To solve this problem, we can set up a system of equations based on the information given.

Let:

• $x$ = the amount of the 12% salt solution (in litres)
• $y$ = the amount of the 2% salt solution (in litres)

From the problem, we know two things:

1. The total volume of the mixture is 6 litres: $x + y = 6$

2. The total amount of salt in the mixture must equal the amount of salt in a 6% solution of 6 litres:

   • Salt in the 12% solution: $0.12x$
   • Salt in the 2% solution: $0.02y$
   • Salt in the 6% solution: $0.06 \times 6 = 0.36$

   Thus, we have: $0.12x + 0.02y = 0.36$

Now we can solve this system of equations.

Step 1: Solve for $y$ in the first equation

From $x + y = 6$: $y = 6 - x$

Step 2: Substitute $y$ in the second equation

Substituting $y$ into the second equation: $0.12x + 0.02(6 - x) = 0.36$ Expanding this: $0.12x + 0.12 - 0.02x = 0.36$ Combining like terms: $0.10x + 0.12 = 0.36$ Subtracting 0.12 from both sides: $0.10x = 0.24$ Dividing by 0.10: $x = 2.4$

Step 3: Find $y$

Substituting $x$ back into the equation for $y$: $y = 6 - 2.4 = 3.6$

Final Answer

The technician must mix:

• 2.4 litres of the 12% solution
• 3.6 litres of the 2% solution

If you need further details or have questions, let me know!

Related Questions

1. How would the solution change if the target concentration were different?
2. What would happen if more of the 12% solution were available?
3. How can this method be applied to other mixtures?
4. What are other ways to calculate concentrations in solutions?
5. How does the volume of the solutions affect the concentration?

Tip: Always check your equations for consistency with the physical constraints of the problem.