Question

A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?

Answer 1

Let the larger tap take x hours to fill the tank completely. 

∴ Part of tank filled by the larger tap in 1 hour = 1/x 

Also, the smaller tap takes (x + 3) hours to fill the tank completely. 

∴ Part of tank filled by the smaller tap in 1 hour = 1/(x+3) 

∴Part of tank filled by both the taps in 1 hour 

= ( 1/x + 1/(x + 3) But, the tank gets filled in 2 hours by both the taps.

∴ Part of tank filled by both the taps in 1 hour =1/2

 According to the given condition,

1/x + 1/(x + 3) = 1/2

∴ (x + 3 + x)/x(x + 3) = 1/2

∴ (2x + 3)/x(x + 3) = 1/2

∴ 2(2x + 3) = x(x + 3) 

∴ 4x + 6 = x² + 3x 

∴ x² + 3x – 4x – 6 = 0 

∴ x² – x – 6 = 0 

∴ x² – 3x + 2x – 6 = 0 

∴ x(x – 3) + 2(x – 3) = 0 

∴ (x – 3)(x + 2) = 0 By using the property, if the product of two numbers is zero, then at least one of them is zero, we get 

∴ x – 3 = 0 or x + 2 = 0

∴ x = 3 or x = -2 

But, time cannot be negative. 

∴ x = 3 and x + 3 = 3 + 3 = 6 

∴ The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.