Mixture Problem Analysis
The question asks us to find the amount of water needed to change the concentration of water in a milk-water mixture. We start with a 30-liter mixture that has milk and water in a 7:3 ratio. The goal is to add only water until the mixture contains 40% water.
Initial Mixture Quantities Calculation
First, let's determine the initial amounts of milk and water in the 30-liter mixture.
- The ratio of milk to water is 7:3.
- Total parts in the ratio = 7 + 3 = 10 parts.
- The total volume of the mixture is 30 liters.
- Volume per part = $\frac{\text{Total Volume}}{\text{Total Parts}} = \frac{30 \text{ liters}}{10 \text{ parts}} = 3 \text{ liters/part}$.
Now we can calculate the initial volumes:
- Initial Milk Volume = 7 parts $\times$ 3 liters/part = 21 liters.
- Initial Water Volume = 3 parts $\times$ 3 liters/part = 9 liters.
We can verify this: 21 liters (milk) + 9 liters (water) = 30 liters (total volume).
Calculating Water Addition for Target Percentage
We need to add water to the mixture. Let the amount of water added be '$x$' liters. Adding water does not change the amount of milk, but it increases the total volume and the amount of water.
- The amount of milk remains constant: 21 liters.
- The new amount of water will be: (9 + $x$) liters.
- The new total volume of the mixture will be: (30 + $x$) liters.
The problem requires the final mixture to be 40% water. We can express this as:
$ \frac{\text{Final Water Volume}}{\text{Final Total Volume}} = \frac{40}{100} = 0.4 $
Substituting the expressions for the final volumes:
$ \frac{9 + x}{30 + x} = 0.4 $
Solving for Water Added (x)
Now, we solve the equation for '$x$':
- Multiply both sides by $(30 + x)$: $9 + x = 0.4 \times (30 + x)$
- Distribute 0.4 on the right side: $9 + x = (0.4 \times 30) + (0.4 \times x)$ $9 + x = 12 + 0.4x$
- Rearrange the terms to group '$x$' terms on one side and constants on the other: $x - 0.4x = 12 - 9$
- Combine like terms: $0.6x = 3$
- Solve for '$x$' by dividing by 0.6: $x = \frac{3}{0.6}$ $x = \frac{30}{6}$ $x = 5$
So, 5 liters of water should be added.
Verification of the Result
Let's check if adding 5 liters of water results in a mixture with 40% water.
- Initial mixture volume = 30 liters.
- Initial water = 9 liters.
- Amount of water added = 5 liters.
- Final water volume = 9 + 5 = 14 liters.
- Final total volume = 30 + 5 = 35 liters.
- Percentage of water in the final mixture = $\frac{\text{Final Water Volume}}{\text{Final Total Volume}} \times 100$ $= \frac{14 \text{ liters}}{35 \text{ liters}} \times 100$ $= \frac{2}{5} \times 100$ $= 0.4 \times 100 = 40\%$
The final mixture indeed contains 40% water, confirming our calculation.
Important Questions from Mixture Problems (Notes)
- A has a container containing 60 litres of pure milk. He takes out 4 litres of milk and replaces it with the same quantity of water. He sells this mixture to B. B sells 30 litres of the mixture and added 5 litres of water in the remaining mixture. The ratio of milk to water in the remaining mixture is:
- 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 ?
Suppose a tap mixes hot water and cold water in a ratio that depends linearly on the proportion of opening. Water out of the tap has temperature 40°C when the tap is half-open, and 30°C when it is three-fourths open. To get water at 50°C, the tap should be_______________.
- 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 $3:2$?
- Alloy A is formed by mixing iron (Fe) and nickel (Ni) in the ratio 3:4, while alloy B is formed by mixing Fe and Ni in the ratio 9:5. If equal quantities of alloys A and B are melted together to form a new alloy C, what will be the ratio of Fe to Ni in the alloy C?
