Question

A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of water and acid in the original solution would have been:

• A. 5:3
• B. 3:5
• C. 7:8
• D. 4:5
• E. 2:3

Hint:

When solving mixture problems, focus on tracking the amount of the component of interest (here, aci(d) and the total volume at each step. This helps in forming clear equations.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Mixture problems involve tracking the volume of specific components (acid/water) through changes in concentration.
Step 2: Key Formula or Approach:
Concentration \( = \frac{\text{Component Volume}}{\text{Total Volume}} \).
Step 3: Detailed Explanation:
Let original acid volume be \( a \) and water be \( w \).
1. After adding 2L water:
Acid \( = a \), Total volume \( = a + w + 2 \).
\( \frac{a}{a+w+2} = 0.5 \Rightarrow 2a = a+w+2 \Rightarrow a - w = 2 \). (i)
2. After adding 15L acid to this:
Acid \( = a+15 \), Total volume \( = (a+w+2) + 15 = a+w+17 \).
\( \frac{a+15}{a+w+17} = 0.8 \Rightarrow a+15 = 0.8(a+w+17) \).
\( 5a + 75 = 4a + 4w + 68 \Rightarrow a - 4w = -7 \). (ii)
Subtracting (ii) from (i):
\( (a-w) - (a-4w) = 2 - (-7) \Rightarrow 3w = 9 \Rightarrow w = 3 \).
Substituting in (i): \( a - 3 = 2 \Rightarrow a = 5 \).
The original ratio of water and acid is \( w : a = 3 : 5 \).
Step 4: Final Answer:
The ratio is 3 : 5.