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:
\[\boxed{{3}:{2}}.\]