A tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.

Answer: The bigger tap takes 10 hours and the smaller tap takes 15 hours.

Let the bigger tap alone fill the tank in \(x\) hours. Then the smaller tap takes \((x + 5)\) hours.

In 1 hour the bigger tap fills \(\dfrac{1}{x}\) and the smaller tap fills \(\dfrac{1}{x + 5}\) of the tank. Together they fill \(\dfrac{1}{6}\) of the tank in 1 hour.

\(\dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{1}{6}\)

\(6(x + 5) + 6x = x(x + 5)\)

\(12x + 30 = x^2 + 5x \Rightarrow x^2 - 7x - 30 = 0\)

\(x^2 - 10x + 3x - 30 = 0 \Rightarrow (x - 10)(x + 3) = 0\)

\(x = 10\) or \(x = -3\). Time cannot be negative, so \(x = 10\).

Bigger tap: 10 hours; smaller tap: \(10 + 5 = 15\) hours.

Check: \(\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3 + 2}{30} = \dfrac{1}{6}\).