A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

13 hrs 15 min
23 hrs 45 min
34 hrs
44 hrs 15 min

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Time taken by one tap to fill half of the tank = 3 h r s . \\ Part filled by the four taps in 1hour \\ = (4 \times \frac{1}{6}) = \frac{2}{3} \\ Remaining part = (1 - \frac{1}{2}) = \frac{1}{2} \\ ∴ \frac{2}{3} : \frac{1}{2} :: 1 : x \\ \Rightarrow x = (\frac{1}{2} \times 1 \times \frac{3}{2}) \\ = \frac{3}{4} h o u r s i . e ., 45 mins . \\ So, total time taken = 3 h r s . 45 m i n s . \end{aligned}

$\eqalign{ & {\text{Time}}\,{\text{taken}}\,{\text{by}}\,{\text{one}}\,{\text{tap}}\,{\text{to}}\,{\text{fill}}\,{\text{half}}\,{\text{of}}\,{\text{the}}\,{\text{tank}} = 3\,hrs. \cr & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,{\text{the}}\,{\text{four}}\,{\text{taps}}\,{\text{in}}\,{\text{1hour}} \cr & = \left( {4 \times \frac{1}{6}} \right) = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{part}} = \left( {1 - \frac{1}{2}} \right) = \frac{1}{2} \cr & \therefore \frac{2}{3}:\frac{1}{2}::1:x \cr & \Rightarrow x = \left( {\frac{1}{2} \times 1 \times \frac{3}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{4}\,hours\,i.e.,\,45\,\operatorname{mins} . \cr & {\text{So,}}\,{\text{total}}\,{\text{time}}\,{\text{taken}} = 3\,hrs.\,45\,mins. \cr}$