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This problem involves mixing solutions of different concentrations. Let's break it down and find the percentage of the solution left in the second bottle.

### Problem Breakdown:
- Bottle 1 has a 37% salt solution.
- Bottle 2 has a 12% salt solution.
- Some portion of the solution in Bottle 1 is replaced with an equal portion from Bottle 2.
- After mixing, the concentration in Bottle 1 becomes 17%.

We need to calculate the concentration that remains in Bottle 2 after some of its solution is moved to Bottle 1.

### Solution Approach:

Let:
- \( V \) be the total volume of each bottle (since both bottles are identical).
- \( xV \) be the portion of Bottle 1's solution that is replaced, where \( 0 < x < 1 \) (fraction of volume replaced).

Initially:
- The mass of salt in Bottle 1 is \( 0.37V \) (since the concentration is 37%).
- The mass of salt in Bottle 2 is \( 0.12V \).

After replacing \( xV \) of Bottle 1's solution with Bottle 2's solution:
- The new mass of salt in Bottle 1 is the remaining salt after removing \( x \) portion of it, plus the salt introduced from Bottle 2.
  
The equation for the salt mass balance in Bottle 1 is:

\[
(1-x)(0.37V) + x(0.12V) = 0.17V
\]

Solving for \( x \), we can determine the fraction of volume replaced. Once we have \( x \), we can determine the remaining concentration in Bottle 2.

Let me calculate this step by step.The fraction of the solution removed from Bottle 1 and replaced by Bottle 2's solution is \( x = 0.8 \). This means 80% of Bottle 1's solution was replaced by the solution from Bottle 2.

Now, since 80% of the solution was removed from Bottle 2 and transferred to Bottle 1, we need to determine the concentration left in Bottle 2 after this process. Initially, the concentration in Bottle 2 was 12%, and removing 80% of it will affect the remaining concentration.

Let's calculate the final concentration in Bottle 2.The concentration of the solution left in Bottle 2, after 80% of its solution was removed, is 2.4%.

Would you like more details or have any further questions?

Here are 5 related questions you can explore:
1. How does changing the volume fraction \( x \) affect the final concentration in Bottle 1?
2. What happens if Bottle 2 had a higher concentration than Bottle 1 initially?
3. How would you calculate the concentration in Bottle 1 if the roles of the bottles were reversed?
4. If both bottles had the same initial concentration, how would that change the final result?
5. What is the effect of mixing unequal volumes instead of identical bottles?

**Tip**: When solving mixture problems, setting up the mass balance equation for the total amount of solute can simplify the solution process significantly.

Two identical bottles are completely filled with a solution of salt and water having different concentrations. The concentration of the solution in the first bottle is 37% and that in the second bottle is 12%. A part of the solution from the first bottle is thrown away and is replaced with an equal volume of the solution from the second bottle. What percentage of solution is left in the second bottle if the concentration of the solution in the first bottle is now 17%?

Solve a mixture problem involving two bottles with salt solutions of different concentrations (37% and 12%). Learn how to calculate the concentration left in Bottle 2 after replacing part of Bottle 1's solution. Suitable for Grades 9-11.

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