Answer

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Hint: One of the multiples of the LCM of the velocities will be the distance travelled by them to meet again. As the path is circular, it is certain that they will meet again otherwise, they will never meet.

Complete step-by-step answer:

Let us assume that the three cyclists meet again after x days from starting.

Therefore, the first cyclist travelled \[48x\] km before the meeting. Similarly, the second and third cyclist travelled \[60x\] and \[72x\] km respectively.

Since the path is circular with circumference 360 km, after travelling 360 km they cross the starting point again.

Therefore, we can write \[48x-360{{k}_{1}}=60x-360{{k}_{2}}=72x-360{{k}_{3}}\] where \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] integers are as they meet finally and after 360km, they travel a whole circle completely.

Now, from first and second equations, we get,

\[12x=360({{k}_{2}}-{{k}_{1}})\]

Or,\[x=30({{k}_{2}}-{{k}_{1}})\]

Similarly, taking second and third equation or third and first equation we get,

\[x=30({{k}_{3}}-{{k}_{2}})\] and \[x=15({{k}_{3}}-{{k}_{1}})\] respectively.

So, \[30({{k}_{2}}-{{k}_{1}})=30({{k}_{3}}-{{k}_{2}})\]

Or, \[{{k}_{2}}-{{k}_{1}}={{k}_{3}}-{{k}_{2}}\]

So, we can see \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] are in Arithmetic Progression.

Hence, we can write \[{{k}_{1}}={{k}_{2}}-d\] and \[{{k}_{3}}={{k}_{2}}+d\] where d is the common difference of the A.P.

So, \[x=30d\]

Now d cannot be 0 or negative as x is positive.

So, the minimum integral value of d is one. Hence for d=1, we get x = 30

Therefore, they meet after 30 days.

Hence, the correct option for the given question is option (d) 30 days.

Note: After completing each complete cycle, they are just passing the starting point. If we consider the LCM of the velocities, we will get it as 720, which can be individually covered by them in 10, 12 and 15 days respectively. Hence, the distance they meet again must be a multiple of 720. If we multiply 30 with the velocities of all of them, we can clearly see that they are all multiples of 720.

Complete step-by-step answer:

Let us assume that the three cyclists meet again after x days from starting.

Therefore, the first cyclist travelled \[48x\] km before the meeting. Similarly, the second and third cyclist travelled \[60x\] and \[72x\] km respectively.

Since the path is circular with circumference 360 km, after travelling 360 km they cross the starting point again.

Therefore, we can write \[48x-360{{k}_{1}}=60x-360{{k}_{2}}=72x-360{{k}_{3}}\] where \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] integers are as they meet finally and after 360km, they travel a whole circle completely.

Now, from first and second equations, we get,

\[12x=360({{k}_{2}}-{{k}_{1}})\]

Or,\[x=30({{k}_{2}}-{{k}_{1}})\]

Similarly, taking second and third equation or third and first equation we get,

\[x=30({{k}_{3}}-{{k}_{2}})\] and \[x=15({{k}_{3}}-{{k}_{1}})\] respectively.

So, \[30({{k}_{2}}-{{k}_{1}})=30({{k}_{3}}-{{k}_{2}})\]

Or, \[{{k}_{2}}-{{k}_{1}}={{k}_{3}}-{{k}_{2}}\]

So, we can see \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] are in Arithmetic Progression.

Hence, we can write \[{{k}_{1}}={{k}_{2}}-d\] and \[{{k}_{3}}={{k}_{2}}+d\] where d is the common difference of the A.P.

So, \[x=30d\]

Now d cannot be 0 or negative as x is positive.

So, the minimum integral value of d is one. Hence for d=1, we get x = 30

Therefore, they meet after 30 days.

Hence, the correct option for the given question is option (d) 30 days.

Note: After completing each complete cycle, they are just passing the starting point. If we consider the LCM of the velocities, we will get it as 720, which can be individually covered by them in 10, 12 and 15 days respectively. Hence, the distance they meet again must be a multiple of 720. If we multiply 30 with the velocities of all of them, we can clearly see that they are all multiples of 720.

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