Let the total volume of solution in each bottle be $V$.

Step 1: Define the discarded volume

Let the volume of the solution discarded from the first bottle (and replaced by the second bottle's solution) be $x$. This volume $x$ is also the amount taken from the second bottle.

Step 2: Calculate the new concentration in the first bottle

The initial salt in the first bottle is $0.4V$, since the concentration is $40\%$. After discarding $x$ volume, the remaining salt is:

$\text{Salt left from the first bottle} = 0.4V - 0.4x$

Replacing $x$ volume with the second bottle's solution (concentration $15\%$) adds $0.15x$ of salt:

$\text{Total salt after replacement} = 0.4V - 0.4x + 0.15x = 0.4V - 0.25x$

The new total volume of the first bottle is still $V$, so the new concentration is:

$\text{New concentration} = \frac{0.4V - 0.25x}{V}$

This concentration is given as $22\%$, or $0.22$. Equating:

$\frac{0.4V - 0.25x}{V} = 0.22$

Simplify:

$0.4V - 0.25x = 0.22V$

$0.18V = 0.25x$

$x = \frac{0.18V}{0.25} = 0.72V$

Thus, $x = 0.72V$.

Step 3: Calculate the remaining solution in the second bottle

Since $x = 0.72V$ volume was taken from the second bottle, the remaining solution in the second bottle is:

$\text{Remaining solution in the second bottle} = V - 0.72V = 0.28V$

Step 4: Convert to percentage

The percentage of solution left in the second bottle is:

$\frac{0.28V}{V} \times 100\% = 28\%$

Final Answer:

The percentage of the solution left in the second bottle is 28%.

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Would you like further clarification or additional examples? Here are 5 related questions for exploration:

1. How does the salt concentration change if the replacement volume differs?
2. What happens to the first bottle’s concentration if the second bottle has $10\%$ concentration instead of $15\%$?
3. How can this problem be generalized to $n$ bottles with varying concentrations?
4. Can we calculate the total salt transferred during the process?
5. How would the answer change if the bottles had unequal volumes?

Tip: Always track the total quantity of the solute (e.g., salt) during such mixing problems to ensure mass conservation!