A bus starts at \(6: 00 \mathrm{pm}\). from starting point at the speed of \(18 \mathrm{~m} / \mathrm{s}\), reached its destination and waited for 40 minutes. And again returned back at the speed of \(28 \mathrm{~m} / \mathrm{s}\). If the time taken in forward journey is same as time taken in reverse and waiting time. What will be the time when it reaches again at its starting point?
A. 8:32 PM B. 9:44 PM C. 7:52 PM D. 9:56 PM E. Data Insufficient
Correct Answer: B
Explanation
Say distance is \(\mathrm{d} \mathrm{m}\). Time taken \(=\) Distance \(/\) Speed Time taken in forward journey \(=\mathrm{d} / 18\) seconds. Time taken in reverse journey \(=\mathrm{d} / 28\) seconds. Given that, \(d / 18=d / 28+(40 \times 60) \quad\) [40 minutes \(=40 \times 60\) seconds] \(\Rightarrow \mathrm{d} \times(5 / 252)=2400\) \(\Rightarrow \mathrm{d}=2400 \times 252 / 5=120960 \mathrm{~m}\) \(\therefore\) Total time taken \(=(\mathrm{d} / 18)+(\mathrm{d} / 28)+2400=13440\) seconds \(\Rightarrow\) Time taken \(=13440 / 60\) minutes \(=224\) minutes \(=3\) hours and 44 minutes \(\therefore\) The bus reaches back at 9:44 PM
