percentages of orange juice

Question

How much of​ a(n) 60% orange juice drink must be mixed with 14 gallons of a 40%

orange juice drink to obtain a mixture that is 50% orange​ juice?

Answer 1

These percentages are telling us the concentration of orange juice in each drink, i.e., the amount of orange juice per unit volume. Essentially we are measuring the density of orange juice in the liquid. Since this is the ratio of orange juice to volume, if we multiply the volume by the concentration we get the total number of orange juice in the solution.

Let C₁= 60% = .6 be the concentration of the first orange juice drink

Let C₂= 40% = .4 be the concentration of the second orange juice drink

Let C₃= 50% = .5 be the concentration of the final orange juice drink

Let V₁, V₂, and V₃ be the volumes of the first second and final orange drinks

We are given that V₂ = 14 Gallons and we know that V₃ = V₁ + V₂ = V₁ +14

We know that the total amount of orange juice the drinks is the same when we mix the first and second drinks into the final drink, hence we can set up an equation as follows

C₁V₁+C₂V₂=C₃V₃

The concentration times the volume equals the amount of orange juice in the drink, which allows us to set up this equation.

Now if we plug in V₂ = 14 and V₃ = 14 + V₁ we get

C₁V₁+14C₂=C₃(14+V₁)

Now we can solve for V₁

C₁V₁+14C₂=14C₃+C₃V₁ By the distributive property

C₁V₁-C₃V₁+14C₂=14C₃

C₁V₁-C₃V₁=14C₃-14C₂

V₁(C₁-C₃)=14(C₃-C₂) By the distributive property

V₁=14(C₃-C₂)/(C₁-C₃)

Now if we plug in our values for C₁, C₂, and C₃ into our equation above we can get the final answer, V₁ = 14 Gallons. Hence, we need 14 Gallons of the 60% orange juice drink to be combined with 14 gallons of the 40% orange drink to get a final mixture that is 50% orange juice.

Please let me know if you have any further questions :)

Answer 2

Let x be the number of gallons to add

0.6x+0.4(14) = 0.5(x+14)

0.6x+5.6 = 0.5x+7

0.1x=7-5.6=1.4

x = 14 gallons