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Question:

An overhead tank, which supplies water to a settlement, is filled by three bore wells. The first two bore wells operating together fill the tank in the same time as taken by the third bore well to fill it. The second bore well fills the tank 10 hours faster than the first one and 8 hours slower than the third one. The time required by the third bore well to fill the tank alone is:

Explanation:

Let the first bore well fill the tank in x hours. 

Hence, the second fills it in (x − 10) hours and the third fills it in (x − 10) − 8 = (x − 18) hours. 

Let the total capacity of the tank to be filled be (x)(x − 10)(x − 18) and let the individual quantity filled by the three respective bore wells be a, b and c. 

∴ a = (x − 10)(x − 18); b = (x)(x − 18); c = (x)(x − 10) 
 
Since the first two bore wells fill the tank in the same time as the third bore well alone, amount of work done by first two bore wells per hour is the same as amount of work done by the third bore well per hour. 

i.e. a + b = c

∴ (x − 10)(x − 18) + (x)(x − 18) = (x)(x − 10)

∴ (x − 18)(2x − 10) = (x)(x − 10)

∴ 2x² − 46x + 180 = x² − 10x

∴ x² − 36x + 180 = 0 

∴ x = 30 or x = 6
 
Since (x − 18) is number of hours, this value has to be positive. Hence, x > 18 i.e. x = 30

∴ Time taken by third borewell = (x − 18) = 12 hours

Hence, option (b).