A vessel contains a mixture of alcohol and water in the ratio of $9 : x$ respectively. When $500 ext{ ml}$ of this mixture is taken out and added to another vessel containing $100 ext{ ml}$ of water, the ratio of alcohol to water in the second vessel becomes $3 : 7$. Find the value of $x$.

Question

A vessel contains a mixture of alcohol and water in the ratio of $9 : x$ respectively. When $500 	ext{ ml}$ of this mixture is taken out and added to another vessel containing $100 	ext{ ml}$ of water, the ratio of alcohol to water in the second vessel becomes $3 : 7$. Find the value of $x$.

Accepted Answer

**Solution:**

1. Let the quantity of alcohol be $9k$ and water be $xk$ in the original vessel.

2. The total mixture quantity taken out is $500 	ext{ ml}$, so the amount of alcohol taken out is:
   $$\frac{9}{9 + x} 	imes 500 = \frac{4500}{9 + x} 	ext{ ml}$$

3. Similarly, the amount of water taken out is:
   $$\frac{x}{9 + x} 	imes 500 = \frac{500x}{9 + x} 	ext{ ml}$$

4. This $500 	ext{ ml}$ mixture is added to another vessel containing $100 	ext{ ml}$ of water only.

5. Therefore, in the second vessel:
   - Alcohol = $\frac{4500}{9 + x} 	ext{ ml}$
   - Water = $100 + \frac{500x}{9 + x} 	ext{ ml}$

6. Given the ratio of alcohol to water in the second vessel is $3 : 7$, so:
   $$\frac{\frac{4500}{9 + x}}{100 + \frac{500x}{9 + x}} = \frac{3}{7}$$

7. Cross-multiplying:
   $$7 	imes \frac{4500}{9 + x} = 3 	imes \left(100 + \frac{500x}{9 + x}ight)$$

8. Multiply both sides by $(9 + x)$ to eliminate denominators:
   $$7 	imes 4500 = 3 	imes \left(100(9 + x) + 500xight)$$

9. Simplify:
   $$31500 = 3 	imes (900 + 100x + 500x) = 3 	imes (900 + 600x) = 2700 + 1800x$$

10. Rearranging:
    $$31500 - 2700 = 1800x$$
    $$28800 = 1800x$$

11. Divide both sides by 1800:
    $$x = \frac{28800}{1800} = 16$$

**Answer:** $16$

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