Q.A solution of $9\%$ acid is to be diluted by adding $3\%$ acid solution to it. The resulting mixture is to be more than $5\%$ but less than $7\%$ acid. If there is $460$ litres of the $9\%$ solution, how many litres of $3\%$ solution will have to be added?

The key idea here is Mixture Concentration. We determine the total amount of solute (acid) and total volume, then use the given concentration range to form inequalities.

1. Let $x$ be the volume (in litres) of the $3\%$ acid solution added.

   The total amount of acid in the mixture will be $0.09(460) + 0.03x$.

   The total volume of the mixture will be $460 + x$.

2. The concentration of the resulting mixture must be more than $5\%$ and less than $7\%$. This gives us the inequality:

3. We solve the left part of the inequality: $0.05(460 + x) < 0.09(460) + 0.03x$ $23 + 0.05x < 41.4 + 0.03x$ …

When we mix two solutions of different concentrations, the resulting concentration is not simply the average of the two. Instead, it's a weighted average, where the "weights" are the volumes of each solution. This makes intuitive sense: if you add a lot of a weak solution to a small amount of strong solution, the final mixture will be closer to the weak solution's concentration. Conversely, if you add a small amount of weak solution to a large amount of strong solution, the final mixture will still be quite strong.

The total amount of acid in the mixture is the sum of the acid contributed by each individual solution. The total volume of the mixture is the sum of the individual volumes. The final concentration is then the total amount of acid divided by the total volume. This fundamental idea allows us to set up an equation for the final concentration and then use inequalities to find the required range for the added volume.

1. Identify Given Quantities and Define the Unknown

   We are given:

   • Volume of the first solution ($9\%$ acid): $V_1 = 460$ litres
   • Concentration of the first solution: $C_1 = 9\% = 0.09$
   • Concentration of the second solution ($3\%$ acid): $C_2 = 3\% = 0.03$

   We need to find the volume of the second solution to be added. Let this be $V_2$ litres.

2. Formulate Total Acid Amount and Total Volume

   The amount of acid in the first solution is $C_1 V_1 = 0.09 \times 460$ litres.

   The amount of acid in the second solution is $C_2 V_2 = 0.03 \times V_2$ litres.

   The total amount of acid in the mixture is the sum of these:

   Total Acid $= (0.09 \times 460) + (0.03 \times V_2)$ litres

   Total Acid $= 41.4 + 0.03 V_2$ litres

   The total volume of the mixture is the sum of the individual volumes:

   Total Volume $= V_1 + V_2 = 460 + V_2$ litres

3. Express the Final Concentration

   The concentration of the resulting mixture, $C_{final}$, is the total amount of acid divided by the total volume:

   $C_{final} = \frac{\text{Total Acid}}{\text{Total Volume}} = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2}$

   Substituting the expressions from Step 2:

4. Set Up the Inequalities The problem states that the resulting mixture is to be more than $5\%$ but less than $7\%$ acid. This translates to the following compound inequality:

Substituting the expression for $C_{final}$:

We can break this into two separate inequalities:
*   Inequality 1: $\frac{41.4 + 0.03 V_2}{460 + V_2} > 0.05$
*   Inequality 2: $\frac{41.4 + 0.03 V_2}{460 + V_2} < 0.07$

Since $V_2$ represents a volume, $V_2 > 0$, which means $460 + V_2$ is always positive. Therefore, we can multiply both sides of the inequalities by $(460 + V_2)$ without changing the direction of the inequality sign.

5. Solve Inequality 1 …