An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be :

The same quantity of heat Q is needed to boil the water each time, and the supply voltage V is the same.
$Q = \dfrac{V^2}{R_1}t_1 = \dfrac{V^2}{R_2}t_2 = \dfrac{V^2}{R}t$, where R is the resistance of the parallel combination.

For parallel coils: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$
From the heat equation, $\dfrac{1}{R} = \dfrac{Q}{V^2 t}$, so $\dfrac{Q}{V^2t} = \dfrac{Q}{V^2t_1} + \dfrac{Q}{V^2t_2}$
$\Rightarrow \dfrac{1}{t} = \dfrac{1}{t_1} + \dfrac{1}{t_2} \Rightarrow t = \dfrac{t_1t_2}{t_1 + t_2}$
$t = \dfrac{10 \times 40}{10 + 40} = 8$ min