Question:medium

From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

Updated On: Jan 15, 2026
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Solution and Explanation

Let \( c \) represent the container's capacity. The objective is to determine the milk remaining after two separate withdrawals of 9 liters each.

Step 1: Milk after the first 9-liter withdrawal

The milk remaining after the initial 9 liters are removed is:

\[ \left( c \times \left( c - \frac{9}{c} \right) \right) \]

Step 2: Milk after the second 9-liter withdrawal

The milk remaining after the subsequent 9 liters are removed is:

\[ \left( c \times \left( c - \frac{9}{c} \right) \times \left( c - \frac{9}{c} \right) \right) \]

Step 3: Milk-to-water ratio

Post-withdrawal, the milk-to-water ratio is 16:9. This implies the fraction of milk in the container is:

\[ \frac{16}{25} \]

Step 4: Equation for remaining milk

The equation representing the remaining milk in terms of \( c \) is:

\[ \frac{c \times \left( \frac{c-9}{c} \right)^2}{c} = \frac{16}{25} \]

Step 5: Equation solution

\[ \left( c - \frac{9}{c} \right)^2 = \frac{16}{25} \] Upon taking the square root of both sides:

\[ c - \frac{9}{c} = \frac{4}{5} \]

Step 6: Solving for \( c \)

\[ c - \frac{9}{c} = 1 \quad \Rightarrow \quad c = 45 \]

Final Answer:

The container's capacity is \( \boxed{45} \) liters.

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