Problem Solving

How many gallons of water must be mixed with 1 gallon of a 15% salt solution to obtain a 10% salt solution?

A. 0.50
B. 0.67
C. 1.00
D. 1.50
E. 2.00

The OA is A.

Let the number of gallons of water to be added

(concentration of A)*(Volume of A)+(concentration of B)*(Volume of B) = (desired concentration in resultant mixture)*(Volume of resultant mixture = A+B)

Water has a salt concentration of 0

(0.15)*(1) + (0)*(x) = (0.1)*(1+x)

0.15 = 0.1 + 0.1x

0.05 = 0.1x

0.05/0.1 = x --> x = 0.5.

Has anyone another approach to solve this PS question? Regards!

AAPL wrote:How many gallons of water must be mixed with 1 gallon of a 15% salt solution to obtain a 10% salt solution?

A. 0.50
B. 0.67
C. 1.00
D. 1.50
E. 2.00

Amount of salt = 15% of 1 gallon = 0.15 gallon.
We can PLUG IN THE ANSWERS, which represent the amount of water that must be added to reduce the salt concentration to 10%.

B: 2/3 gallon
Salt concentration = (amount of salt)/(new total volume) = 0.15/(1 + 2/3) = 0.15/(5/3) = (0.15 * 3)/5 = (15*3)/500 = (3*3)/100 = 9/100 = 9%.
The salt concentration is TOO LOW.
To increase the salt concentration, we must add LESS WATER.

The correct answer is A.

A: 0.5 gallon
Salt concentration = (amount of salt)/(new total volume) = 0.15/(1 + 0.5) = 0.15/1.5= 15/150 = 1/10 = 10%.
Success!

AAPL wrote:How many gallons of water must be mixed with 1 gallon of a 15% salt solution to obtain a 10% salt solution?

A. 0.50
B. 0.67
C. 1.00
D. 1.50
E. 2.00

Salt percentage in the original solution: 15%.
Salt percentage in the added water: 0%.
Salt percentage in the mixture: 10%.

Let S = the original 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 (15% and 0%) on the ends and the goal percentage (10%) in the middle.
S 15%------------10%--------------0% W

Step 2: Calculate the distances between the percentages.
S 15%-----5-----10%-----10-----0% W

Step 3: Determine the ratio in the mixture.
The required ratio of original solution to added water is equal to the RECIPROCAL of the distances in red.
S:W = 10:5 = 1 : 0.5.

The ratio in blue indicates that the 1 gallon of original solution must be combined with 0.5 gallon of pure water.

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 solution, the amount of salt = 0.15(1) = 0.15 gallon.
After the water is added, the 0.15 gallon of salt must constitute 10% of the final mixture:
0.15= 0.1x
15 = 10x.
x = 15/10 = 1.5 gallons.
Since the volume of the final mixture is 1.5 gallons, and the volume of the original solution is 1 gallon, the volume of the added water = 1.5 - 1 = 0.5 gallon.