Math Mixture Problem!?

Question

One canned juice is 30% orange juice; another is 10% orange juice. How many liters of each should be mixed together in order to obtain a 20L of 19% orange juice?

Answer 1

Let
x = amount of 30% orange juice
20 - x = amount of 10% orange juice

x(.3) + (20 - x)(.1) = 20(.19)
.3x + 2 - .1x = 3.8
.2x = 1.8
x = 1.8/.2 = 9

20 - x = 20 - 9 = 11

The mixture should be 9 liters of 30% orange juice and 11 liters of 10% orange juice.

Answer 2

x=30 % orange juice needed
20-x=10% orange juice needed

0.3x+0.1(20-x)=20*0.19
0.3x+2-0.1x=3.8
0.2x=1.8
x=9

20-x=10% orange juice needed
20-9
11

9 L of 30% orange juice are needed and 11 L of 10% orange juice are needed.

Answer 3

Let the liters of 30% juice be x
Therefore, the liters of 10% juice is (20-x)
30% of x and 10% of (20-x) must give 19% of 20
.3x+.1(20-x)=.19(20)
.3x+2-.1x=3.8
.2x=3.8-2
.2x=1.8
2x=18
x=9
9 liters of 30% juice, 11 liters of 10% juice.