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:
10
Solving Mixture Ratio Problems: Acid and Alcohol
This problem involves a mixture of acid and alcohol where the ratio changes after adding more of one component. We need to find the initial quantity of acid in the mixture.
Let's represent the initial quantities based on the given ratio.
- The initial ratio of acid to alcohol is 3 : 2.
- Let the quantity of acid in the original mixture be \(3x\) litres.
- Let the quantity of alcohol in the original mixture be \(2x\) litres.
Now, consider the change made to the mixture:
- 10 litres of alcohol are added to the mixture.
- The quantity of acid remains \(3x\) litres.
- The new quantity of alcohol becomes \(2x + 10\) litres.
After adding the alcohol, the new ratio of acid to alcohol is given as 3 : 5.
We can set up a proportion (an equation where two ratios are equal) based on the new quantities and the new ratio:
\(\text{New ratio of acid to alcohol} = \frac{\text{Quantity of Acid}}{\text{Quantity of Alcohol}}\)
\(\frac{3x}{2x + 10} = \frac{3}{5}\)
Now, we solve this equation for \(x\) to find the value of the common multiplier.
Step-by-Step Calculation:
- Start with the equation: \(\frac{3x}{2x + 10} = \frac{3}{5}\)
- Cross-multiply: \(3x \times 5 = 3 \times (2x + 10)\)
- Simplify both sides: \(15x = 6x + 30\)
- Subtract \(6x\) from both sides: \(15x - 6x = 30\)
- Combine like terms: \(9x = 30\)
- Divide both sides by 9 to find \(x\): \(x = \frac{30}{9}\)
- Simplify the fraction: \(x = \frac{10}{3}\)
The value of \(x\) is \(\frac{10}{3}\).
The question asks for the quantity of acid in the original mixture. The quantity of acid was represented as \(3x\).
Substitute the value of \(x\) back into the expression for the quantity of acid:
\(\text{Quantity of Acid} = 3x = 3 \times \frac{10}{3}\)
\(\text{Quantity of Acid} = 10\)
So, the quantity of acid in the original mixture was 10 litres.
Revision Table: Mixture Ratio Problem
| Component | Initial Quantity (litres) | Change | Final Quantity (litres) | Ratio Component |
|---|---|---|---|---|
| Acid | \(3x\) | None | \(3x\) | 3 |
| Alcohol | \(2x\) | Add 10 | \(2x + 10\) | 5 |
The ratio of the final quantities is \(\frac{3x}{2x+10}\), which is equal to \(\frac{3}{5}\).
Additional Information: Understanding Ratios and Proportions
A ratio is a comparison of two quantities by division. In this problem, the ratio 3:2 means for every 3 parts of acid, there are 2 parts of alcohol.
A proportion is an equation stating that two ratios are equal. We used a proportion \(\frac{3x}{2x + 10} = \frac{3}{5}\) to solve the problem. Solving proportions often involves cross-multiplication.
Mixture problems frequently involve setting up equations based on the quantities of components and how they change or relate to each other in different states of the mixture (initial, after adding/removing components, etc.). Keeping track of which quantity corresponds to which part of the ratio is crucial.
In problems where a component is added or removed, only the quantity of that specific component changes, while the quantity of other components remains constant unless stated otherwise. This was the case here, where only the amount of alcohol changed.
Important Questions from To Make a Mixture from Two Mixtures
In a mixture of 75 liters, the ratio of milk to water is 3 : 2. If the ratio is to be 1 : 2, how much of water should be added?
There are two containers Xand Y. Xcontains 100 ml of milk and Ycontains 100 ml of water. 20 ml of milk from Xis transferred to Y. After mixing well, 20 ml of the mixture in Yis transferred back to X. If mdenotes the proportion of milk in Xand ndenotes the proportion of water in Y, then which one of the following is correct?
Two vessels P and Q contain liquid A and liquid B in the ratio $4 : 3$ and $5 : 4$ respectively. In what ratio must the mixtures from vessel P and vessel Q be combined to obtain a new mixture in vessel R containing liquid A and liquid B in the ratio $11 : 8$?
In a vessel, a mixture of milk and water is in ratio $9 : 5$, while in another vessel mixture of milk and water is in ratio $3 : 8$. In what ratio mixture of both the vessels should be mixed together so that in the resultant mixture ratio of milk and water becomes $13 : 19$?
30 litres of salt solution contains 5% salt. How many litres of water must be added so as to get a resulted solution containing 3% salt?
A. 20 litres
B. 25 litres
C. 30 litres
D. 35 litres
