A radiator contains 25 quarts of a water and anti-freeze, of which 60% is antifreeze. How much if this solution should be drained and replaced with water for the new solution to be 40% antifreeze?
Consider just the amount of anti-freeze present. If x quarts of 60% solution are drained and replaced with water (0% solution), then you have
0.60(25-x) + 0.00(x) = 0.40(25)
Well, draining a radiator sounds like a real "coolant" idea. Let's calculate the amount of solution that needs to be drained, shall we?
Currently, the radiator contains 60% antifreeze, which means it has 40% water. If we drain x quarts of the solution, the remaining 25 - x quarts will have the same concentration.
After draining, we have (25 - x) quarts of solution with 40% antifreeze and 60% water. To get the desired 40% antifreeze concentration, we need an equal ratio of water and antifreeze.
So, we want to have 40% antifreeze concentration in (25 - x) quarts of solution. That means we'll have 40% water concentration as well. This implies that (25 - x) quarts will also contain 40% water.
Now, we set up the equation: 40% of (25 - x) quarts = 60% of x quarts.
0.4(25 - x) = 0.6x
Now we can solve for x:
10 - 0.4x = 0.6x
10 = 1x
x = 10
So, you need to drain and replace 10 quarts of the solution with water for the new solution to be 40% antifreeze. Hope that helps - and remember, don't let it go over your head, or you may find yourself in a "cooling" situation!
To solve this problem, we need to determine how much of the solution should be drained and replaced with water in order to achieve a 40% antifreeze concentration.
Let's denote the amount to be drained and replaced with water as 'x'.
Initially, there are 25 quarts of solution in the radiator, and 60% of it is antifreeze. This means that there are 0.60 * 25 = 15 quarts of antifreeze in the radiator initially.
After draining 'x' quarts of the solution, the amount of antifreeze remaining in the radiator will be 15 - 0.60 * x.
The total volume of the diluted solution after 'x' quarts are drained and replaced with water will still be 25 quarts, as we are replacing only a portion of the original solution.
Therefore, the amount of antifreeze in the diluted solution will be 0.40 * 25 = 10 quarts.
Setting up the equation: 15 - 0.60 * x = 10
Simplifying the equation: -0.60 * x = 10 - 15
-0.60 * x = -5
Dividing both sides by -0.60: x = -5 / -0.60
x = 8.33
We cannot drain a fraction of a quart, so we need to round up to the nearest whole number.
Therefore, approximately 9 quarts of the solution should be drained and replaced with water to achieve a 40% antifreeze concentration.
To solve this problem, let's break it down step by step:
Step 1: Determine the amount of anti-freeze present in the original solution
Given that there are 25 quarts of a water and anti-freeze solution in the radiator, and 60% of this solution is antifreeze, we can calculate the amount of antifreeze present:
Amount of antifreeze = 25 quarts * 60% = 15 quarts
Step 2: Determine the desired amount of antifreeze in the new solution
Since we want the new solution to contain 40% antifreeze, we need to determine the amount of antifreeze required in this new solution. We want to keep the same volume, so the amount of antifreeze needed will be:
Amount of antifreeze needed = 25 quarts * 40% = 10 quarts
Step 3: Determine the amount of solution that needs to be drained
To find the amount of solution that needs to be drained (let's call it x quarts), we subtract the amount of antifreeze needed from the amount of antifreeze present in the original solution:
Amount of solution drained = 15 quarts - 10 quarts = 5 quarts
Therefore, 5 quarts of the original solution should be drained.