Word problem:

Radiator in a certain make needs to contain 50 liters of 40% antifreeze. The radiator now contains 50 liters of 20% antifreeze. How many liters of this solution must be drained and replaced with 100% antifreeze to get the desired strength?

You want to double the concentration?

If you remove X liters, you remove .2X antifreeze, and .8X water.

You have left in the radiator after draining X, 10-.2X liters of pure antifreeze. You need a total of 20 liters to get 40percent antifreeze.

So you add 10+.2X of antifreeze.

You drain out 10-2X of solution.

X=10+.2X
.8X=10
x=12.5 liters

Let X be the number of liters that must be drained and replaced with 100% antifreeze.

[0.2(50-X)+X]/50 = 0.4

Solve that for X.

To solve this problem, we can follow these steps:

Step 1: Find the amount of antifreeze in the current solution.
The radiator initially contains 50 liters of a 20% antifreeze solution. The amount of antifreeze in this solution can be calculated by multiplying the volume of the solution by the concentration of antifreeze as a decimal.
Amount of antifreeze = 50 liters * 0.20 = 10 liters of antifreeze

Step 2: Determine the desired concentration of antifreeze.
The radiator needs to contain a 40% antifreeze solution, which means the final concentration of antifreeze needs to be 0.40 as a decimal.

Step 3: Let's assume x liters need to be drained and replaced with 100% antifreeze.
After x liters are drained, the amount of antifreeze in the remaining solution is (10 - 0.20x) liters. The total volume of the solution remains constant at 50 liters.

Step 4: Calculate the amount of pure antifreeze added.
If we add x liters of 100% antifreeze, the amount of pure antifreeze added will be x liters.

Step 5: Set up an equation based on the concentration of antifreeze in the final solution.
The concentration of antifreeze in the final solution can be calculated by dividing the amount of antifreeze by the total volume of the solution.
Final concentration of antifreeze = (10 - 0.20x + x) / 50 = 0.40

Step 6: Solve the equation for x.
(10 - 0.20x + x) / 50 = 0.40
10 - 0.20x + x = 0.40 * 50
10 + 0.80x = 20
0.80x = 20 - 10
0.80x = 10
x = 10 / 0.80
x = 12.5

Therefore, 12.5 liters of the 20% antifreeze solution need to be drained and replaced with 100% antifreeze to achieve the desired concentration of 40%.

To solve this word problem, we need to find out how many liters of the 20% antifreeze solution need to be drained and replaced with 100% antifreeze solution to achieve the desired concentration of 40%.

Let's break down the problem into steps:

Step 1: Determine the amount of antifreeze in the initial solution
The radiator initially contains 50 liters of a 20% antifreeze solution. The term "20% antifreeze" means that 20% of the solution is antifreeze. Therefore, we can calculate the amount of antifreeze in the initial solution by multiplying the volume by the percentage.

Amount of antifreeze = 50 liters × 20% = 50 liters × (20/100) = 10 liters

So, the initial solution contains 10 liters of antifreeze.

Step 2: Determine the amount of antifreeze needed in the final solution
The desired concentration is 40% antifreeze, and the total volume remains at 50 liters. We need to calculate the amount of antifreeze required for the final solution.

Amount of antifreeze needed = 50 liters × 40% = 50 liters × (40/100) = 20 liters

So, the final solution should contain 20 liters of antifreeze.

Step 3: Determine how much of the initial solution needs to be replaced
We need to find out how much of the initial 20% antifreeze solution needs to be drained and replaced with 100% antifreeze solution to reach the desired quantity of antifreeze (20 liters).

Let's assume we need to drain x liters of the initial solution and replace it with 100% antifreeze solution. The remaining solution will be (50 - x) liters.

The amount of antifreeze in the drained solution would be x liters × 20% = x liters × (20/100) = (1/5)x liters.

The amount of antifreeze in the added pure antifreeze solution would be x liters × 100% = x liters.

Now, let's calculate the amount of antifreeze in the final solution:

Amount of antifreeze in the final solution = Antifreeze from drained solution + Antifreeze from added solution
= (1/5)x + x
= (1/5 + 5/5)x
= (6/5)x

Step 4: Set up an equation to solve for x
Since the final solution should contain 20 liters of antifreeze, we can set up an equation:

(6/5)x = 20

Step 5: Solve for x
To solve the equation, multiply both sides by (5/6):

x = 20 × (5/6)
x = 100/6
x ≈ 16.67 liters (rounded to two decimal places)

Therefore, approximately 16.67 liters of the initial 20% antifreeze solution must be drained and replaced with pure antifreeze solution to achieve the desired strength of 40% antifreeze.