Understanding the Mixing Problem
This problem asks us to determine the ratio in which two mixtures, each containing spirit and water, must be combined to obtain a final mixture with a specific spirit-to-water ratio. We are given the initial ratios in two separate bottles and the desired final ratio.
Initial Setup
Let's denote the quantities of spirit and water in the bottles.
- Bottle 1: Contains spirit and water in the ratio \(1:4\). This means for every 1 part spirit, there are 4 parts water. The total parts are \(1+4=5\). So, the fraction of spirit is \(\frac{1}{5}\) and the fraction of water is \(\frac{4}{5}\).
- Bottle 2: Contains spirit and water in the ratio \(4:1\). This means for every 4 parts spirit, there is 1 part water. The total parts are \(4+1=5\). So, the fraction of spirit is \(\frac{4}{5}\) and the fraction of water is \(\frac{1}{5}\).
- Target Mixture: We want to obtain a new mixture where the ratio of spirit to water is \(1:3\). The total parts are \(1+3=4\). So, the desired fraction of spirit is \(\frac{1}{4}\) and the desired fraction of water is \(\frac{3}{4}\).
Solving Using Alligation Method
The alligation method is a quick way to solve such mixture problems. We focus on the proportion of one component, typically the one that is present in all mixtures (in this case, spirit).
We calculate the percentage (or fractional) concentration of spirit in each bottle and the target mixture:
- Concentration of Spirit in Bottle 1 = \(\frac{1}{1+4} = \frac{1}{5}\)
- Concentration of Spirit in Bottle 2 = \(\frac{4}{4+1} = \frac{4}{5}\)
- Desired Concentration of Spirit in the Final Mixture = \(\frac{1}{1+3} = \frac{1}{4}\)
Now, we set up the alligation:
| Concentration in Bottle 1 (\(\frac{1}{5}\)) | Concentration in Bottle 2 (\(\frac{4}{5}\)) | |
| Desired Concentration (\(\frac{1}{4}\)) | ||
| Difference (\(|\frac{4}{5} - \frac{1}{4}|\)) | Difference (\(|\frac{1}{5} - \frac{1}{4}|\)) |
Let's calculate the differences:
- Difference 1 = \(|\frac{4}{5} - \frac{1}{4}| = |\frac{16 - 5}{20}| = \frac{11}{20}\)
- Difference 2 = \(|\frac{1}{5} - \frac{1}{4}| = |\frac{4 - 5}{20}| = |-\frac{1}{20}| = \frac{1}{20}\)
The ratio in which the mixtures should be mixed is the inverse ratio of these differences:
Ratio (Bottle 1 : Bottle 2) = Difference 1 : Difference 2
Ratio = \(\frac{11}{20} : \frac{1}{20}\)
To simplify, we can multiply both sides by 20:
Ratio = \(11 : 1\)
Alternative: Algebraic Method
Let the quantity of mixture taken from Bottle 1 be \(x\) units and from Bottle 2 be \(y\) units. The total quantity of the new mixture will be \((x+y)\) units.
- Amount of Spirit from Bottle 1 = \(x \times \frac{1}{5} = \frac{x}{5}\)
- Amount of Spirit from Bottle 2 = \(y \times \frac{4}{5} = \frac{4y}{5}\)
- Total amount of Spirit in the new mixture = \(\frac{x}{5} + \frac{4y}{5}\)
The fraction of spirit in the new mixture is \(\frac{\text{Total Spirit}}{\text{Total Volume}} = \frac{\frac{x}{5} + \frac{4y}{5}}{x+y}\).
We want this fraction to be equal to the desired spirit fraction, which is \(\frac{1}{4}\).
\( \frac{\frac{x}{5} + \frac{4y}{5}}{x+y} = \frac{1}{4} \)
Multiply both sides by \(4(x+y)\) to clear the denominators:
\( 4 \left( \frac{x}{5} + \frac{4y}{5} \right) = 1 (x+y) \)
\( \frac{4x}{5} + \frac{16y}{5} = x+y \)
Multiply the entire equation by 5:
\( 4x + 16y = 5(x+y) \)
\( 4x + 16y = 5x + 5y \)
Rearrange the terms to group \(x\) and \(y\) terms:
\( 16y - 5y = 5x - 4x \)
\( 11y = x \)
\( \frac{x}{y} = \frac{11}{1} \)
Therefore, the ratio \(x:y\) is \(11:1\).
Conclusion
Both the alligation method and the algebraic method show that the mixtures from the two bottles must be mixed in the ratio \(11:1\) to achieve a final mixture with a spirit-to-water ratio of \(1:3\).
Similar Questions
- There are two containers A and B. In container A, the ratio of milk and water is 1:3 and in container B, the ratio of milk and water is \(m:n\). If the mixture in the containers A and B are mixed in the ratio 2:3 to get 20 litres of a mixture having milk and water in the ratio 3:7, then what is the value of \(\frac{m}{n}\)?
A mixture of 100 L contains kerosene and turpentine oil in the ratio 3:2. What is the minimum quantity of kerosene in litres (whole number) that should be mixed in the mixture so that the resulting mixture has 20% of kerosene?
The mean weight of 150 students in a class is 60 kg. The mean weight of boys in the class is 70 kg and that of girls is 55 kg. What is the ratio of number of boys to number of girls ?
- X and Y are two alloys of copper (Cu) and zinc (Zn). Alloy X is prepared by mixing Cu and Zn in the ratio 5:4, and alloy Y is prepared by mixing Cu and Zn in the ratio 5:13 respectively. If equal quantities of alloys X and Y are melted to form a third alloy Z, then what is the ratio of Cu to Zn in Z?
An alloy (X) of gold and silver is taken in the ratio 1 : 2 and another alloy (Y) of gold and silver is taken in the ratio 2 : 3. How many parts of X and Y must be taken to obtain a new alloy Z consisting of gold and silver in the ratio 4 : 7?
Important Questions from Mixture and Alligation
A bottle contains 20 litres of liquid A. 4 litres of liquid A is taken out of it d replaced by same quantity of liquid B. Again 4 litre of the mixture is taken out and replaced by same quantity of liquid B. What is the ratio of quantity of liquid A to that of liquid B in the final mixture?
The average score of a batsman after his 50th innings was 46.4. After 60th innings, his average Score increases by 2.6. What was his average score in the last ten innings?
If 1 litre of water weighs 1 kg, then how many cubic millimeters of water will weigh 0.1 gm?
A vessel full of water weighs 4o kg. If it is one-third filled, its weight becomes 20 kg. What is the weight of the empty vessel?
Two equal glasses of same type are respectively 1/3 and 1/4 full of milk. They are then filled up with water and the contents are mixed in a pot. What is the ratio of milk and water in the pot?
