Ten liters of pure water are added to a 50-liter pot of soup that is [tex]$50\%$[/tex] broth. Fill in the missing parts of the table.

[tex]\[
\begin{array}{|l|c|c|c|}
\hline & \text{Original} & \text{Added} & \text{New} \\
\hline \text{Amount of broth} & 25 & 0 & c \\
\hline \text{Total solution} & 50 & 10 & d \\
\hline
\end{array}
\][/tex]



Answer :

To solve this problem, let's analyze each component step-by-step:

### Step 1: Understanding Initial Quantities
- Initial amount of broth: 25 liters
- Initial total solution: 50 liters

### Step 2: Added Components
- Added broth: 0 liters (since we are adding pure water)
- Added total solution: 10 liters of pure water

### Step 3: Calculating the New Quantities
- New amount of broth:
Since no broth is being added, the amount of broth remains the same:
[tex]\[ \text{New amount of broth} = 25 \text{ liters} \][/tex]
- New total solution:
The total solution increases by the amount of pure water added:
[tex]\[ \text{New total solution} = 50 \text{ liters (initial solution)} + 10 \text{ liters (added water)} = 60 \text{ liters} \][/tex]

### Step 4: Filling in the Table
- The new amount of broth (c) is: 25 liters
- The new total solution (d) is: 60 liters

So, the completed table looks like this:

[tex]\[ \begin{array}{|l|c|c|c|} \hline & \text{Original} & \text{Added} & \text{New} \\ \hline \begin{array}{l} \text{Amount} \\ \text{of broth} \end{array} & 25 & 0 & 25 \\ \hline \begin{array}{l} \text{Total} \\ \text{solution} \end{array} & 50 & 10 & 60 \\ \hline \end{array} \][/tex]