salt and the second solution is 45% salt

20/100x + 45/100 y= 30/100 × 20

0.2x + 0.45y= 0.3×20

0.2x + 0.45y= 6equation 2

Substitute 20-y for x in equation 2

0.2(20-y) + 0.45y= 6

4-0.2y+ 0.45y= 6

4 + 0.25y= 6

0.25y= 6-4

0.25y= 2

y= 2/0.25

y= 8

Substitute y for 8 in equation 1

x + y= 20

x + 8=20

x= 20-8

x= 12

Hence 8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt.