Question

A container has a mixture of acid and water. If 2 liters of water are added, the acid content reaches 50%. After adding 18 liters of acid, the acid content increases to 80%. What was the initial acid to water ratio in the container?

• A. 4:3
• B. 3:2
• C. 4:2
• D. 1:2

Answer 1

The Correct Option is B

Let the initial amounts of acid and water in the container be denoted by $A$ liters and $W$ liters, respectively.
Step 1: After adding 2 liters of water
When 2 liters of water are added, the total volume of the mixture becomes $A + W + 2$.
The acid content is $50\%$, so:
$$\frac{A}{A + W + 2} = 0.50.$$
This gives the first equation:
$$A = 0.50(A + W + 2).$$
Simplify this equation:
$$A = 0.50A + 0.50W + 1,$$
$$A - 0.50A = 0.50W + 1,$$
$$0.50A = 0.50W + 1,$$
$$A = W + 2. \quad \text{(Equation 1)}$$
Step 2: After adding 18 liters of acid
When 18 liters of acid are added, the total volume of the mixture becomes $A + W + 2 + 18 = A + W + 20$.  
The acid content is $80\%$, so:
$$\frac{A + 18}{A + W + 20} = 0.80.$$
This gives the second equation:
$$A + 18 = 0.80(A + W + 20).$$
Simplify this equation:
$$A + 18 = 0.80A + 0.80W + 16$$
$$A - 0.80A = 0.80W + 16 - 18,$$
$$0.20A = 0.80W - 2,$$
$$A = 4W - 10. \quad \text{(Equation 2)}$$
Step 3: Solve the system of equations
Now, solve the system of equations (1) and (2).  
From Equation (1):
$$A = W + 2.$$
Substitute this into Equation (2):
$$W + 2 = 4W - 10.$$
Solve for $W$:
$$2 + 10 = 4W - W,$$
$$12 = 3W,$$
$$W = 4.$$
Step 4: Find $A$
Substitute $W = 4$ into Equation (1):
$$A = W + 2 = 4 + 2 = 6.$$
Step 5: Initial Acid to Water Ratio
The initial acid to water ratio is:
$$\frac{A}{W} = \frac{6}{4} = \frac{3}{2}.$$
Final Answer
The initial acid to water ratio is:
$$\boxed{{3}:{2}}.$$