Mixture and Solution Problems

The video tutorial explains how a chemist can mix three different acid solutions to create a 100-liter mixture with a 45% acid concentration. The chemist must use twice as much of the 10% solution as the 40% solution due to limited supply. The tutorial sets up a system of equations to solve for the amounts of each solution needed, using substitution and back substitution methods. The final solution involves 25 liters of 10% solution, 12.5 liters of 40% solution, and 62.5 liters of 60% solution.

What is the desired concentration of the final acid mixture?

• 60%

• 10%

• 40%

• 45%

Why must twice as much 10% solution be used compared to the 40% solution?

• The 40% solution is in limited supply.

• The 40% solution is more expensive.

• The 10% solution is cheaper.

• The 10% solution is more effective.

What variable is used to represent the amount of 60% solution?

• Y

• Z

• W

• X

Which equation represents the total volume of the mixture?

• Y + Z = 100

• X = 2Y

• 0.1X + 0.4Y + 0.6Z = 45

• X + Y + Z = 100

What method is used to solve the system of equations?

• Graphical method

• Matrix method

• Trial and error

• Substitution and elimination

What is the value of Y, the amount of 40% solution used?

• 10 liters

• 12.5 liters

• 25 liters

• 62.5 liters

How many liters of the 10% solution are used?

• 10 liters

• 25 liters

• 62.5 liters

• 12.5 liters

What is the total amount of 60% solution used in the mixture?

• 62.5 liters

• 25 liters

• 12.5 liters

• 10 liters

Which equation helps to find the value of Z?

• Z = 100 - 3Y

• X = 2Y

• 0.6Y + 0.6Z = 45

• X + Y + Z = 100

What is the final step in solving the problem?

• Writing a report

• Calculating the cost

• Verifying the solution

• Mixing the solutions