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

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Re: Amar and Rajesh are running on a circular track. Their 8th point of me [#permalink]
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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]
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]
Can Somebody explain this . Not understanding how 12 is a factorof X+1. It is equal to X+1­
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Amar and Rajesh are running on a circular track. Their 8th point of me [#permalink]
Can Somebody explain this . Not understanding how 12 is a factorof X+1. It is equal to X+1­

­I solved this question but found it tricky (this topic is a bit of an improvement area for me!). My method was a bit lengthy - sharing the essence of it here in case it helps.

Let C be the circumference (length) of the track.

(1) When will Amar and Rajesh first meet?

t = C/(x+1)

(2) When will they met again?

After another "t". They meet at t,2t,3t, etc...

(3) When they will meet for the 8th time?

At 8t = 8C/(x+1)

(4) When will they meet for the 20th time?

At 20t = 20C/(x+1)

(5) How much distance have Amar and Rajesh covered by the time they meet for the 8th time?

Amar -> 8C(x)/(x+1)
Rajesh -> 8C(1)/(x+1)

(6) How much distance have Amar and Rajesh covered by the time they meet for the 20th time?

Amar -> 20C(x)/(x+1)
Rajesh -> 20C(1)/(x+1)

(7) What does it mean to say that the 8th meeting and the 20th meeting happened at the same spot?

While the total distance travelled by Rajesh (or Amar) varies between 8th and 20th meetings, the distance of the spot from the starting point will be same in the case of both meetings.

Example to understand this: Consider the movement of minutes hand of a clock (cirumference = 60).

A clock that has covered 15 min worth of distance -> (0)(60) + 15
A clock that covered 75 min worth of distance -> (1)(60) + 15

The distance covered varies (15 vs. 75) but the position on the clock with regard to starting position remains the same (the "15 min" mark).

Total Distance = (number of complete revolutions) (distance/revolution) + Remainder = nC+15 (in this case).

(8) Translating the fact that 8th meeting and 20th meeting happened at the same spot in a similar fashion

Let's pick Rajesh's journey (easier as there is one less "x" to deal with when looking at distance covered).

Distance covered by the time of 8th meeting = 8C(1)/(x+1) = (n1)(C) + Some remainder
Distance covered by the time of 20th meeting = 20C(1)/(x+1) = (n2)(C) + Some remainder

Same spot => The remainder is the same in both cases

Subtracting one from the other ->

12C / (x+1) = (n2-n1)C
=> 12 = (n2-n1)(x+1)

n1, n2 are integers (number of full revolutions). So, n2-n1 is also an integer. Then (x+1) is also an integer.

(9) Considering the possible scenarios for 12 = (n2-n1)(x+1)

(n2-n1) x (x+1) || x = <>
1 x 12 || x = 11
2 x 6 || x = 5
3 x 4 || x = 3
4 x 3 || x = 2
6 x 2 || x = 1
12 x 1 || x = 0.

Check the choices -> 5,2,11,3 are present. The only one in the choices which is not a possible value of x is 4 (Choice C).

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