A 40 - litre mixture contains 25% alcohol and 75% water. If 10 litres of water are added to the mixture, the percentage of alcohol in the new mixture is:
20%
Analyzing the Initial Mixture Composition
We begin with a 40-litre mixture that contains both alcohol and water. The composition is given as percentages:
- Alcohol: 25%
- Water: 75%
To solve this mixture problem, the first step is to determine the actual quantity (in litres) of alcohol and water in the initial 40-litre mixture.
Let's calculate the amount of alcohol:
\(\text{Amount of alcohol} = 25\% \text{ of } 40 \text{ litres}\)
\(\text{Amount of alcohol} = \frac{25}{100} \times 40 = \frac{1}{4} \times 40 = 10 \text{ litres}\)
Now, let's calculate the amount of water:
\(\text{Amount of water} = 75\% \text{ of } 40 \text{ litres}\)
\(\text{Amount of water} = \frac{75}{100} \times 40 = \frac{3}{4} \times 40 = 30 \text{ litres}\)
We can check that the total initial volume is \(10 \text{ litres (alcohol)} + 30 \text{ litres (water)} = 40 \text{ litres}\), which matches the given initial volume.
Modifying the Mixture by Adding Water
The problem states that 10 litres of water are added to the original 40-litre mixture. It's important to note that only water is added; the amount of alcohol remains unchanged.
Let's find the new total volume of the mixture after adding the water.
\(\text{New total volume} = \text{Initial total volume} + \text{Amount of water added}\)
\(\text{New total volume} = 40 \text{ litres} + 10 \text{ litres} = 50 \text{ litres}\)
The amount of alcohol in the new mixture is still the initial amount because no alcohol was added or removed.
\(\text{Amount of alcohol in new mixture} = 10 \text{ litres}\)
The amount of water in the new mixture will be the initial amount plus the added water:
\(\text{Amount of water in new mixture} = 30 \text{ litres} + 10 \text{ litres} = 40 \text{ litres}\)
The new total volume is \(10 \text{ litres (alcohol)} + 40 \text{ litres (water)} = 50 \text{ litres}\), confirming our calculation.
Calculating the New Percentage of Alcohol
We now have the composition of the new 50-litre mixture: 10 litres of alcohol and 40 litres of water. We need to find the percentage of alcohol in this new mixture.
The formula for percentage is:
\(\text{Percentage of component} = \left(\frac{\text{Amount of component}}{\text{Total amount of mixture}}\right) \times 100\%\)
Using this formula for alcohol:
\(\text{Percentage of alcohol} = \left(\frac{\text{Amount of alcohol in new mixture}}{\text{New total volume}}\right) \times 100\%\)
\(\text{Percentage of alcohol} = \left(\frac{10 \text{ litres}}{50 \text{ litres}}\right) \times 100\%\)
\(\text{Percentage of alcohol} = \left(\frac{1}{5}\right) \times 100\%\)
\(\text{Percentage of alcohol} = 0.2 \times 100\%\)
\(\text{Percentage of alcohol} = 20\%\)
Therefore, the percentage of alcohol in the new mixture is 20%.
Summary of Steps
- Calculate initial alcohol amount: 25% of 40 L = 10 L.
- Calculate initial water amount: 75% of 40 L = 30 L.
- Add water: 10 L added to the mixture.
- Calculate new total volume: 40 L + 10 L = 50 L.
- Alcohol amount remains constant: 10 L.
- Calculate new alcohol percentage: (\(\frac{10}{50}\)) * 100% = 20%.
Revision Table: Mixture Percentage Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Percentage Composition | The proportion of a component in a mixture expressed as a percentage of the total mixture volume or weight. | \(\frac{\text{Amount of component}}{\text{Total amount}} \times 100\%\) |
| Amount of Component | Calculating the actual quantity of a component if its percentage and the total mixture amount are known. | Percentage \(\times\) Total amount |
| Mixture Problems | Problems involving combining or altering mixtures with different compositions to find the resulting composition or quantities. | Require careful tracking of component amounts and total amounts. |
Additional Information: Dilution and Concentration
This problem is an example of a dilution scenario. When you add more of the solvent (water) to a solution (alcohol and water mixture), the concentration of the solute (alcohol) decreases, even though the amount of solute stays the same. This is because the total volume of the mixture increases.
Concentration is often expressed as a percentage, as in this problem. Understanding how adding or removing components affects the total volume and the amount of each component is key to solving mixture problems.
In this case, adding water diluted the alcohol, reducing its percentage in the overall mixture from 25% to 20%. If alcohol had been added instead of water, the percentage of alcohol would have increased (assuming the added alcohol was 100% alcohol, or a higher concentration than the original mixture).
Important Questions from To Make a Mixture from Two Mixtures
One cup has juice and water in the ratio 5 ∶ 2, while another cup of the same capacity has them in the ratio 7 ∶ 4, respectively. If contents of both the cups (when full) are poured in a vessel, then what will be the final ratio of water to juice in the vessel?
A and B are solutions of acid and water. The ratios of water and acid in A and B are 4 : 5 and 1 : 2 respectively. If x liters of A is mixed with y liters of B, then the ratio of water and acid in the mixture becomes 8 : 13 What is x : y?
A drink of chocolate and milk contains 8% pure chocolate by volume. If 10 litres of pure milk are added to 50 litres of this drink, the percentage of chocolate in the new drink is:
Mixture A contains chocolate and milk in the ratio 4 ∶ 3 and mixture B contains chocolate and milk in the ratio 5 ∶ 2. A and B are taken in the ratio 5 ∶ 6 and mixed to form a new mixture. The percentage of chocolate in the new mixture is closest to:
If 80 litres of milk solution has 60% milk in it, then how much milk should be added to make milk 80% in the solution?
