A 9 percent salt-water solution is mixed with 4 ounces of an 18 percent salt-water solution in order to obtain a 15 percent salt-water solution. How much of the first solution should be used?
1 ounce
2 ounces
3 ounces
4 ounces
Explain your answer.
let the amount to be added be x ounces
.09x + .18(4) = .15(x+4)
I suggest multiplying by 100, then it is easy to solve for x
To solve this problem, we can use the concept of the weighted average. The key idea is that the amount of salt in the final solution should be equal to the sum of the amounts of salt in the two initial solutions.
Let's denote the amount of the 9 percent salt-water solution we need to use as "x". Therefore, the amount of salt in this solution is 0.09x.
The amount of the 18 percent salt-water solution that we need to use is given as 4 ounces. Therefore, the amount of salt in this solution is 0.18 * 4 = 0.72 ounces.
When these two solutions are mixed, the total amount of salt in the final mixture is 0.09x + 0.72.
According to the problem, we want the final mixture to be a 15 percent salt-water solution. This means that the amount of salt in the mixture should be 0.15 times the total volume of the mixture. The total volume of the mixture is obtained by adding the amounts of the two initial solutions: x + 4.
Therefore, we can set up the equation: 0.09x + 0.72 = 0.15(x + 4).
Now, we can solve this equation to find the value of x.
0.09x + 0.72 = 0.15x + 0.6
0.15x - 0.09x = 0.72 - 0.6
0.06x = 0.12
x = 0.12 / 0.06
x = 2
So, we need to use 2 ounces of the 9 percent salt-water solution in order to obtain a 15 percent salt-water solution. Therefore, the correct answer is 2 ounces.