Tough Question on hw

Question

A chemist has three different acid solutions. The first acid solution contains 20%20% acid, the second contains 30%30% and the third contains 75%75%. They want to use all three solutions to obtain a mixture of 8484 liters containing 40%40% acid, using 22 times as much of the 75%75% solution as the 30%30% solution. How many liters of each solution should be used?

Answer 1

I need to set up a system of equations based on the given constraints.

Let me define variables:

1. Let x = liters of 20% solution
2. Let y = liters of 30% solution
3. Let z = liters of 75% solution

Setting up the equations:

Equation 1 (Total volume): x + y + z = 84

Equation 2 (Relationship between solutions): "Using 2 times as much of the 75% solution as the 30% solution" z = 2y

Equation 3 (Acid content): The final mixture should contain 40% acid 0.20x + 0.30y + 0.75z = 0.40(84) 0.20x + 0.30y + 0.75z = 33.6

Solving the system:

From Equation 2: z = 2y

Substitute into Equation 1: x + y + 2y = 84 x + 3y = 84 x = 84 - 3y

Substitute both into Equation 3: 0.20(84 - 3y) + 0.30y + 0.75(2y) = 33.6 16.8 - 0.6y + 0.30y + 1.5y = 33.6 16.8 + 1.2y = 33.6 1.2y = 16.8 y = 14

Now find x and z:

1. z = 2y = 2(14) = 28 liters
2. x = 84 - 3y = 84 - 3(14) = 84 - 42 = 42 liters

Answer:

1. 42 liters of 20% acid solution
2. 14 liters of 30% acid solution
3. 28 liters of 75% acid solution

Verification:

1. Total: 42 + 14 + 28 = 84 ✓
2. z = 2y: 28 = 2(14) ✓
3. Acid content: 0.20(42) + 0.30(14) + 0.75(28) = 8.4 + 4.2 + 21 = 33.6 = 0.40(84) ✓