how many gallons of a 90% antifreeze solution must be mixed with 100 gallons of 30% antifreeze to get a mixture that is 80% antifreeze?

450

To solve this problem, we can use the concept of mixing two solutions with different concentrations to obtain a desired concentration. Let's break it down step by step:

Step 1: Set up the equation
Let x be the number of gallons of the 90% antifreeze solution.
The concentration of antifreeze in the 90% solution is 90% (or 0.9).
The concentration of antifreeze in the 30% solution is 30% (or 0.3).
We want to find the total volume of the mixture that is 80% antifreeze.

Step 2: Express the equation using the information given
The amount of antifreeze in the 90% solution can be expressed as 0.9x.
The amount of antifreeze in the 30% solution can be expressed as 0.3 * 100 (since there are 100 gallons of the 30% solution).

Step 3: Set up the equation based on the desired concentration
We want the final mixture to have a concentration of 80% antifreeze (or 0.8).
The amount of antifreeze in the final mixture can be expressed as 0.8 * (x + 100) (since the total volume of the mixture will be the sum of the 90% solution and the 30% solution).

Step 4: Solve the equation
Now we can set up the equation and solve for x:
0.9x + 0.3 * 100 = 0.8 * (x + 100)

Step 5: Solve the equation for x
0.9x + 30 = 0.8x + 80
0.9x - 0.8x = 80 - 30
0.1x = 50
x = 50 / 0.1
x = 500

Step 6: Answer
To obtain a mixture that is 80% antifreeze, 500 gallons of the 90% antifreeze solution must be mixed with 100 gallons of the 30% antifreeze solution.

To find out how many gallons of a 90% antifreeze solution must be mixed with 100 gallons of 30% antifreeze to obtain a mixture that is 80% antifreeze, we can use the concept of the concentration of a solution.

Let's assume that x gallons of the 90% antifreeze solution are needed to be mixed with the 100 gallons of 30% antifreeze solution.

The total volume after mixing will be 100 gallons + x gallons = (100 + x) gallons.

To calculate the concentration of antifreeze in the new mixture, we can look at the amount of pure antifreeze in both the 30% and 90% solutions.

In the 100 gallons of 30% antifreeze, we have 0.30 * 100 = 30 gallons of pure antifreeze.

In the x gallons of 90% antifreeze, we have 0.90 * x = 0.9x gallons of pure antifreeze.

In the final mixture (100 + x) gallons at 80% antifreeze, we have 0.80 * (100 + x) gallons of pure antifreeze.

Since the amount of pure antifreeze is the same in both the initial and final mixtures, we can equate them:

30 + 0.9x = 0.80 * (100 + x)

Let's solve this equation to find the value of x:

30 + 0.9x = 80 + 0.80x

0.9x - 0.80x = 80 - 30

0.1x = 50

x = 50 / 0.1

x = 500

Therefore, you need to mix 500 gallons of the 90% antifreeze solution with 100 gallons of the 30% antifreeze solution in order to achieve a mixture that is 80% antifreeze.

let the amount of 90% antifreeze to be added be x gallons

.9x + .3(100) = .8(100+x)
times 10
9x + 3(100) = 8(100+x)
9x+300 = 800+8x
x = 500

500 gallons of the 90% must be added.