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: Define Variables:
Let the original amount of acid be $a$ litres and water be $w$ litres. Total volume = $a+w$.
Step 2: Form Equation from First Addition:
Add 2 litres of water. New total = $a + w + 2$. Acid concentration = $\frac{a}{a+w+2} = 0.5$. So, $a = 0.5(a+w+2) \implies 2a = a+w+2 \implies a = w+2$. (1)
Step 3: Form Equation from Second Addition:
To the new solution (after first addition), add 15 litres of acid. Acid amount becomes $a + 15$. Water amount remains $w+2$. Total volume = $a + w + 2 + 15 = a + w + 17$. Final acid concentration = $\frac{a+15}{a+w+17} = 0.8 = \frac{4}{5}$. So, $5(a+15) = 4(a+w+17) \implies 5a+75 = 4a+4w+68 \implies a = 4w - 7$. (2)
Step 4: Solve the System:
From (1) and (2): $w+2 = 4w - 7 \implies 3w = 9 \implies w = 3$. Then $a = w+2 = 5$.
Step 5: Find the Ratio:
Original water : original acid = $w : a = 3 : 5$.
Step 6: Final Answer:
The ratio of water to acid in the original solution is $3:5$.