Two water taps together can fill a tank in 17/8 hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Given two taps can fill a tank in 17/8=15/8 hours
Amount of water filled by two taps in 1 hour is 1/(15/8)=8/15
Let the time taken by longer diameter tap to fill the tank be x hrs
the time taken by other tap to fill the tank be y hrs
from the given condition x = y-2
Amount of water filled by longer diameter tap in 1 hour is 1/x
Amount of water filled by smaller diameter tap in 1 hour is 1/y
Amount of water filled by both taps in 1 hour is 1/x + 1/y = 8/15
1/y-2 + 1/y = 8/15
(2y-2)/(y(y-2)) = 8/15
15(2y-2) = 8y(y-2)
30y - 30 = 8y² - 16y
8y² - 46y +30 = 0
4y² - 23y + 15 = 0
4y² - 20y -3y +15 = 0
4y(y-5) -3(y-5) = 0
(4y-3)(y-5) = 0
y = 3/4, 5
x = y-2
If y = 3/4, x = 3/4 -2 = -5/4 (not possible)
If y = 5, x = 5-2 = 3
the time taken by longer diameter tap to fill the tank be 3 hrs
the time taken by other tap to fill the tank be 5 hrs