Question

The radius of a circular track is 63m. Two cyclists X and Y start together from the same position at the same time and in the same direction with speeds of 33 m/min and 44 m/min respectively. After how many minutes will they meet again at the starting point?

Answer 1

Hint: Assume the given radius of the circular track as ‘r’ and \[{{v}_{1}}\] and \[{{v}_{2}}\] as the speeds of cyclists X and Y respectively. Now, apply the formula: - $\text{Speed} = \dfrac{\text{(Distance)}}{\text{(Time)}}$ to calculate time taken by the two cyclists to complete one round of the track. Assume \[{{t}_{1}}\] and \[{{t}_{2}}\] as the time taken by cyclists X and Y respectively. Take L.C.M of these two times to find the time after which X and Y will meet again at the starting point.

Complete step-by-step solution
Here, we have been provided with a circular track of radius 63m. We have to find the time after which the two cyclists will meet at the starting point given that they have started from the same position in the same direction at different speeds.
Now, let us assume the radius of the circular track as ‘r’, the velocities or speeds of cyclists X and Y as \[{{v}_{1}}\] and \[{{v}_{2}}\] respectively and time is taken by X and Y as \[{{t}_{1}}\] and \[{{t}_{2}}\] respectively, to complete one round. So, when they will complete one round they will cover a distance equal to the circumference of the circular track. So, we have,
1. For cyclist X: -
Speed = \[{{v}_{1}}\] = 33 m/min
Time taken to complete 1 round = \[{{t}_{1}}\]
Distance travelled = circumference = \[2\pi r\], where r = 63m
Applying the formula: - Speed = (distance / time), we get,
\[\begin{align}
  & \Rightarrow {{v}_{1}}=\dfrac{2\pi r}{{{t}_{1}}} \\
 & \Rightarrow {{t}_{1}}=\dfrac{2\pi r}{{{v}_{1}}} \\
\end{align}\]
Substituting the given values and \[\pi =\dfrac{22}{7}\], we get,
\[\Rightarrow {{t}_{1}}=\dfrac{2\times \dfrac{22}{7}\times 63}{33}\]
\[\Rightarrow {{t}_{1}}=12\] minutes
So, the time taken by cyclist X to complete one round is 12 minutes.
2. For cyclist Y: -
Speed = \[{{v}_{2}}=44\] m/min
Time taken to complete one round = \[{{t}_{2}}\]
Distance travelled = circumference = \[2\pi r\], where r = 63m
Applying the formula: - Speed = (distance / time), we get,
\[\begin{align}
  & \Rightarrow {{v}_{2}}=\dfrac{2\pi r}{{{t}_{2}}} \\
 & \Rightarrow {{t}_{2}}=\dfrac{2\pi r}{{{v}_{2}}} \\
\end{align}\]
Substituting the given values and \[\pi =\dfrac{22}{7}\], we get,
\[\Rightarrow {{t}_{2}}=\dfrac{2\times \dfrac{22}{7}\times 63}{44}\]
\[\Rightarrow {{t}_{1}}=9\] minutes
So, the time taken by cyclist Y to complete one road is 9 minutes.
Now, the time after which they will meet again at the starting point will be the L.C.M of \[{{t}_{1}}\] and \[{{t}_{2}}\]. Here, \[{{t}_{1}}\] and \[{{t}_{2}}\] are 12 minutes and 9 minutes respectively. To find L.C.M let us write the multiples of 12 and 9.
\[\Rightarrow \] Multiples of 12 = 12, 24, 36, 48, ……
\[\Rightarrow \] Multiples of 9 = 9, 18, 27, 36, ……..
Clearly, we can see that the first common multiple is 36. Hence, cyclists, X, and Y will meet at the starting point again after 36 minutes.

Note: One may note that here we do not have to calculate H.C.F of 12 and 9 to get the required time. If we do so then we will get a time lower than both 12 minutes and 9 minutes which can never be the meeting time. Note that we can also apply the process of prime factorization to calculate the L.C.M of 12 and 9. But here the process of finding multiple was easier because the numbers were small. You must remember the distance-time formula to solve the question.