We've got 0.4 gallons of OJ and 0.6 gallons of water. When we add x gallons of OJ, we'll have (x + 0.4) gallons of OJ and 0.6 gallons of water.

The % of OJ at that point will be

OJ / Total =>

(x + 0.4) / (x + 0.4 + 0.6)

We want that to equal 60%:

(x + 0.4) / (x + 1) = .6

x + 0.4 = .6 * (x + 1)

x + 0.4 = 0.6x + 0.6

0.4x = 0.2

x = 0.5

ardz24 wrote:One gallon of soft drink is made of 40% orange juice and 60% water, how many additional gallons of orange juice must be mixed in to make the orange juice 60% of the soft drink?

A. 0.5
B. 1
C. 1.25
D. 1.5
E. 2

Juice percentage in the soft drink: 40%.
Juice percentage in the added orange juice: 100%.
Juice percentage in the mixture: 60%.

Let S = the soft drink and J = the added juice.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for the soft drink and the added juice (40% and 100%) on the ends and the percentage for the mixture (60%) in the middle.
S 40%------------60%-------------100% J

Step 2: Calculate the distances between the percentages.
S 40%----20-----60%----40-----100% J

Step 3: Determine the ratio in the mixture.
The required ratio of soft drink to added juice is equal to the RECIPROCAL of the distances in red.
S:J = 40:20 = 2:1 = 1 : (1/2).

Since S:J = 1 : (1/2), 1 gallon of soft drink must be combined with 1/2 gallon of orange juice.

The correct answer is A.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

An alternate approach:

In the original soft drink, the amount of water = (0.6)(1) = 0.6 gallon.
After orange juice is added, this 0.6 gallon of water must constitute 40% of the final mixture:
0.6 = 0.4x
x = (0.6)/(0.4) = 6/4 = 3/2.
Since the volume of the final mixture is 1.5 gallons, and the volume of the original soft drink is 1 gallon, the volume of the added juice = 1 .5 - 1 = 0.5 gallon.

ardz24 wrote:One gallon of soft drink is made of 40% orange juice and 60% water, how many additional gallons of orange juice must be mixed in to make the orange juice 60% of the soft drink?

A. 0.5
B. 1
C. 1.25
D. 1.5
E. 2

We are given that one gallon of soft drink contains 0.4 gallons of OJ and 0.6 gallons of water. To make OJ 60%, we can let n = the number of gallons of additional OJ and create the following equation:

(0.4 + n)/(1 + n) = 0.6

0.4 + n = 0.6 + 0.6n

0.4n = 0.2

n = 1/2

Answer: A