a radiator holds 5 gallons. how much of the 30% antifreeze solution should be drained and replaced with pure water to reduce the solution to 9% antifreeze?
If x gallons are drained, then 30% of what remains is antifreeze. So, since water is 0% antifreeze,
.3(5-x) + 0x = .09*5
0.3
To solve this problem, we need to calculate the amount of the 30% antifreeze solution that needs to be drained and replaced with pure water.
Let's start by defining the variables:
Let x be the amount of the 30% antifreeze solution to be drained (in gallons).
Step 1: Calculate the amount of antifreeze in the radiator before draining.
The initial amount of antifreeze in the radiator is given by 30% of 5 gallons:
Antifreeze before draining = 0.30 * 5 = 1.5 gallons
Step 2: Calculate the amount of antifreeze remaining after draining.
The amount of antifreeze remaining after draining x gallons of the 30% antifreeze solution is:
Antifreeze after draining = 1.5 - (0.30x)
Step 3: Determine the amount of water needed to be added.
The total amount of liquid in the radiator after draining can be expressed as 5 - x gallons. Since we want the solution to be 9% antifreeze, the amount of antifreeze after adding water should be 9% of the total amount in the radiator:
Antifreeze after adding water = 0.09 * (5 - x)
Step 4: Set up the equation.
The amount of antifreeze after draining (step 2) should be equal to the amount of antifreeze after adding water (step 3):
1.5 - (0.30x) = 0.09 * (5 - x)
Step 5: Solve the equation for x.
Let's solve the equation to find the value of x:
1.5 - 0.30x = 0.45 - 0.09x
0.30x - 0.09x = 0.45 - 1.5
0.21x = 1.05
x = 1.05 / 0.21
x ≈ 5
Step 6: Answer the question.
Approximately 5 gallons of the 30% antifreeze solution should be drained and replaced with pure water to reduce the solution to 9% antifreeze.
To solve this problem, let's break it down step by step.
Step 1: Find out how much antifreeze is currently in the radiator.
- Since the radiator holds 5 gallons and the concentration of the antifreeze is 30%, the amount of antifreeze in the solution can be found by multiplying 5 gallons by 0.30 (or 30% as a decimal).
- 5 gallons * 0.30 = 1.5 gallons of antifreeze.
Step 2: Determine the target amount of antifreeze in the new solution.
- We want to reduce the concentration of the antifreeze to 9%, so we need to calculate the desired amount of antifreeze in the new solution.
- Let's call the amount of the solution to be drained and replaced with water as "x" gallons.
- The amount of antifreeze in the new solution after draining and replacing will be 1.5 gallons minus the amount drained and replaced with water.
- Since there will still be 5 gallons of solution in the radiator, the amount of antifreeze in the new solution will be (1.5 - x) gallons.
Step 3: Set up the equation and solve it.
- We need to set up an equation using the information we have so far. The equation will represent the ratio of antifreeze to the total solution.
- The equation can be written as: (1.5 - x) / 5 = 0.09.
- This equation states that the antifreeze in the new solution divided by the total solution is equal to 0.09 (or 9% as a decimal).
- Now we can solve the equation for x, which represents the amount of solution to be drained and replaced.
Let's solve the equation:
(1.5 - x) / 5 = 0.09
Multiply both sides by 5 to get rid of the denominator:
1.5 - x = 0.09 * 5
1.5 - x = 0.45
Subtract 1.5 from both sides to isolate x:
-x = 0.45 - 1.5
-x = -1.05
Multiply both sides by -1 to solve for x:
x = -1.05 * -1
x = 1.05
Therefore, you should drain and replace approximately 1.05 gallons of the 30% antifreeze solution with pure water to reduce the solution to 9% antifreeze.