How does "How many gallons of a 12% indicator solution must be mixed with a 20% indicator solution to get 10 gal of a 14% solution?" turn into ".12x + .2(10+ -1x) = .14(10)"?? Please, I need to understand how it works! Thank you.
Let X be the number of 12%-solution gallons added. 10-X will be the number of 20%-solution gallons.
0.14*10 = 0.20(10 -X) + 0.12 X
That is also the equation you wrote, and it says that the amount of solute in the mix is the sum of the amounts in the two mixed solutions.
Just solve it for X.
1.4 = 2 - 0.2 X + 0.12 X
0.6 = 0.08 X
X = 7.5 gallons
To understand how the equation .12x + .2(10 - x) = .14(10) is derived from the given problem, we need to break down the problem into simpler steps.
Step 1: Let's assume that x represents the number of gallons of the 12% indicator solution that we need to mix. Since the total volume required is 10 gallons, the quantity of the 20% indicator solution will be (10 - x) gallons.
Step 2: The next step is to calculate the amount of indicator in the 12% solution. The 12% solution contains 12% indicator per gallon. So, the number of indicator units in x gallons of the 12% solution is 0.12x.
Step 3: Now, let's determine the amount of indicator in the 20% solution. The 20% solution contains 20% indicator per gallon. So, the number of indicator units in (10 - x) gallons of the 20% solution is 0.2(10 - x).
Step 4: To find the total amount of indicator in the mixed solution, we add the indicator units in the 12% and 20% solutions. Thus, the total indicator units in the mixed solution is 0.12x + 0.2(10 - x).
Step 5: According to the problem, the resulting solution should be a 14% indicator solution. Therefore, the total indicator units in the mixed solution should be 14% of the total volume (10 gallons). So, we have 0.14(10) indicator units.
Step 6: Finally, we can set up the equation based on step 4 and step 5:
0.12x + 0.2(10 - x) = 0.14(10)
This equation represents the relationship between the indicator units in the 12% solution, the indicator units in the 20% solution, and the total indicator units in the resulting 14% solution.
To solve the problem of mixing different concentrations of solution, we need to set up a mathematical equation based on the given information. Let's break down the steps involved in converting the given problem into the equation ".12x + .2(10 - x) = .14(10)".
Step 1: Assign variables
Let's assign variables to the unknown quantities in the problem:
- x: represents the number of gallons of the 12% indicator solution to be mixed
- (10 - x): represents the number of gallons of the 20% indicator solution to be mixed (since we need a total of 10 gallons, and we already have x gallons of the 12% solution)
Step 2: Determine the equation for the concentration of the solution
The concentration of a solution is given by the ratio of the amount of indicator (in gallons) to the total volume of the solution (in gallons). In this problem, we want to get a 10-gallon solution with a concentration of 14%.
Since we have two different solutions being mixed, the equation for the concentration of the resulting solution is:
(.12x + .2(10 - x))/10 = .14
Step 3: Simplify the equation
To simplify the equation, we need to distribute and multiply the terms inside the parentheses:
(.12x + 2 - .2x)/10 = .14
Combining like terms, we get:
(.12x - .2x + 2)/10 = .14
Simplifying further, we obtain:
(.12x - .2x)/10 + 2/10 = .14
Combine the terms:
-.08x/10 + 2/10 = .14
Simplify:
-.008x + .2 = .14
Step 4: Multiply both sides of the equation by 10
To eliminate the fraction, we can multiply both sides of the equation by 10:
10(-.008x + .2) = 10(.14)
This gives us:
-.08x + 2 = 1.4
Step 5: Move constants to one side and the variable term to the other side
Now, let's isolate the variable term by moving the constant term to the other side of the equation:
-.08x = 1.4 - 2
Simplifying:
-.08x = -0.6
Step 6: Divide both sides of the equation by -0.08
To solve for x, divide both sides of the equation by (-0.08):
x = -0.6 / (-0.08)
Simplifying:
x = 7.5
Thus, to obtain a 14% solution with a total volume of 10 gallons, you need to mix 7.5 gallons of the 12% indicator solution with 2.5 gallons of the 20% indicator solution.