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Amar and Rajesh are running on a circular track. Their 8th point of me

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Amar and Rajesh are running on a circular track. Their 8th point of me  [#permalink]

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New post Updated on: 31 Mar 2020, 11:15
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Amar and Rajesh are running on a circular track. Their 8th point of meeting is same as their 20th me.If they are running in the opposite direction and the ratio of their speeds is x:1 (x is a natural number), which cannot be value of x?

a. 5
b. 2
c. 4
d. 11
e. 3

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Please highlight if you found this helpful, Thanks in advance

Originally posted by zs2 on 31 Mar 2020, 11:13.
Last edited by Bunuel on 31 Mar 2020, 11:15, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Amar and Rajesh are running on a circular track. Their 8th point of me  [#permalink]

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New post 31 Mar 2020, 21:06
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Total number of distinct meetings = x+1

Since Their 8th point of meeting is same as their 20th, 8 and 20 will leave same remainder when divided by x+1.

8 = a*(x+1) + k, where a and k are positive integers.

20= b*(x+1) +k, where b is a positive integer.

Subtract both equations

12 = (b-1)*(x+1)

or x+1 is a factor of 12

x+1 can be 1, 2 ,3 ,4 ,6 or 12

x can be 0(rejected), 1,2,3,5 or 11.


zs2 wrote:
Amar and Rajesh are running on a circular track. Their 8th point of meeting is same as their 20th me.If they are running in the opposite direction and the ratio of their speeds is x:1 (x is a natural number), which cannot be value of x?

a. 5
b. 2
c. 4
d. 11
e. 3
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Re: Amar and Rajesh are running on a circular track. Their 8th point of me  [#permalink]

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New post 31 Mar 2020, 21:11
nick1816 can you please explain "total number of distinct meetings = x+1 " ?
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Re: Amar and Rajesh are running on a circular track. Their 8th point of me  [#permalink]

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New post 31 Mar 2020, 21:32
Suppose circumference of the track = C

Time taken by them to meet first time= \(\frac{C}{x+1 }\) {Since they're both moving in opposite direction)

Time taken by them to meet at the starting point for the first time =\( LCM(\frac{C}{x}, \frac{C}{1}) = \frac{LCM(C,C)}{HCF(x,1) }= C\)

Number of times they will meet before meeting at the starting point = [C]/[C/x+1] = x+1


preetamsaha wrote:
nick1816 can you please explain "total number of distinct meetings = x+1 " ?
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Re: Amar and Rajesh are running on a circular track. Their 8th point of me  [#permalink]

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New post 01 Apr 2020, 11:22
nick1816 thanks for the explanation.
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Re: Amar and Rajesh are running on a circular track. Their 8th point of me   [#permalink] 01 Apr 2020, 11:22

Amar and Rajesh are running on a circular track. Their 8th point of me

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