Charise made 8text{ liters} of a 40% solution of Chemical X. If she began with 2text{ liters} of a 20%...

1. Translate the problem requirements: Charise starts with 2L of 20% solution, adds p liters of 10% solution and q liters of 50% solution to create 8L total of 40% solution. We need to find q.
2. Set up the volume constraint: Since we know the total volume and some components, establish the relationship between p and q.
3. Apply the mixture principle: Track the actual amount of pure Chemical X from each source and set it equal to the pure chemical in the final mixture.
4. Solve the system efficiently: Use substitution to find q directly from the chemical amount equation.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what's happening in plain English first. Charise is like a chemist mixing different solutions together. She starts with one solution, adds two more different solutions, and ends up with a final mixture.

Here's what we know:

• She starts with 2 liters of a 20% Chemical X solution
• She adds p liters of a 10% Chemical X solution
• She adds q liters of a 50% Chemical X solution
• Her final mixture is 8 liters total of a 40% Chemical X solution
• We need to find the value of q

Process Skill: TRANSLATE - Converting the mixture problem into mathematical relationships

2. Set up the volume constraint

Since liquids don't disappear when we mix them, the total volume must add up. Think of it like pouring different colored liquids into a container - the final volume equals the sum of what you poured in.

Starting volume + Added volume 1 + Added volume 2 = Final volume
\(2 + p + q = 8\)

Therefore: \(p + q = 6\)

This means \(p = 6 - q\). We can use this relationship to eliminate one variable.

3. Apply the mixture principle

Now here's the key insight: when you mix solutions, the amount of pure chemical from each source must equal the pure chemical in the final mixture.

Let's think about this step by step:

• From 2 liters of 20% solution: \(2 \times 0.20 = 0.4\) liters of pure Chemical X
• From p liters of 10% solution: \(p \times 0.10 = 0.1p\) liters of pure Chemical X
• From q liters of 50% solution: \(q \times 0.50 = 0.5q\) liters of pure Chemical X
• In the final 8 liters of 40% solution: \(8 \times 0.40 = 3.2\) liters of pure Chemical X

Setting up our equation:
Pure chemical from source 1 + Pure chemical from source 2 + Pure chemical from source 3 = Pure chemical in final mixture

\(0.4 + 0.1p + 0.5q = 3.2\)

Process Skill: APPLY CONSTRAINTS - Using the principle that chemical amounts must balance

4. Solve the system efficiently

Now we substitute \(p = 6 - q\) from our volume constraint into our chemical equation:

\(0.4 + 0.1(6 - q) + 0.5q = 3.2\)

Let's solve this step by step:
\(0.4 + 0.6 - 0.1q + 0.5q = 3.2\)
\(1.0 + 0.4q = 3.2\)
\(0.4q = 2.2\)
\(q = 2.2 ÷ 0.4 = 5.5\)

Let's verify: if \(q = 5.5\), then \(p = 6 - 5.5 = 0.5\)
Check volume: \(2 + 0.5 + 5.5 = 8\) ✓
Check chemical: \(0.4 + 0.1(0.5) + 0.5(5.5) = 0.4 + 0.05 + 2.75 = 3.2\) ✓

Final Answer

The value of q is 5.5 liters.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding percentage concentration

Confusing 40% by volume with other interpretations leads to wrong pure chemical calculations.

2. Omitting the chemical balance constraint

Using only the volume equation yields one equation with two unknowns.

3. Misreading the final mixture requirement

Leads to incorrect target values.

Errors while executing the approach

1. Decimal arithmetic mistakes

Errors in calculating pure chemical amounts compound through the solution.

2. Substitution errors

Distributing 0.1(6 - q) incorrectly changes coefficients.

3. Sign errors combining like terms

Leads to wrong q coefficient.

Errors while selecting the answer

No likely errors; q=5.5 matches an answer choice directly.