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Question 6 : A Container contains X Litres of Milk. A thief stole 50 Litres of Milk and replaced it with the same quantity of water. He repeated the same process further two times. And thus Milk in the container is only X-122 litres. Then what is the quantity of water in the final mixture?

- 122 Litre
- 124 Litre
- 128 Litre
- 250 Litre

122 Litre

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Each time the thief steals 50 liters from the container of volume x liters and replaces it with water, the concentration of the mixture becomes \\frac{x-50}{x}\\) of the original concentration.

Since, initially, we had pure milk, the initial concentration will be 100% or 1.

After doing this replacement for 3 times in total, the concentration becomes, (\\frac{x-50}{x}\\)) * (\\frac{x-50}{x}\\)) * (\\frac{x-50}{x}\\)) * 1

= (\\frac{x-50}{x}\\))^{3}.

It is also told that at the end the volume of milk in the container = x-122.

So the final concentration of milk is also equal to \\frac{x-122}{x}\\)

So, (\\frac{x-50}{x}\\))^{3} = \\frac{x-122}{x}\\)

\\frac{(x-50)^{3}}{x^{3}}\\) = \\frac{x-122}{x}\\)

(x-50)^{3} = x^{2} * (x-122)

x^{3}-3*x*50(x-50)-50^{3} = x^{3}-122x^{2}

x^{3}-150 x^{2} -7500 x-125000 = x^{3}-122x^{2}

-28 x^{2} -7500x -125000 = 0

Solving which we get, either x = 250 or x = 17.85

Since there are x-122 liters of milk at the end x has to be greater than 122. Therefore x=250.

Therefore, Quantity of milk in final mixture = x-122 = 250-122 = 128 liters.

The question is **"A Container contains X Litres of Milk. A thief stole 50 Litres of Milk and replaced it with the same quantity of water. He repeated the same process further two times. And thus Milk in the container is only X-122 litres. Then what is the quantity of water in the final mixture?" **

Choice A is the correct answer

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