Solved on Jan 20, 2024

Determine the amount of $100\%$ antifreeze needed to replace the $20\%$ solution in a radiator to achieve a $40\%$ antifreeze solution.

STEP 1

Assumptions
1. The radiator currently contains 70 liters of a 20% antifreeze solution.
2. We want to achieve a 40% antifreeze solution in the radiator.
3. We will drain a certain amount of the 20% solution and replace it with 100% antifreeze to reach the desired concentration.
4. The total volume of the solution in the radiator remains 70 liters after the process.

STEP 2

Let's denote the amount of solution to be drained and replaced with 100% antifreeze as $x$ liters.

STEP 3

The amount of antifreeze in the original 20% solution is $0.20 \times 70$ liters.
$Antifreeze_{original} = 0.20 \times 70$

STEP 4

Calculate the amount of antifreeze in the original solution.
$Antifreeze_{original} = 0.20 \times 70 = 14\ \text{liters}$

STEP 5

When $x$ liters of the 20% solution are drained, we lose $0.20 \times x$ liters of antifreeze.

STEP 6

After draining $x$ liters of the 20% solution, the remaining antifreeze in the radiator is $14 - 0.20 \times x$ liters.

STEP 7

We then add $x$ liters of 100% antifreeze to the radiator.

STEP 8

The total amount of antifreeze in the radiator after adding $x$ liters of 100% antifreeze is $(14 - 0.20 \times x) + x$ liters.

STEP 9

We want the final solution to be 40% antifreeze in a 70-liter solution.
$Antifreeze_{final} = 0.40 \times 70$

STEP 10

Calculate the amount of antifreeze needed in the final solution.
$Antifreeze_{final} = 0.40 \times 70 = 28\ \text{liters}$

STEP 11

Set up the equation to find $x$ by equating the antifreeze in the final solution to the total amount of antifreeze after the replacement.
$(14 - 0.20 \times x) + x = 28$

STEP 12

Simplify the equation by combining like terms.
$14 - 0.20x + x = 28$
$14 + 0.80x = 28$

STEP 13

Subtract 14 from both sides of the equation to isolate the terms with $x$ .
$0.80x = 28 - 14$
$0.80x = 14$

STEP 14

Divide both sides of the equation by 0.80 to solve for $x$ .
$x = \frac{14}{0.80}$

STEP 15

Calculate the value of $x$ .
$x = \frac{14}{0.80} = 17.5$

STEP 16

The amount of the 20% antifreeze solution that needs to be drained and replaced with 100% antifreeze to obtain a 40% solution in the radiator is 17.5 liters.
The correct answer is $17.5 \mathrm{~L}$ .

Was this helpful?

Determine the amount of $100\%$ antifreeze needed to replace the $20\%$ solution in a radiator to achieve a $40\%$ antifreeze solution.

STEP 1

STEP 2

Let's denote the amount of solution to be drained and replaced with 100% antifreeze as $x$ liters.

STEP 3

The amount of antifreeze in the original 20% solution is $0.20 \times 70$ liters.
$Antifreeze_{original} = 0.20 \times 70$

STEP 4

Calculate the amount of antifreeze in the original solution.
$Antifreeze_{original} = 0.20 \times 70 = 14\ \text{liters}$

STEP 5

When $x$ liters of the 20% solution are drained, we lose $0.20 \times x$ liters of antifreeze.

STEP 6

After draining $x$ liters of the 20% solution, the remaining antifreeze in the radiator is $14 - 0.20 \times x$ liters.

STEP 7

We then add $x$ liters of 100% antifreeze to the radiator.

STEP 8

The total amount of antifreeze in the radiator after adding $x$ liters of 100% antifreeze is $(14 - 0.20 \times x) + x$ liters.

STEP 9

We want the final solution to be 40% antifreeze in a 70-liter solution.
$Antifreeze_{final} = 0.40 \times 70$

STEP 10

Calculate the amount of antifreeze needed in the final solution.
$Antifreeze_{final} = 0.40 \times 70 = 28\ \text{liters}$

STEP 11

Set up the equation to find $x$ by equating the antifreeze in the final solution to the total amount of antifreeze after the replacement.
$(14 - 0.20 \times x) + x = 28$

STEP 12

Simplify the equation by combining like terms.
$14 - 0.20x + x = 28$
$14 + 0.80x = 28$

STEP 13

Subtract 14 from both sides of the equation to isolate the terms with $x$ .
$0.80x = 28 - 14$
$0.80x = 14$

STEP 14

Divide both sides of the equation by 0.80 to solve for $x$ .
$x = \frac{14}{0.80}$

STEP 15

Calculate the value of $x$ .
$x = \frac{14}{0.80} = 17.5$

STEP 16

The amount of the 20% antifreeze solution that needs to be drained and replaced with 100% antifreeze to obtain a 40% solution in the radiator is 17.5 liters.
The correct answer is $17.5 \mathrm{~L}$ .

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Determine the amount of 100%100\%100% antifreeze needed to replace the 20%20\%20% solution in a radiator to achieve a 40%40\%40% antifreeze solution.

STEP 1

STEP 2

Let's denote the amount of solution to be drained and replaced with 100% antifreeze as xxx liters.

STEP 3

The amount of antifreeze in the original 20% solution is 0.20×700.20 \times 700.20×70 liters.Antifreezeoriginal=0.20×70Antifreeze_{original} = 0.20 \times 70Antifreezeoriginal​=0.20×70

STEP 4

Calculate the amount of antifreeze in the original solution.Antifreezeoriginal=0.20×70=14 litersAntifreeze_{original} = 0.20 \times 70 = 14\ \text{liters}Antifreezeoriginal​=0.20×70=14 liters

STEP 5

When xxx liters of the 20% solution are drained, we lose 0.20×x0.20 \times x0.20×x liters of antifreeze.

STEP 6

After draining xxx liters of the 20% solution, the remaining antifreeze in the radiator is 14−0.20×x14 - 0.20 \times x14−0.20×x liters.

STEP 7

We then add xxx liters of 100% antifreeze to the radiator.

STEP 8

The total amount of antifreeze in the radiator after adding xxx liters of 100% antifreeze is (14−0.20×x)+x(14 - 0.20 \times x) + x(14−0.20×x)+x liters.

STEP 9

We want the final solution to be 40% antifreeze in a 70-liter solution.Antifreezefinal=0.40×70Antifreeze_{final} = 0.40 \times 70Antifreezefinal​=0.40×70

STEP 10

Calculate the amount of antifreeze needed in the final solution.Antifreezefinal=0.40×70=28 litersAntifreeze_{final} = 0.40 \times 70 = 28\ \text{liters}Antifreezefinal​=0.40×70=28 liters

STEP 11

Set up the equation to find xxx by equating the antifreeze in the final solution to the total amount of antifreeze after the replacement.(14−0.20×x)+x=28(14 - 0.20 \times x) + x = 28(14−0.20×x)+x=28

STEP 12

Simplify the equation by combining like terms.14−0.20x+x=2814 - 0.20x + x = 2814−0.20x+x=2814+0.80x=2814 + 0.80x = 2814+0.80x=28

STEP 13

Subtract 14 from both sides of the equation to isolate the terms with xxx.0.80x=28−140.80x = 28 - 140.80x=28−140.80x=140.80x = 140.80x=14

STEP 14

Divide both sides of the equation by 0.80 to solve for xxx.x=140.80x = \frac{14}{0.80}x=0.8014​

STEP 15

Calculate the value of xxx.x=140.80=17.5x = \frac{14}{0.80} = 17.5x=0.8014​=17.5

STEP 16

The amount of the 20% antifreeze solution that needs to be drained and replaced with 100% antifreeze to obtain a 40% solution in the radiator is 17.5 liters.The correct answer is 17.5 L17.5 \mathrm{~L}17.5 L.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Determine the amount of $100\%$ antifreeze needed to replace the $20\%$ solution in a radiator to achieve a $40\%$ antifreeze solution.

Let's denote the amount of solution to be drained and replaced with 100% antifreeze as $x$ liters.

The amount of antifreeze in the original 20% solution is $0.20 \times 70$ liters.
$Antifreeze_{original} = 0.20 \times 70$

Calculate the amount of antifreeze in the original solution.
$Antifreeze_{original} = 0.20 \times 70 = 14\ \text{liters}$

When $x$ liters of the 20% solution are drained, we lose $0.20 \times x$ liters of antifreeze.

After draining $x$ liters of the 20% solution, the remaining antifreeze in the radiator is $14 - 0.20 \times x$ liters.

We then add $x$ liters of 100% antifreeze to the radiator.

The total amount of antifreeze in the radiator after adding $x$ liters of 100% antifreeze is $(14 - 0.20 \times x) + x$ liters.

We want the final solution to be 40% antifreeze in a 70-liter solution.
$Antifreeze_{final} = 0.40 \times 70$

Calculate the amount of antifreeze needed in the final solution.
$Antifreeze_{final} = 0.40 \times 70 = 28\ \text{liters}$

Set up the equation to find $x$ by equating the antifreeze in the final solution to the total amount of antifreeze after the replacement.
$(14 - 0.20 \times x) + x = 28$

Simplify the equation by combining like terms.
$14 - 0.20x + x = 28$
$14 + 0.80x = 28$

Subtract 14 from both sides of the equation to isolate the terms with $x$ .
$0.80x = 28 - 14$
$0.80x = 14$

Divide both sides of the equation by 0.80 to solve for $x$ .
$x = \frac{14}{0.80}$

Calculate the value of $x$ .
$x = \frac{14}{0.80} = 17.5$

The amount of the 20% antifreeze solution that needs to be drained and replaced with 100% antifreeze to obtain a 40% solution in the radiator is 17.5 liters.
The correct answer is $17.5 \mathrm{~L}$ .