Two water taps together can fill a tank in
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 time taken by the tap of larger diameter be t and smaller diameter be t + 10
Part of tank filled in 1 hour by the tap of larger diameter ![]()
Part of tank filled in 1 hour by the tap of smaller diameter ![]()
Part of tank filled in 1 hour when both the taps were working together ![]()
So we can say:
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⇒ 150t + 750 = 8(t2 + 10t)
⇒ 8t2 - 70t - 750 = 0
⇒ 4t2 - 35t - 375 = 0
Performing factorization we get:
⇒ 4t2 - 60t + 25t - 375 = 0
⇒ 4t(t - 15) + 25(t - 15) = 0
⇒ (4t + 25)(t - 15) = 0
t ![]()
Since time is always a positive quantity, so
Time taken by the tap of larger diameter = 15 h
Time taken by the tap of smaller diameter = (15 + 10) = 25 h
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