Question

An 8-litre cylinder contains a mixture of oxygen and nitrogen, the volume of oxygen being 16% of total volume. A few litres of the mixture is released and an equal amount of nitrogen is added. Then the same amount of the mixture as before is released and replaced by nitrogen for the second time. As result, the oxygen content becomes 9% of total volume. How many litres of mixture is released each time?

• A. 7 litres
• B. 5 litres
• C. 2 litres
• D. None of these

Hint:

Use the concept of percentage decrease for such problems, where the mixture is replaced in steps.

Answer 1

The Correct Option is C

Let the amount of mixture released each time be \( x \) litres. The total volume of the cylinder is 8 litres, and initially, the oxygen content is 16% of the total volume. Therefore, the initial volume of oxygen is: \[ \text{Initial oxygen volume} = \frac{16}{100} \times 8 = 1.28 \text{ litres} \] When \( x \) litres of the mixture is released, the volume of oxygen is reduced, and an equal amount of nitrogen is added. After the first replacement, the volume of oxygen left is: \[ \text{Oxygen volume after first replacement} = \left( 1 - \frac{x}{8} \right) \times 1.28 \] Now, the same amount of the mixture is removed again. After the second replacement, the volume of oxygen is reduced further, and the volume of oxygen left becomes 9% of the total volume. Therefore: \[ \text{Final oxygen volume} = \frac{9}{100} \times 8 = 0.72 \text{ litres} \] Using the formula for the oxygen volume after both replacements, we get: \[ \left( 1 - \frac{x}{8} \right)^2 \times 1.28 = 0.72 \] Solving this equation: \[ \left( 1 - \frac{x}{8} \right)^2 = \frac{0.72}{1.28} = \frac{9}{16} \] \[ 1 - \frac{x}{8} = \frac{3}{4} \] \[ \frac{x}{8} = \frac{1}{4} \] \[ x = 2 \text{ litres} \] Thus, the amount of mixture released each time is \( 2 \) litres.