The radiator in a certain make of car needs to contain 20 liters of 40% antifreeze. The radiator now contains 20 liters of 20% antifreeze. How many liters of this solution must be drained and replaced with 100% antifreeze to get the desired strength?
(4x-12)/(x²+4x+3)*(x²-1)/(x-3)
To solve this problem, we need to find out how many liters of the current solution must be drained and replaced with 100% antifreeze to achieve the desired strength of 40%.
Let's denote the volume of the solution to be drained as 'x'. Since x liters of the current solution will be drained, the remaining solution will be (20 - x) liters.
The amount of antifreeze in the original solution is given by:
20 liters * 20% = 4 liters antifreeze
When 'x' liters of the original solution are drained, the amount of antifreeze drained will be:
x liters * (20% antifreeze) = 0.2x liters antifreeze
When 'x' liters of 100% antifreeze are added, the amount of antifreeze added will be:
x liters * (100% antifreeze) = 1x liters antifreeze
After draining 'x' liters and adding 'x' liters of 100% antifreeze, the total amount of antifreeze in the radiator will be:
(4 liters antifreeze - 0.2x liters antifreeze) + 1x liters antifreeze
Now, we want this total amount of antifreeze to be 40% of the total volume (20 - x) liters, so we can set up the following equation:
((4 - 0.2x) + x) / (20 - x) = 40%
Simplifying this equation:
(4 - 0.2x + x) / (20 - x) = 40%
(4 + 0.8x) / (20 - x) = 40%
Now, we can cross-multiply and solve for 'x':
(4 + 0.8x) * 100 = 40 * (20 - x)
400 + 80x = 800 - 40x
120x = 400
x = 400 / 120
x = 3.33 (approximately)
Therefore, approximately 3.33 liters of the current solution must be drained and replaced with 100% antifreeze to achieve the desired strength of 40%.
To solve this problem, we need to use the concept of mixtures and apply a formula that represents the amount of an ingredient in a mixture.
Let's break down the problem step by step:
1. Determine the initial amount of antifreeze in the radiator:
The radiator contains 20 liters of a 20% antifreeze solution. To find the amount of antifreeze in this solution, we multiply the volume (20 liters) by the concentration as a decimal (20% = 0.20):
Initial amount of antifreeze = 20 liters * 0.20 = 4 liters
2. Determine the desired amount of antifreeze in the radiator:
The radiator needs a 40% antifreeze solution with a volume of 20 liters. To find the desired amount of antifreeze in this solution, we multiply the volume (20 liters) by the concentration as a decimal (40% = 0.40):
Desired amount of antifreeze = 20 liters * 0.40 = 8 liters
3. Set up the equation to represent the problem:
Let "x" represent the amount of the 20% antifreeze that needs to be drained and replaced with 100% antifreeze.
The amount of antifreeze drained would be x liters, and the amount of antifreeze added would also be x liters.
In the initial solution, the antifreeze content is:
4 liters (initial amount of antifreeze) - x liters (drained) + x liters (added) = 4 - x + x = 4 liters
In the desired solution, the antifreeze content is:
8 liters (desired amount of antifreeze)
4. Solve the equation for "x":
4 liters = 8 liters
4 - x + x = 8
4 = 8
x = 4 liters
The solution shows that you need to drain 4 liters of the 20% antifreeze solution and replace it with 4 liters of 100% antifreeze in order to obtain the desired 40% antifreeze concentration.
20 litres of 40% contains 20*0.4 = 8 litres of 100% antifreeze.
If we drain x litres and replace with 100% antifreeze, then
0.4(20-x)+1.0x = 8
solve for x.