Initial Solution State

The initial solution has a total volume of $100 \, \text{L}$ with a salt concentration of $60\%$.

• Amount of salt = $60\% \times 100 \, \text{L} = 0.60 \times 100 \, \text{L} = 60 \, \text{L}$.
• Amount of water = $100 \, \text{L} - 60 \, \text{L} = 40 \, \text{L}$.

Part 1: Reducing Concentration to 40% by Adding Water

Let $W$ be the volume of water added (in litres).

• The amount of salt remains the same: $60 \, \text{L}$.
• The new total volume becomes $100 + W \, \text{L}$.
• The desired concentration is $40\%$.

Set up the equation for the new concentration:

$ \frac{\text{Amount of Salt}}{\text{New Total Volume}} = \text{Desired Concentration} $

$ \frac{60}{100 + W} = 40\% $

$ \frac{60}{100 + W} = 0.40 $

Solve for $W$:

$ 60 = 0.40 \times (100 + W) $

$ 60 = 40 + 0.40W $

$ 60 - 40 = 0.40W $

$ 20 = 0.40W $

$ W = \frac{20}{0.40} = 50 \, \text{L} $

Therefore, $50 \, \text{L}$ of water must be added.

The new volume is $100 \, \text{L} + 50 \, \text{L} = 150 \, \text{L}$.

The salt amount is still $60 \, \text{L}$ ($60 \, \text{L} / 150 \, \text{L} = 0.40 = 40\%$).

Part 2: Increasing Concentration to 50% by Adding Pure Salt

Currently, the solution has $150 \, \text{L}$ total volume and $60 \, \text{L}$ of salt.

Let $S$ be the volume of pure salt added (in litres).

• The new amount of salt becomes $60 + S \, \text{L}$.
• The new total volume becomes $150 + S \, \text{L}$.
• The desired concentration is $50\%$.

Set up the equation for the new concentration:

$ \frac{\text{New Amount of Salt}}{\text{New Total Volume}} = \text{Desired Concentration} $

$ \frac{60 + S}{150 + S} = 50\% $

$ \frac{60 + S}{150 + S} = 0.50 $

Solve for $S$:

$ 60 + S = 0.50 \times (150 + S) $

$ 60 + S = 75 + 0.50S $

$ S - 0.50S = 75 - 60 $

$ 0.50S = 15 $

$ S = \frac{15}{0.50} = 30 \, \text{L} $

Therefore, $30 \, \text{L}$ of pure salt must be added.

Final Answer: Add $50 \, \text{L}$ of water, then $30 \, \text{L}$ of salt.