A vessel contained a solution of acid and water, in which water was 64%. Four litres of the solution was taken out of the vessel and the same quantity of water was added. If the resulting solution contains 30% acid, the quantity (in litres) of the water in the solution, at the beginning in the vessel, was:
15.36
Understanding the Mixture Problem
This problem involves a mixture of acid and water in a vessel. We start with a certain composition, remove a portion, and then add a different substance (water in this case). The composition changes, and we are given the final composition to find the initial quantity of water.
Initial State Analysis
The problem states that initially, the solution contains 64% water. This means the remaining percentage must be the acid.
- Initial Water percentage: 64%
- Initial Acid percentage: 100% - 64% = 36%
Let the initial total quantity of the solution in the vessel be \(V\) litres.
- Initial quantity of acid = \(36\%\) of \(V = \frac{36}{100} \times V = 0.36V\) litres
- Initial quantity of water = \(64\%\) of \(V = \frac{64}{100} \times V = 0.64V\) litres
Step-by-Step Process: Removal and Addition
Four litres of the solution are taken out. When a portion of a solution is removed, the concentration (or percentage) of each component remains the same as in the original solution. The amount of each component removed is proportional to its percentage in the solution.
- Quantity of solution removed = 4 litres
- Quantity of acid removed = 36% of 4 litres = \(0.36 \times 4 = 1.44\) litres
- Quantity of water removed = 64% of 4 litres = \(0.64 \times 4 = 2.56\) litres
After removing 4 litres, the quantity of the remaining solution is \(V - 4\) litres.
- Quantity of acid remaining = Initial acid - Acid removed = \(0.36V - 1.44\) litres
- Quantity of water remaining = Initial water - Water removed = \(0.64V - 2.56\) litres
Next, the same quantity of water (4 litres) is added to the remaining solution.
- Quantity of water added = 4 litres
- Quantity of acid in the new solution = Acid remaining = \(0.36V - 1.44\) litres (Acid quantity doesn't change when water is added)
- Quantity of water in the new solution = Water remaining + Water added = \((0.64V - 2.56) + 4\) litres
- Quantity of water in the new solution = \(0.64V + 1.44\) litres
The total quantity of the new solution is the quantity remaining after removal plus the quantity added: \((V - 4) + 4 = V\) litres. The total volume is back to the original volume.
Final State Analysis and Equation Setup
The problem states that the resulting solution contains 30% acid.
- Quantity of acid in the new solution = 30% of the new total volume
- New total volume = \(V\) litres
- Quantity of acid in the new solution = \(0.30 \times V = 0.30V\) litres
We have two expressions for the quantity of acid in the new solution. We can set them equal to each other to form an equation and solve for \(V\), the initial total volume.
Equation: \(0.36V - 1.44 = 0.30V\)
Solving for the Initial Volume
Let's solve the equation for \(V\):
\(0.36V - 1.44 = 0.30V\)
Subtract \(0.30V\) from both sides:
\(0.36V - 0.30V - 1.44 = 0\)
\(0.06V - 1.44 = 0\)
Add \(1.44\) to both sides:
\(0.06V = 1.44\)
Divide by \(0.06\):
\(V = \frac{1.44}{0.06}\)
To simplify the division, we can multiply the numerator and denominator by 100:
\(V = \frac{1.44 \times 100}{0.06 \times 100} = \frac{144}{6}\)
\(V = 24\)
So, the initial total quantity of the solution in the vessel was 24 litres.
Calculating the Initial Quantity of Water
The question asks for the initial quantity of water in the solution. We know that initially, the solution contained 64% water, and the initial total volume was 24 litres.
Initial quantity of water = 64% of 24 litres
Initial quantity of water = \(0.64 \times 24\)
Calculation:
\(0.64 \times 24 = (64 \times 24) \div 100\)
\(64 \times 24 = 64 \times (20 + 4) = 64 \times 20 + 64 \times 4\)
\(64 \times 20 = 1280\)
\(64 \times 4 = 256\)
\(1280 + 256 = 1536\)
Initial quantity of water = \(1536 \div 100 = 15.36\) litres.
Conclusion
The quantity of water in the solution at the beginning in the vessel was 15.36 litres.
| Stage | Total Volume (litres) | Acid Quantity (litres) | Water Quantity (litres) |
|---|---|---|---|
| Beginning | \(V = 24\) | \(0.36V = 0.36 \times 24 = 8.64\) | \(0.64V = 0.64 \times 24 = 15.36\) |
| After removing 4L | \(V - 4 = 20\) | \(0.36(V-4) = 0.36 \times 20 = 7.2\) | \(0.64(V-4) = 0.64 \times 20 = 12.8\) |
| After adding 4L water | \((V-4) + 4 = 24\) | \(0.36(V-4) = 7.2\) | \((0.64(V-4)) + 4 = 12.8 + 4 = 16.8\) |
Wait, let's re-check the quantity of water in the final mixture based on the final acid percentage. The final mixture is 30% acid, so it must be 70% water.
Final Water percentage = 70%
Final total volume = \(V = 24\) litres
Final quantity of water = 70% of 24 litres = \(0.70 \times 24\)
\(0.70 \times 24 = (7 \times 24) \div 10\)
\(7 \times 24 = 168\)
Final quantity of water = \(168 \div 10 = 16.8\) litres.
This matches the calculation \((0.64V - 2.56) + 4 = (0.64 \times 24 - 2.56) + 4 = (15.36 - 2.56) + 4 = 12.8 + 4 = 16.8\) litres.
The steps and calculations are consistent, and the initial quantity of water was 15.36 litres.
Revision Table: Key Concepts
| Concept | Explanation |
|---|---|
| Percentage Composition | Represents the proportion of a component in a mixture (e.g., 64% water means 64 parts water out of 100 parts solution). |
| Removal of Solution | When a part of the solution is removed, the remaining solution has the same percentage composition as the original. The quantity of each component removed is proportional to its percentage. |
| Addition of a Component | Adding a pure component (like water) changes the overall volume and the concentration of all components. The quantity of the added component increases, while others remain unchanged (unless they react). |
| Setting up Equations | Comparing the quantity of a component (like acid) in the final mixture, calculated in two ways (from the process steps and from the final percentage), allows solving for unknown initial quantities. |
Additional Information: Mixture Problems
Mixture problems are common in quantitative aptitude and mathematics. They often involve combining different substances or changing the composition of an existing mixture through removal or addition.
- Direct Mixing: Two or more mixtures with different compositions are mixed together. The final volume is the sum of individual volumes, and the quantity of each component in the final mixture is the sum of its quantities in the initial mixtures.
- Removal and Replacement: A part of the mixture is removed, and replaced with a pure substance or another mixture. This process can change the composition significantly, often decreasing the concentration of the original components if replaced by something else. The problem in this solution is a classic example of removal and replacement with a pure substance (water).
- Concentration: Often expressed as a percentage, ratio, or proportion. It tells us how much of a particular substance is present in a given amount of the mixture or solution.
Solving these problems usually involves:
- Identifying the components and their initial concentrations.
- Tracking the quantity of each component through each step of the process (removal, addition, etc.).
- Using the information about the final composition to set up an equation.
- Solving the equation to find the unknown quantity, often the initial volume or the quantity of a component.
It's important to be careful with calculations and ensure you are tracking the quantities of individual components, not just the total volume or percentages.
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