(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 \( Q \).

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**.