Problem Overview: Mixing Container Contents
This problem involves determining the precise ratio at which mixtures from two identical containers, labeled P and Q, should be combined to achieve a specific final milk-to-water ratio. Container P holds a mixture with a milk-to-water ratio of 1:1, while container Q contains a mixture with a ratio of 3:1. The goal is to find the mixing ratio (from P to Q) that results in a final mixture with a milk-to-water ratio of 5:4.
Analyzing Container Compositions
First, let's determine the proportion (fraction) of milk and water in each container:
| Container | Milk : Water Ratio | Milk Fraction | Water Fraction |
|---|---|---|---|
| P | 1 : 1 | $\frac{1}{1+1} = \frac{1}{2}$ | $\frac{1}{1+1} = \frac{1}{2}$ |
| Q | 3 : 1 | $\frac{3}{3+1} = \frac{3}{4}$ | $\frac{1}{3+1} = \frac{1}{4}$ |
| Target Mixture | 5 : 4 | $\frac{5}{5+4} = \frac{5}{9}$ | $\frac{4}{5+4} = \frac{4}{9}$ |
Setting Up the Mixing Ratio Equation
Let's assume we need to take '$x$' units of the mixture from container P and '$y$' units of the mixture from container Q. The total volume of the final mixture will be '$x + y$' units.
We can calculate the total amount of milk in the final mixture by summing the milk contributions from both containers:
- Amount of milk from P = $x \times \text{Milk fraction in P} = x \times \frac{1}{2}$
- Amount of milk from Q = $y \times \text{Milk fraction in Q} = y \times \frac{3}{4}$
Total milk in the final mixture = $\frac{x}{2} + \frac{3y}{4}$.
The problem states that the final mixture should have a milk-to-water ratio of 5:4. This means the fraction of milk in the final mixture should be $\frac{5}{9}$.
We can set up an equation based on the milk concentration:
$$ \frac{\text{Total milk in final mixture}}{\text{Total volume of final mixture}} = \text{Target milk fraction} $$
$$ \frac{\frac{x}{2} + \frac{3y}{4}}{x+y} = \frac{5}{9} $$
Solving for the Required Ratio (x:y)
Now, we solve this equation to find the ratio '$x:y$'.
- Simplify the numerator of the fraction: To combine $\frac{x}{2}$ and $\frac{3y}{4}$, find a common denominator, which is 4. $$ \frac{x}{2} + \frac{3y}{4} = \frac{2x}{4} + \frac{3y}{4} = \frac{2x + 3y}{4} $$
- Substitute this back into the equation: $$ \frac{\frac{2x + 3y}{4}}{x+y} = \frac{5}{9} $$
- Simplify the left side: $$ \frac{2x + 3y}{4(x+y)} = \frac{5}{9} $$
- Cross-multiply to eliminate the denominators: $$ 9(2x + 3y) = 5 \times 4(x+y) $$
- Expand both sides of the equation: $$ 18x + 27y = 20(x+y) $$ $$ 18x + 27y = 20x + 20y $$
- Rearrange the terms to group '$x$' terms on one side and '$y$' terms on the other: $$ 27y - 20y = 20x - 18x $$
- Simplify the equation: $$ 7y = 2x $$
- Express the ratio '$x:y$'. Divide both sides by '$y$' and then by 2: $$ \frac{x}{y} = \frac{7}{2} $$
Thus, the mixture should be taken in the ratio 7:2 from container P to container Q.
Verifying the Solution Using Water Concentration
We can verify this result by considering the water concentration. The total amount of water in the final mixture is:
- Amount of water from P = $x \times \text{Water fraction in P} = x \times \frac{1}{2}$
- Amount of water from Q = $y \times \text{Water fraction in Q} = y \times \frac{1}{4}$
Total water in the final mixture = $\frac{x}{2} + \frac{y}{4}$.
The target fraction of water in the final mixture is $\frac{4}{9}$.
The equation for water concentration is:
$$ \frac{\frac{x}{2} + \frac{y}{4}}{x+y} = \frac{4}{9} $$
Simplifying the numerator gives $\frac{2x+y}{4}$. So the equation becomes:
$$ \frac{2x + y}{4(x+y)} = \frac{4}{9} $$
Cross-multiplying:
$$ 9(2x + y) = 16(x+y) $$
$$ 18x + 9y = 16x + 16y $$
Rearranging terms:
$$ 18x - 16x = 16y - 9y $$
$$ 2x = 7y $$
$$ \frac{x}{y} = \frac{7}{2} $$
This confirms that the required ratio is indeed 7:2.
Final Answer: Required Ratio
The mixtures should be taken out from containers P and Q in the ratio 7:2 to achieve the desired milk-to-water ratio of 5:4 in the final mixture.
Important Questions from To Make a Mixture from Two Mixtures
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In a milk and water solution, the ratio of milk to water is 1 : 4. M litres of milk is added in it and ratio become 1 : 3. Again N litres of water is added in it and ratio become 2 : 7. If N – M = 4, then what is the initial quantity of solution?
A mixture contains acid and alcohol in the ratio of 3 : 2. On adding 10 litres of alcohol in mixture, the ratio of acid to alcohol becomes 3 : 5. The quantity of acid (in litres) in the original mixture was:
How many litres of acid are there in 12 litres of a 20% solution?
A mixture of fog contains water and air in the ratio 5 ∶ 3. On adding 8 liters of air, the ratio of water and air becomes 1 ∶ 1. The quantity of water in the original fog is:
