GATE CS 2011 Q.65 (GA Q.65): Quantitative Aptitude

A container originally contains 10 litres of pure spirit. From this container 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this process is repeated one more time. How much spirit is now left in the container?

7.58 litres

7.84 litres

7 litres

7.29 litres

Official answer and solution

Official answer: D
From: Official GATE 2011 answer key

Solution

Apply the standard formula for successive dilutions where a fixed volume of mixture is repeatedly replaced by water.

Let $V = 10$ litres be the initial total capacity containing pure spirit, and $x = 1$ litre be the quantity drawn out and replaced with water in each operation.

The dilution operation is performed 3 times in total:

Step 1: 1 litre replaced with water.
Step 2: 1 litre mixture replaced with water.
Step 3: Repeated one more time.

The amount of pure spirit remaining after $n$ successive operations is given by:

\text{Remaining spirit} = V \left(1 - \frac{x}{V}\right)^n

For $V = 10$ , $x = 1$ , and $n = 3$ :

\text{Remaining spirit} = 10 \left(1 - \frac{1}{10}\right)^3 = 10 \times (0.9)^3 = 10 \times 0.729 = 7.29 \text{ litres}

Step-by-step verification:

Initial: 10 L spirit, 0 L water.
After operation 1: $10 - 1 = 9$ L spirit, 1 L water.
After operation 2: $1$ L removed contains $0.9$ L spirit $\implies 9 - 0.9 = 8.1$ L spirit, 1.9 L water.
After operation 3: $1$ L removed contains $0.81$ L spirit $\implies 8.1 - 0.81 = 7.29$ L spirit, 2.71 L water.

Hence, exactly 7.29 litres of spirit is left in the container.

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2011 Computer Science and Information Technology (CS)
Question: 65 of 65, printed as GA Q.65
Section: General Aptitude

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