Question

Tom, Dick and Harry started out a 100km journey. Tom and Harry went by automobile at the rate of \[25\text{ km/hr}\], while Dick walked at the rate of \[5\text{ km/hr}\]. After a certain distance, Harry got off and walked on at \[5\text{ km/hr}\], while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:
A. 5
B. 6
C. 7
D. 8

Answer 1

Hint: The statement “a person moves or travels at a rate of \[\text{x km/hr}\]” can be interpreted as the person can travel x kilometres in 1 hour as it is \[\text{x km/hr}\]. Rate can also be termed as velocity or speed depending upon whether the displacement or distance is given. Here there isn’t any mentioning about the direction so we will consider it as the speed of the person. \[\text{Speed(s)=}\dfrac{\text{Distance travelled(d)}}{\text{Time elapsed(T)}}\].

Complete step by step answer:
We have been given that the total length of the journey is 100km. Initially, Tom and Harry go at a rate of \[25\text{ km/hr}\] on a motorcycle, while Dick walks at a rate of \[5\text{ km/hr}\].
Let us assume that Tom and Harry travelled a distance of x km, then as the speed of Dick is one fifth of the speed of Tom, Dick would have travelled a distance of \[\dfrac{x}{5}\text{ km}\].
Distance travelled by Tom and Harry before Harry split up in time t is equal to \[\text{x km}\].
Distance travelled by Dick in time t is equal to \[\dfrac{x}{5}\text{ km}\].
Now, at time t Tom comes back for Dick and Harry continues to walk at a rate of \[5\text{ km/hr}\]. Here, the distance of \[\left( x-\dfrac{x}{5} \right)\text{ km}\] would be travelled by Tom and Dick for the same time interval.
Let the distance travelled by Tom be A and by Dick be B, the speed of Tom is \[25\text{ km/hr}\] as he is on a motorcycle and that of Dick is \[5\text{ km/hr}\].
\[\Rightarrow A+B=\left( x-\dfrac{x}{5} \right)=\dfrac{4x}{5}\text{ km}\]
Dick and Tom meet at the same time. And, we know that
\[\Rightarrow \text{Speed(s)=}\dfrac{\text{Distance travelled(d)}}{\text{Time elapsed(T)}}\to \text{Time elapsed(T)=}\dfrac{\text{Distance travelled(d)}}{\text{Speed(s)}}\]
Both meet at the same time. So, \[\dfrac{A}{5}=\dfrac{B}{25}\].
\[\Rightarrow 5A=B\].
Substituting the value of B as 5A in \[A+B=\dfrac{4x}{5}\], we get
\[\begin{align}
  & \Rightarrow A+5A=\dfrac{4x}{5} \\
 & \Rightarrow A=\dfrac{4x}{30}=\dfrac{2x}{15} \\
\end{align}\]
Since \[B=5A\],
\[\Rightarrow B=5\left( \dfrac{2x}{15} \right)=\dfrac{10x}{15}=\dfrac{2x}{3}\]
Finally, Tom and Dick start to move at \[25\text{ km/hr}\] to complete 100km and Harry is also moving at a rate of \[5\text{ km/hr}\] to complete the 100km.
All three of them meet at the same time. So, time taken by Tom and Dick to meet each other and the time taken by Tom and Dick to reach the end is equal to the time taken by Harry.
The time taken by Tom and Dick to meet each other is \[\dfrac{1}{5}\left( \dfrac{2x}{15} \right)=\dfrac{2x}{75}\] and the time taken by Tom and Dick together is equal to \[\dfrac{1}{25}\left( 100-x+\dfrac{2x}{3} \right)\] and the time taken by harry is equal to \[\dfrac{1}{5}(100-x)\]. Thus, we can write
\[\begin{align}
  & \Rightarrow \dfrac{2x}{75}+\dfrac{1}{25}\left( 100-x+\dfrac{2x}{3} \right)=\dfrac{1}{5}\left( 100-x \right) \\
 & \Rightarrow \dfrac{2x}{75}+\dfrac{1}{25}\left( 100-\dfrac{x}{3} \right)=\dfrac{1}{5}\left( 100-x \right) \\
 & \Rightarrow \dfrac{2x}{75}+\dfrac{300-x}{75}=\dfrac{100-x}{5} \\
 & \Rightarrow \dfrac{2x+300-x}{75}=\dfrac{100-x}{5} \\
 & \Rightarrow 300+x=\dfrac{75}{5}\left( 100-x \right)=15(100-x) \\
 & \Rightarrow 300+x=1500-15x \\
 & \Rightarrow x+15x=1500-300 \\
 & \Rightarrow 16x=1200 \\
 & \Rightarrow x=75\text{ km} \\
\end{align}\]
\[\therefore \] Total time taken is equal to \[\dfrac{75}{25}+\dfrac{25}{5}=3+5=8\text{ hrs}\]
\[\therefore \] The correct option is D.

Note: Students often make mistakes by considering the given rate as velocity and then they tend to calculate the displacement instead of distance. But it should be observed clearly that the direction of any of the three travellers isn’t specified so it should be considered as distance and the given speeds should be noted down precisely while calculating.