Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 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 tap of smaller diameter to fill the tank be x hours

The time taken by tap of larger diameter to fill the tank = x – 9 hours

Volume of tank be V

Volume of tank filled by tap of smaller diameter in x hours = V

Volume of tank filled by tap of smaller diameter in 1 hour = V/x

Volume of tank filled by tap of smaller diameter in 6 hours =

Volume of tank filled by tap of larger diameter in 6 hours =

Volume of tank filled by tap of smaller diameter in 6 hours + Volume of tank filled by tap of larger diameter in 6 hours = V

12x – 54 = x^{2} – 9x

x^{2} – 21x + 54 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = – 21 c = 54

= 1.54 = 54

And either of their sum or difference = b

= – 21

Thus the two terms are – 18 and – 3

Difference = – 18 – 3 = – 21

Product = – 18. – 3 = 54

x^{2} – 18x – 3x + 54 = 0

x(x – 18) – 3(x – 18) = 0

(x – 18)(x – 3) = 0

(x – 18) = 0 (x – 3) = 0

x = 18 or x = 3

For x = 3 time taken by tap of larger diameter is negative which is not possible

Thus x = 18

Hence the time taken by tap of smaller diameter to fill the tank be 18 hours

The time taken by tap of larger diameter to fill the tank = 18 – 9 = 9hours

57