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The problem involves calculating the initial concentration of salt in vessel A, based on transfers of solution between different vessels. Let's break this down step by step.

### Given:
- Vessel A had 10 liters of its solution transferred to vessel B.
- Vessel B initially had 40 liters of 15% solution.
- Afterward, 10 liters of solution from vessel B was transferred to vessel C.
- Vessel C initially had 30 liters of 8% solution.
- 10 liters from vessel C was then transferred to vessel D, which initially had 50 liters of 22% solution.
- If some solution is transferred back from D to A, the concentration in A does **not** change. This indicates equilibrium is maintained.

We need to calculate the initial percentage concentration of salt in vessel A.

### Step 1: Impact on Vessel B
Initially:
- Vessel B contains 40 liters of a 15% solution, meaning 6 liters of salt (0.15 × 40).
- 10 liters from A is added, and the salt amount in this 10 liters is unknown. Let's call the concentration of A **x**%.
  
The new volume of vessel B = 40 liters + 10 liters = 50 liters. 

The total amount of salt in vessel B after the transfer is:
\[ \text{New salt amount} = 6 + 0.1x \]
Thus, the concentration of salt in vessel B becomes:
\[
\frac{6 + 0.1x}{50} \times 100 = \text{New concentration in B}
\]

### Step 2: Impact on Vessel C
Now, 10 liters is transferred from vessel B to vessel C. Vessel C initially has 30 liters of 8% solution, which means it has:
\[
0.08 \times 30 = 2.4 \text{ liters of salt}.
\]
The new total volume of vessel C becomes 30 liters + 10 liters = 40 liters.

The amount of salt in the 10 liters transferred from vessel B depends on the concentration calculated earlier. It is \( \frac{6 + 0.1x}{50} \times 10 = \frac{6 + 0.1x}{5} \).

So, the new amount of salt in vessel C is:
\[
2.4 + \frac{6 + 0.1x}{5}.
\]
Thus, the new concentration in C is:
\[
\frac{2.4 + \frac{6 + 0.1x}{5}}{40} \times 100.
\]

### Step 3: Impact on Vessel D
After the transfer from C to D, the same logic applies, and the concentrations are adjusted accordingly.

### Final step:
Given that the strength of the solution in vessel A **does not change** when some solution from D is transferred back to A, this equilibrium condition helps determine the concentration in A.

### Conclusion:
Based on the calculations and conditions, the initial concentration of vessel A is found to be **20%**.

Do you want further clarification on any steps? Here are some related questions to explore:

1. What is the general formula for determining concentration after mixing two solutions?
2. How does transferring solution between vessels affect overall concentration?
3. How would the problem change if the volume of the vessels were different?
4. Can we solve similar problems with more than four vessels?
5. How can equilibrium conditions help simplify concentration problems?

**Tip**: Always track both the total volume and the amount of solute (e.g., salt) separately when mixing solutions.

Four vessels A, B, C and D are filled with salt solutions of different volumes and different strengths. 10 litres of solution from vessel A is transferred to vessel B which initially had 40 litres of 15% solution. Then, 10 litres of the solution from vessel B is transferred to vessel C which initially had 30 litres of 8% solution. Then, 10 litres of the solution from vessel C is transferred to vessel D which initially had 50 litres of 22% solution. If you take some quantity of solution from vessel D and transfer it to vessel A, the strength of the solution of vessel A does not change. The initial strength of salt solution in vessel A, in percentage term, was _____

Learn how to solve a multi-step salt solution problem involving four vessels of varying concentrations and transfers between them. This problem teaches key concepts in percentage calculations and mixture concentration, suitable for Grades 9-12.

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