Two water taps together can fill a tank in 9(3/8) 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.

Let, the time taken by large tap to fill the tank = x hours
So, the time taken by smaller tap to fill the tank = x + 10 hours
Therefore, in 1 hour,
tank filled by larger tap = 1/x+10
According to question,
1/x + 1/x + 10 = 1/(9 (3/8))

⇒ x + 10 + x/x(x + 10) = 8/75
⇒ 75(2x + 10) = 8x(x + 10)
⇒ 150x +750 = 8x² + 80x
⇒ 8x² – 70x – 750 = 0
⇒ 4x² – 35x – 375 = 0
⇒ 4x² – 60x + 25x – 375 = 0
⇒ 4x(x – 15) + 25 (x – 15) = 0
⇒ (x – 15) (x + 25) = 0
⇒ (x – 15) = 0 or (x + 25) = 0
Either x = 15 or x = – 25
But, x ≠ – 25, as x is the time taken to fill the tank which can’t be negative. So, x
Hence, the time taken by larger top to fill the tank = 15 hours
and the time by smaller tap to fill the tank = 15 + 10 = 25 hours.