. A big container has a mixture of Aand B , in ratio 3 : 2 . The total quantity of the mixture was increased 1.6 times by adding B. Next , 41.6 litres of the solution in the Container was replaced by B , which changed the ratio of A to B to 3 : 7 , then find the initial volume of mixture ( in litres )

Let the initial volume of the mixture be \( x \) liters. Since the mixture of A and B is in the ratio 3:2, the quantities of A and B initially are:

\[
A = \frac{3}{5}x \quad \text{and} \quad B = \frac{2}{5}x
\]

Now, the total quantity of the mixture was increased by 1.6 times by adding B. This means the new total quantity is \( 1.6x \), and the amount of B added is:

\[
1.6x - x = 0.6x
\]

So the new quantity of A remains \( \frac{3}{5}x \), while the new quantity of B becomes:

\[
\frac{2}{5}x + 0.6x = \frac{2}{5}x + \frac{3}{5}x = x
\]

After this, 41.6 liters of the solution is replaced by B, which changes the ratio of A to B to 3:7. Let’s denote the amount of A remaining after removing 41.6 liters as \( A' \) and the amount of B as \( B' \).

Since 41.6 liters of the solution was removed and replaced by B, the volume of A removed would be proportional to the amount of A in the solution. Before removal, the total volume of the mixture is still \( 1.6x \), and the fraction of A in this mixture is:

\[
\frac{\frac{3}{5}x}{1.6x} = \frac{3}{8}
\]

So, the amount of A removed is:

\[
\frac{3}{8} \times 41.6 = 15.6 \, \text{liters}
\]

Thus, the remaining quantity of A is:

\[
A' = \frac{3}{5}x - 15.6
\]

Similarly, the amount of B removed is:

\[
41.6 - 15.6 = 26 \, \text{liters}
\]

So, the remaining quantity of B after removing 41.6 liters is:

\[
B' = x - 26 + 41.6 = x + 15.6
\]

Now, according to the new ratio of A to B (3:7), we have the equation:

\[
\frac{A'}{B'} = \frac{3}{7}
\]

Substituting the values of \( A' \) and \( B' \):

\[
\frac{\frac{3}{5}x - 15.6}{x + 15.6} = \frac{3}{7}
\]

Cross-multiplying gives:

\[
7 \left( \frac{3}{5}x - 15.6 \right) = 3 \left( x + 15.6 \right)
\]

Simplifying:

\[
7 \times \frac{3}{5}x - 7 \times 15.6 = 3x + 3 \times 15.6
\]
\[
\frac{21}{5}x - 109.2 = 3x + 46.8
\]

Multiplying through by 5 to eliminate the fraction:

\[
21x - 546 = 15x + 234
\]

Bringing like terms together:

\[
21x - 15x = 234 + 546
\]
\[
6x = 780
\]

Solving for \( x \):

\[
x = \frac{780}{6} = 130
\]

Thus, the initial volume of the mixture is \( \boxed{130} \) liters.

Would you like further details or clarifications on any part of this solution?

Here are 5 related questions to explore:
1. What would happen if the initial ratio of A to B was different, say 2:3?
2. How would the final ratio of A to B change if we replaced a different amount of the mixture with B?
3. If only part of the removed mixture was replaced with B, how would that affect the final ratio?
4. Can this method be applied to mixtures with more than two components? How?
5. How would the solution change if we added A instead of B?

**Tip:** When solving ratio problems, always ensure that you carefully track the total quantities and changes step by step to avoid confusion.

A big container has a mixture of A and B, in a ratio of 3:2. The total quantity of the mixture was increased 1.6 times by adding B. Next, 41.6 liters of the solution in the container was replaced by B, which changed the ratio of A to B to 3:7. Find the initial volume of the mixture (in liters).

This math problem involves a container with a mixture of A and B in a ratio of 3:2, and how adding more B changes the mixture's volume and ratio. Learn how to calculate the initial volume and understand key algebraic and proportion concepts. Perfect for students in Grades 9-10.

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