Arithmetic

Mixture & Alligation

Hard

applied

From a container full of orange juice, 7.5 liters was drawn out and replaced by soda water. This process is repeated 5 more time. The ratio of quantity of orange juice and soda water left in the container is 4:5. How much liter of orange juice did the container originally had? [(Use:0.44) 1/6 = 0.802]

20.44 litres

30.802 litres

37.9 litres

17.5 litres

✅ Correct Option: 3

A container full of orange juice has 7.5 liters drawn out and replaced by soda water. This process is repeated 5 more times (6 times total). The final ratio of orange juice to soda water is 4:5.

Let $V$ be the original volume of orange juice in liters.

After each removal and replacement, the fraction of orange juice remaining is $\frac{V - 7.5}{V}$ .

After 6 such operations, the amount of orange juice left is:

$V \times \left(\frac{V - 7.5}{V}\right)^6$

The final ratio of orange juice to soda water is 4:5, meaning orange juice forms $\frac{4}{9}$ of the total volume.

Amount of orange juice remaining $= \frac{4}{9} \times V$

Equating the two expressions:

$V \times \left(\frac{V - 7.5}{V}\right)^6 = \frac{4}{9} \times V$

$\left(\frac{V - 7.5}{V}\right)^6 = \frac{4}{9}$

$\left(1 - \frac{7.5}{V}\right)^6 = 0.44$

Taking the 6th root of both sides using the given value $0.44^{1/6} = 0.802$ :

$1 - \frac{7.5}{V} = 0.802$

$\frac{7.5}{V} = 1 - 0.802$

$\frac{7.5}{V} = 0.198$

$V = \frac{7.5}{0.198}$

$V = 37.878...$

$V \approx 37.9$ liters

Therefore, the container originally had 37.9 liters of orange juice.

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Solution

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