40 litres of a mixture has 10% alcohol and the rest water. If 4 litres of water be mixed in it, the percentage of alcohol in the new mixture will be:
Initial Mixture Analysis
The total volume of the initial mixture is 40 litres.
The percentage of alcohol in the initial mixture is 10%.
The percentage of water in the initial mixture is $100\% - 10\% = 90\%$.
Calculating Initial Component Volumes
Volume of Alcohol:
- $10\%$ of 40 litres = $\frac{10}{100} \times 40 = 4$ litres.
Volume of Water:
- $90\%$ of 40 litres = $\frac{90}{100} \times 40 = 36$ litres.
New Mixture Composition
4 litres of water are added to the mixture.
- New volume of water = Initial water volume + Added water = $36 + 4 = 40$ litres.
- Volume of alcohol remains unchanged = 4 litres.
- New total volume of the mixture = New water volume + Alcohol volume = $40 + 4 = 44$ litres.
Calculating New Alcohol Percentage
The percentage of alcohol in the new mixture is calculated as:
- Percentage Alcohol = $(\frac{\text{Volume of Alcohol}}{\text{New Total Volume}}) \times 100$
- Percentage Alcohol = $(\frac{4}{44}) \times 100$
- Percentage Alcohol = $(\frac{1}{11}) \times 100$
- Percentage Alcohol = $\frac{100}{11}\%$
Converting the improper fraction to a mixed number:
- $\frac{100}{11} = 9\frac{1}{11}$
Therefore, the percentage of alcohol in the new mixture is $9\frac{1}{11}\%$.
Important Questions from Mixture Problems
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