Mixure problem(Quite confusing ..!

[rkiran9]

A substance is shipped in the concentrated form of 75 percent substance per gallon of water.To be usable the solution has to be diluted with water to achieve 15 percent concentration.how many gallons of water does one need to add to the gallons of solution to obtain the usefull concentration?

options:
4.00,4.67,5.00,5.50,7.00

[faraz_jeddah]

A substance is shipped in the concentrated form of 75 percent substance per gallon of water.To be usable the solution has to be diluted with water to achieve 15 percent concentration.how many gallons of water does one need to add to the gallons of solution to obtain the usefull concentration?

options:
4.00,4.67,5.00,5.50,7.00

[spoiler]Is the answer C?[/spoiler]

[Brent@GMATPrepNow]

A substance is shipped in the concentrated form of 75 percent substance per gallon of water.To be usable the solution has to be diluted with water to achieve 15 percent concentration.how many gallons of water does one need to add to the gallons of solution to obtain the usefull concentration?

options:
4.00,4.67,5.00,5.50,7.00

There's not enough information to answer this question. We need to know how many gallons of solutions we are beginning with.
For example, if we have 1 million gallons of solution, we'll need to add millions of gallons of water. Conversely, if we have 1 drop of solution, we'll need to add a very tiny amount of water.

Cheers,
Brent

[Brent@GMATPrepNow]

A substance is shipped in the concentrated form of 75 percent substance per gallon of water.To be usable the solution has to be diluted with water to achieve 15 percent concentration.how many gallons of water does one need to add to one gallon of solution to obtain the useful concentration?

options:
A) 4.00
B) 4.67
C) 5.00
D) 5.50
E) 7.00

NOTE: Given the answer choices, I believe the original question meant to begin with one gallon of solution. So, I have edited the question accordingly.

I find that it often helps to solve mixture problems by sketching each solution with the parts separated. This makes it easy to see what happens when you combine solutions.

Let's let x = the # of gallons of water we need to add to the 1 gallon of solution. We get:


At this point, the resulting mixture contains 0.75 gallons of substance, and the total volume is (1 + x) gallons. Since this new mixture has a 15% concentration, we can write:
(0.75)/(1 + x) = 15%
In other words: (0.75)/(1 + x) = 15/100
Or . . . (0.75)/(1 + x) = 3/20
To solve for x, first cross multiply to get: 3(1 + x) = (20)(0.75)
3 + 3x = 15
3x = 12
x = 4

Answer = A

Cheers,
Brent

[snigdha1605]

Original amount of substance = 3/4 *1 gallon = 3/4 gallons

Given,

15/100 = Amount of Substance / 1 gallon + Amount of additional gallons
15/100 = ( 3/4 )/(1+x)

x= 4

Additional gallons added = 4

Regards,
Snigdha

------------------------------------------
Hit the Thank Button if you like my post!

[GMATGuruNY]

A substance is shipped in the concentrated form of 75 percent substance per gallon of water. To be usable the solution has to be diluted with water to achieve 15 percent concentration. How many gallons of water must be added to one gallon of the shipped solution to obtain the useful concentration?

4.00,4.67,5.00,5.50,7.00

The words in red above reflect the intent of the problem.

Substance percentage in the shipped solution: 75%.
Substance percentage in the added water: 0%.
Substance percentage in the mixture: 15%.

Let S = the shipped solution and W = the added water.
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 two starting percentages (75% and 0%) on the ends and the goal percentage (15%) in the middle.
S 75%------------15%-------------0% W

Step 2: Calculate the distances between the percentages.
S 75%----60-----15%----15-----0% W

Step 3: Determine the ratio in the mixture.
The required rate of shipped solution to added water is the RECIPROCAL of the distances in red.
S:W = 15:60 = 1:4.

Since S:W = 1:4, to every gallon of the shipped solution, 4 gallons of water must be added.

The correct answer is A.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html
https://www.beatthegmat.com/mixture-prob ... 18206.html

An alternate approach:

In the original solution, the amount of substance = .75(1) = .75 gallons.
After the water is added, these .75 gallons of substance must constitute 15% of the final mixture:
.75 = .15x
x = 5.
Since the volume of the final mixture is 5 gallons, and the volume of the shipped solution is 1 gallon, the volume of the added water = 5-1 = 4 gallons.