Question

Gopal sells fruit juice mixture using orange juice and pineapple juice. Gopal prepares this mixture by drawing out a jug of orange juice from a 10 litre container filled with orange juice, and replacing it with pineapple juice. If Gopal draws out another jug of the resultant mixture and replaces it with pineapple juice, the container will have equal volumes of orange juice and pineapple juice. The volume of the jug in litres, is

• A. 2
• B. \(<\) 2 and \(\leq 2.5\)
• C. 2.5
• D. > 2.5 and \(\leq 3\)
• E. \(\geq 3\)

Hint:

- Always use the principle of mixture and proportion: when removing and replacing, reduce each component proportionally.
- Quadratic equations often arise in repeated replacement problems.

Answer 1

The Correct Option is D

Step 1: Initial situation
The container has 10 litres of orange juice initially. Let the jug capacity be \(x\) litres. 
Step 2: After the first replacement 
- Orange juice removed = \(x\). Remaining orange juice = \(10 - x\). 
- Pineapple juice added = \(x\). 
So, mixture now: Orange = \(10 - x\), Pineapple = \(x\). Total = 10 litres. 
Step 3: After the second replacement 
When another jug of \(x\) litres is removed, the fraction of orange juice in the mixture is: \[ \frac{10 - x}{10}, \quad \text{and fraction of pineapple juice} = \frac{x}{10}. \] So, in \(x\) litres removed: - Orange removed = \(\frac{10-x}{10} \times x = \frac{x(10-x)}{10}\). 
- Pineapple removed = \(\frac{x}{10} \times x = \frac{x^2}{10}\). 
Step 4: Remaining after removal 
Remaining orange = \((10 - x) - \frac{x(10-x)}{10} = (10-x)\left(1 - \frac{x}{10}\right) = \frac{(10-x)^2}{10}\). 
Remaining pineapple = \(x - \frac{x^2}{10} = \frac{x(10-x)}{10}\). 
Step 5: After refilling with pineapple juice 
Pineapple increases by \(x\). 
So, pineapple = \(\frac{x(10-x)}{10} + x = \frac{x(10-x) + 10x}{10} = \frac{x(20-x)}{10}\). 
Step 6: Condition for equality 
We need: \[ \frac{(10-x)^2}{10} = \frac{x(20-x)}{10}. \] \[ (10-x)^2 = x(20-x). \] \[ 100 - 20x + x^2 = 20x - x^2. \] \[ 2x^2 - 40x + 100 = 0. \] \[ x^2 - 20x + 50 = 0. \] Step 7: Solve quadratic 
\[ x = \frac{20 \pm \sqrt{400 - 200}}{2} = \frac{20 \pm \sqrt{200}}{2} = \frac{20 \pm 10\sqrt{2}}{2}. \] \[ x = 10 \pm 5\sqrt{2}. \] Numerical values: - \(x = 10 - 5\sqrt{2} \approx 10 - 7.07 = 2.93\). - \(x = 10 + 5\sqrt{2} \approx 17.07\) (not possible since jug < 10 litres). So, valid jug size \(x \approx 2.93\). Step 8: Compare with options 
This lies in the range \(> 2.5\) and \(\leq 3\). Hence, the correct option is (D). \[ \boxed{2.93 \, \text{litres (approx.)}} \]