Question
A milkman adds 10 litres of water to 90 litres of milk. After selling $1/5^{th}$ of the total quantity, he adds water equal to the quantity he has sold. The proportion of water to milk he sells now would be
The correct answer is
29:12:00
This problem involves calculating the final proportion of water to milk in a mixture after a series of additions and subtractions.
Initial Mixture Analysis
- Initial quantity of milk = 90 litres.
- Initial quantity of water = 10 litres.
- Total initial quantity = $90 + 10 = 100$ litres.
Sold Quantity Calculation
- The milkman sells $1/5^{th}$ of the total quantity.
- Quantity sold = $\frac{1}{5} \times 100 = 20$ litres.
- Amount of milk sold = $\frac{1}{5} \times 90 = 18$ litres.
- Amount of water sold = $\frac{1}{5} \times 10 = 2$ litres.
Remaining Mixture Calculation
- Milk remaining after selling = $90 - 18 = 72$ litres.
- Water remaining after selling = $10 - 2 = 8$ litres.
Final Mixture Calculation
- The milkman adds water equal to the quantity he sold.
- Quantity of water added = 20 litres.
- Final quantity of milk = 72 litres (no milk was added).
- Final quantity of water = $8 + 20 = 28$ litres.
- Total final quantity = $72 + 28 = 100$ litres.
Water Milk Proportion Result
- The question asks for the proportion of water to milk in the mixture.
- Final proportion of Water : Milk = $28 : 72$.
- To simplify the ratio, divide both numbers by their greatest common divisor, which is 4.
- Simplified ratio = $\frac{28}{4} : \frac{72}{4} = 7 : 18$.
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Important Questions from Mixture Problems (Notes)
- In 80 litres mixture of milk and water, the ratio of amount of milk to that of amount of water is 7 : 3. In order to make this ratio 2 : 1, how many litres of water should be added ?
- A $595$ litre of mixture contains milk and water in the ratio $17:18$. How much milk must be added to the mixture so that it contains milk and water in the proportion of $3:2$?
- How many liters of water should be added to a 30-liter mixture containing milk and water in the ratio 7:3 such that the resultant mixture has 40% water in it?
Two solutions $X$ and $Y$ containing ingredients $A, B$ and $C$, in proportions $a:b:c$ and $c:b:a$, respectively, are mixed. For the resultant mixture to have $A, B$ and $C$ in $1:1:1$ proportion, it is necessary that $a:b:c$ is
- A 360 ml aqueous solution contains 40% alcohol. How much will be the approximate percentage of alcohol if 3600 ml of water is added to the solution?
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