(c) Container $$ P $$ has 360 g of flour $$ \frac{3}{7} $$ of the flour in container $$ Q $$ is poured into container $$ P $$. Then $$ \frac{4}{9} $$ of flour in container $$ P $$ is poured into the empty container, $$ R $$. Now, container $$ R $$ has 400 g of flour. Calculate, in $$ g $$, the mass of the original flour in container $$ Q $$.

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Let's break down the problem step by step to find the mass of the original flour in container $$ Q $$.

1. **Initial Amounts:**
   - Container $$ P $$ initially has $$ 360 $$ g of flour.
   - Let $$ Q $$ be the original amount of flour in container $$ Q $$.

2. **Transferring Flour from $$ Q $$ to $$ P $$:**
   - $$ \frac{3}{7} $$ of the flour in $$ Q $$ is poured into $$ P $$.
   - The amount poured from $$ Q $$ to $$ P $$ is $$ \frac{3}{7}Q $$.
   - After this transfer, the total flour in $$ P $$ becomes:
     $$
     360 \text{ g} + \frac{3}{7}Q
     $$

3. **Transferring Flour from $$ P $$ to $$ R $$:**
   - $$ \frac{4}{9} $$ of the flour in $$ P $$ is poured into the empty container $$ R $$.
   - The amount poured from $$ P $$ to $$ R $$ is:
     $$
     \frac{4}{9} \left( 360 + \frac{3}{7}Q \right)
     $$
   - We are given that container $$ R $$ ends up with $$ 400 $$ g of flour:
     $$
     \frac{4}{9} \left( 360 + \frac{3}{7}Q \right) = 400
     $$

4. **Solving for $$ Q $$:**
   - Multiply both sides of the equation by $$ \frac{9}{4} $$ to isolate the expression inside the parentheses:
     $$
     360 + \frac{3}{7}Q = 400 \times \frac{9}{4} = 900
     $$
   - Subtract $$ 360 $$ from both sides:
     $$
     \frac{3}{7}Q = 900 - 360 = 540
     $$
   - Multiply both sides by $$ \frac{7}{3} $$ to solve for $$ Q $$:
     $$
     Q = 540 \times \frac{7}{3} = 1260 \text{ g}
     $$

**Answer:**  
The original mass of flour in container $$ Q $$ was **1260 g**.
The original mass of flour in container $$ Q $$ was **1260 g**.

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