How much of a(n)70% orange juice drink must be mixed with 9 gallons of a 10% orange juice drink to obtain a mixture that is 50% orange juice?

Answer 1

Hi Ghost K

18 gallons of 70%

We are given the information below

9 gallons of orange drink at 10% to start with

Some orange drink at 70% to use let that be x

To make an orange drink mixture of 50%

Since you know the concentrations of each component, the amount you are starting with and where you want to get to

We already know that the 9 gallons we start with contains

9(.10) = 0.9 gallons of orange juice and what ever we add to increase this increases the size of the entire mixture therefore we can let

0.9 + 0.7x = increase of orange juice in the new mixture

9 + x = increase size of the mixture

an equation can be set up as follows

(0.9 + 0.7x)/(9 + x) = 0.5

Cross multiply

0.9 + 0.7x = 0.5(9 + x)

0.9 + 0.7x = 4.5 + 0.5x

Combine like terms

0.7x - 0.5x = 4.5 - 0.9

0.2x = 3.6

x = 3.6/0.2

x = 18

Checking

(0.9 + 0.7*18)/9 + 18

(0.9 + 12.6)/27= 13.5/27 = 0.5

I hope this helps.

Answer 2

.7x + .1(9) = (9+x).5

.7x - .5x = .45 -.9

.2x = .36

x= 36/2= 18 gallons 70% ,plus

9 gal 10%

yields 27 gal 50%

70-50=20

50-10=40

40/20=2

twice as much 70% needed as 10%

x/9=2

x=18 gallons of 70%

+9 gall 10%

yields 27 gall 50%