Help!! I do not understand how to work these types of problems. My instructor gives us really long formulas and charts for these and i don't get it!! Im good at algebra, I have an A in the class, so this is irritating to me!
The radiator in a certain make of car needs to contain 70 liters of 40% antifreeze. the radiator now contains 70 liters of 20% antifreeze. How many liters of this solvent must be drained and replaced with 00% antifreeze to get the desired strength?
what you have to do is track the amount of solute. If you add up all the smaller amounts, they must total the total amount.
Now, 5 gal of a 12% solution contains 5*.12 = .6 gal of solute. Thinking along those lines,
70L of 40% antifreeze contains 28L of antifreeze
70L of 20% antifreeze contains 14L of antifreeze
nL of 100% antifreeze contains nL of antifreeze
Now, if we drain nL of 20% solution, we end up with (70-n)L
(70-n)*.2 + n = 70*.4
14 - .2n + n = 28
.8n = 14
n=17.5
so, if we add 17.5L of 100% antifreeze to 52.5L of 20%, we end up with
52.5*.2 + 17.5 = 10.5+17.5 = 28L of antifreeze in 70L, or 40% antifreeze
To solve this problem, you'll need to set up an equation using the concept of mixtures. Don't worry, it's not as complicated as it may seem!
Let's break the problem down step by step:
Step 1: Understand the problem
We have a radiator that currently contains 70 liters of a 20% antifreeze mixture. We need to drain some of it and replace it with a 100% antifreeze solution to achieve a desired strength of 40% antifreeze.
Step 2: Set up the equation
Let's assign variables to the unknown quantities in this problem.
Let 'x' represent the number of liters that need to be drained and replaced with 100% antifreeze.
To set up the equation, we'll focus on the amount of antifreeze in the radiator before and after the draining/replacing process.
Amount of antifreeze before = Amount of antifreeze after
Step 3: Calculate the amount of antifreeze in the radiator before draining
The radiator initially contains 70 liters of a 20% antifreeze mixture.
To find the amount of antifreeze in the mixture, we multiply the total volume (70 liters) by the percentage of antifreeze (20% or 0.20):
Amount of antifreeze before = 70 liters * 0.20 = 14 liters
Step 4: Calculate the amount of antifreeze in the radiator after draining and replacing
After draining 'x' liters and replacing them with 100% antifreeze, we'll still have 70 liters of mixture in the radiator.
The amount of antifreeze in the mixture after draining and replacing can be calculated as follows:
Amount of antifreeze after = (70 - x) liters * 0.40 + x liters * 1.00
The first part of the equation represents the amount of antifreeze in the remaining mixture after draining, and the second part represents the amount of antifreeze in the replacement.
Step 5: Set up the equation
Now, set up the equation by equating the amount of antifreeze before and after the draining/replacing process:
14 liters = (70 - x) liters * 0.40 + x liters * 1.00
Step 6: Solve for 'x'
Simplify the equation by distributing the 0.40 and solve for 'x':
14 = 28 - 0.40x + x
Combine like terms:
0.60x = 14 - 28
0.60x = -14
Divide both sides by 0.60:
x = -14 / 0.60
x ≈ -23.33
Since negative liters doesn't make sense in this context, we can discard this solution.
Step 7: Interpret the result
Since we cannot have negative liters, we conclude that no liquid should be drained and replaced with 100% antifreeze. The radiator already contains the desired strength of 40% antifreeze.
Therefore, you don't need to do anything to the radiator to achieve the desired strength!
I hope this step-by-step explanation helps you understand how to approach and solve mixture problems like this one.