In what ratio must 25% of alcohol be mixed with 50% of alcohol to get a mixture of 40% strength alcohol?

Options

A. `1:2`
B. `2:1`
C. `2:3`
D. `3:2`

Answer

To solve the problem of mixing 25% alcohol with 50% alcohol to achieve a 40% alcohol mixture, we can use the method of alligation. Here’s a step-by-step solution:

### Step 1: Identify the strengths of the alcohol solutions
- Let \( A = 25\% \) (the strength of the first alcohol solution)
- Let \( B = 50\% \) (the strength of the second alcohol solution)
- Let \( X = 40\% \) (the desired strength of the mixture)

### Step 2: Apply the alligation formula
According to the alligation method, we find the difference between the strengths of the solutions and the desired strength.

- Difference between \( B \) and \( X \):
\[ B - X = 50\% - 40\% = 10\% \]

- Difference between \( A \) and \( X \):
\[ X - A = 40\% - 25\% = 15\% \]

### Step 3: Set up the ratio
Now, we can set up the ratio of the two solutions based on the differences calculated:

- The ratio of the quantities of the two solutions is given by:
\[ \text{Ratio} = \frac{X - A}{B - X} = \frac{15\%}{10\%} \]

### Step 4: Simplify the ratio
Now, simplify the ratio:
\[ \frac{15}{10} = \frac{3}{2} \]

### Conclusion
Thus, the ratio in which 25% alcohol must be mixed with 50% alcohol to obtain a 40% alcohol mixture is:
\[ \text{Ratio} = 3:2 \]