Solving Salt Solution Percentage Problems

This problem asks us to find out how much water needs to be added to a salt solution to decrease its concentration. The key idea is that the amount of salt in the solution remains constant; only the total volume changes when water is added.

Understanding the Initial Salt Solution

We start with a salt solution with the following properties:

• Total volume = 30 litres
• Salt concentration = 5%

First, let's calculate the actual amount of salt present in this initial solution. The amount of salt is 5% of the total volume.

Using mathematical notation:

Salt amount $= 5\% \text{ of } 30 \text{ litres}$

To calculate 5% of 30, we can convert the percentage to a decimal or a fraction:

Salt amount $= \frac{5}{100} \times 30$

Salt amount $= 0.05 \times 30$

Salt amount $= 1.5 \text{ litres}$

So, the initial solution contains 1.5 litres of salt.

Adding Water to Change Concentration

Water is added to this solution. We are told that the new, resulted solution should contain 3% salt. Let's assume 'x' litres of water are added to the original 30 litres of solution.

• Amount of water added = x litres
• New total volume of the solution = Original volume + Water added = $(30 + x)$ litres
• New salt concentration = 3%

The crucial point here is that adding water does not change the amount of salt. The amount of salt is still 1.5 litres in the new solution.

Setting Up the Equation

In the new solution, the amount of salt (1.5 litres) is 3% of the new total volume $(30 + x)$ litres.

We can write this as an equation:

Amount of salt in new solution $= 3\% \text{ of new total volume}$

$1.5 = \frac{3}{100} \times (30 + x)$

$1.5 = 0.03 \times (30 + x)$

Solving for the Amount of Water Added

Now, we need to solve this equation for 'x'.

Divide both sides by 0.03:

$\frac{1.5}{0.03} = 30 + x$

To divide 1.5 by 0.03, we can multiply both numbers by 100 to remove the decimals:

$\frac{1.5 \times 100}{0.03 \times 100} = \frac{150}{3} = 50$

So, the equation becomes:

$50 = 30 + x$

Now, isolate 'x' by subtracting 30 from both sides:

$x = 50 - 30$

$x = 20$

Therefore, 20 litres of water must be added to the solution.

Verifying the Solution

Let's check if adding 20 litres of water results in a 3% salt solution.

• Original volume = 30 litres
• Water added = 20 litres
• New total volume = $30 + 20 = 50$ litres
• Amount of salt = 1.5 litres (remains constant)
• New salt concentration = $\frac{\text{Amount of salt}}{\text{New total volume}} \times 100\%$
• New salt concentration = $\frac{1.5}{50} \times 100\%$
• New salt concentration = $\frac{1.5 \times 100}{50}\%$
• New salt concentration = $\frac{150}{50}\%$
• New salt concentration = $3\%$

The calculation confirms that adding 20 litres of water results in a solution containing 3% salt.

From the table, we set up the equation based on the constant amount of salt:

$ (30 + x) \times 0.03 = 1.5 $

Solving this equation gives $x = 20$ litres.

Revision Table: Salt Solution Concentration

Additional Information: Mixture Problems

Mixture problems like this one are common in quantitative aptitude. They typically involve mixing solutions with different concentrations or adding pure solute or solvent to a solution to achieve a desired concentration. The fundamental principle is often based on conserving the amount of the solute or the total volume, depending on what is being added or removed.

• When adding pure solvent (like water), the amount of solute remains constant.
• When adding pure solute (like salt powder), the amount of solvent remains constant (unless the volume change due to adding solute is considered).
• When mixing two solutions, the total amount of solute in the final mixture is the sum of the solute amounts in the initial solutions, and the total volume is the sum of the initial volumes.

These types of problems can often be solved by setting up equations based on the conserved quantity (either solute amount or solvent amount) or by using methods like alligation, although the equation method is straightforward for problems involving adding only water or salt.