Two water taps together can fill a tank in 9 ⅜ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Now, according to the question, we can write as

• 1/x + 1/x - 10 = 75/8
• (x - 10 + x)/ x (x - 10) = 8/75
• 75 (2x - 10) = 8x (x - 10)
• 150x - 750 = 8x2 - 80x

On rearranging we get

• 8x2 - 80x - 150x + 750 = 0
• 8x2 - 230x + 750 = 0
• 4x2 - 115x + 375 = 0
• 4x2 - 100x - 15x + 375 = 0
• 4x (x - 25) - 15 (x - 25) = 0
• (4x - 15) (x - 25) = 0

On simplification we get

• 4x - 15 = 0 x - 25 = 0
• 4x = 15x = 25
• x = 15/4

There are currently two x values.

Consequently, there will be two cases.

• Case (a) - When x = 15/4 hours, then
• (x - 10) = 15/4 - 10 = -25/4 hours

Time cannot be negative, therefore. Consequently, this scenario is not feasible.

• Case (b) - When x = 25 hours
• (x - 10) = 25 - 10 = 15 hours

As a result, the smaller tap can independently fill the tank in 25 hours, while the larger tap needs 15 hours to do so.

Summary:

Two water taps together can fill a tank in 9 ⅜ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Two water taps together can fill a tank in 9 ⅜ hours. The larger diameter tap fills the tank independently in 10 hours less time than the smaller one. Each tap can fill the tank independently for 25 hours for smaller diameters and 15 hours for larger diameters.