Let's solve this using the method of mixture problems with concentrations.

Let:

• $x$ = amount (in ml) of the 14% solution
• $y$ = amount (in ml) of the 44% solution

We know:

1. The total volume of the mixture is 72 ml: $x + y = 72$
2. The resulting concentration is 25% Him, which means: $0.14x + 0.44y = 0.25 \times 72$

Now, we can simplify and solve these equations:

Step 1: Solve the first equation for $y$

$y = 72 - x$

Step 2: Substitute $y = 72 - x$ into the second equation

$0.14x + 0.44(72 - x) = 18$

Expanding and simplifying: $0.14x + 31.68 - 0.44x = 18$ $-0.3x = -13.68$ $x = \frac{-13.68}{-0.3} = 45.6$

Step 3: Substitute $x = 45.6$ back into the first equation to find $y$

$y = 72 - 45.6 = 26.4$

Solution

The amounts required are: $(x, y) = (45.6, 26.4)$

Therefore, you need 45.6 ml of the 14% solution and 26.4 ml of the 44% solution.

Would you like further details on any steps?

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1. How do we set up equations for mixture problems?
2. What is the significance of balancing the concentration in mixture problems?
3. Can this problem be solved using a different method?
4. How does changing the target concentration affect the amounts of each solution needed?
5. What happens if the target concentration is outside the range of given solutions?

Tip: Always double-check the units and concentrations when setting up equations in mixture problems to avoid calculation errors.