Water is filled in a container in such a manner that its volume doubles after every five minutes. If takes 30 minutes for the container to be full, in how much time will it be one-fourth full ?

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Q: 142 (IAS/2001)
Water is filled in a container in such a manner that its volume doubles after every five minutes. If takes 30 minutes for the container to be full, in how much time will it be one-fourth full ?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,4,2,2,0,4,0

keywords: 

{'volume': [0, 0, 1, 0], 'minutes': [0, 0, 1, 1], 'seconds': [3, 3, 8, 6], 'container': [1, 0, 2, 2], 'water': [65, 15, 80, 129], 'much time': [1, 0, 3, 3], 'fourth': [1, 0, 6, 4]}

We know that the volume of water doubles every five minutes. Let`s represent the initial volume of the container as V0 and the time it takes for the container to be full as T.

After 5 minutes, the volume doubles, so it becomes 2V0.

After 10 minutes, it doubles again, becoming 4V0.

After 15 minutes, it doubles once more, becoming 8V0.

After 20 minutes, it doubles again, becoming 16V0.

After 25 minutes, it doubles once more, becoming 32V0.

After 30 minutes, it doubles for the final time, becoming 64V0, which is the capacity of the container.

We want to find the time it takes for the container to be one-fourth full. This means we need to find the time at which the volume is 1/4 of the capacity, or 1/4 * 64V0 = 16V0.

From the above calculations, we can see that after 20 minutes, the volume is 16V0, which is one-fourth of the capacity. Therefore, it will take 20 minutes for the container to be one-fourth full.

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