One gallon of soft drink is made of 40% orange juice and 60%

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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

OA: A

What's wrong with options B and D?

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by Matt@VeritasPrep » Thu Sep 28, 2017 4:28 pm
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

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by GMATGuruNY » Thu Sep 28, 2017 5:33 pm
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.
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by Jeff@TargetTestPrep » Tue Oct 03, 2017 3:13 pm
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

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by Brent@GMATPrepNow » Tue Oct 03, 2017 3:33 pm
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 can also just apply some logic.

IMPORTANT CONCEPT: when we combine EQUAL amounts of two mixtures, the concentration of the resulting mixture will be the AVERAGE of the two mixtures.
For example, if we combine 1 liter of a 10% salt solution with 1 liter of a 40% salt solution, the RESULTING solution will be 25% salt, since (10+40)/2 = 25.

So, If we take 1 gallon of soft drink with 40% orange juice and add an EQUAL amount (1 gallon) of 100% orange juice, then the resulting mixture will be 70% orange juice.
We want the resulting mixture to be 60% orange juice. So, we need to add LESS THAN 1 gallon of 100% orange juice

Check the answer choices . . . only one option (A) is less than 1 gallon.

Answer: A

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