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Question: Two persons start from opposite ends of a $90 \,\text{km}$ straight track and run to and fro between the two ends. The speed of the first person is $108 \; \text{km/hour}$ while that of the second person is $75\; \text{km/hour}$. They continue their motion for $10$ hours. How many times do they pass each other?

My Attempt: It is easy see that for the first time they will meet after $\frac{90}{108+75}$ hours, after that they collectively would travel $2 \cdot 90$ km further at the speed of $183$ km/h to meet for the second time. Using this we can say that they meet for the $n^{\text{th}}$ time after $$\frac{(2n-1) \cdot 90}{108+75} \text{hours}$$ By putting in some values we see that to meet for the $10^{\text{th}}$ time, the two guys would require $ \approx9.344$ hours and to meet for the $11^{\text{th}}$, they would require $ \approx 10.327$ hours. This ultimately means that they meet $10$ times during their motion.

However, my friend used another method and got $12$ as the answer. He said that in $10$ hours the first person would have travelled $1080$ km which ultimately means that he would have traversed the distance of $90$ km track $12$ times and hence would have met the other guy $12$ times.

I want to know why our answers are differing and which one is right?

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Intuitively I expected you to be correct. However after some messy graphical analysis it turns out your friend is.

The crossings are at [0.49, 1.48, 2.46, 2.73, 3.44, 4.43, 5.41, 6.39, 7.38, 8.18, 8.36, 9.34] hrs (2 d.p.). There are 12 in total. Your equation misses Crossings 4, 10 and 12 at 2.73, 8.18 and 9.34 respectively.

I think the reason for this is that sometimes they do not have to cover 180 miles for their next meeting.

I expect the distance (90 km) being an even factor of the total time (10 hours) multiplied by the faster speed (108 kmph) is also a reason why your friends solution works. (1080/90 = even whole number). However I'm not sure.

The fourth crossing

If you want to make this yourself the inputs were

u(x)=90+ (-1)^(m+1)floor((m)/(2))180+ (-1)^(m)108x

and

t(x)= (-1)^(n)floor((n)/(2))180+ (-1)^(n+1)75x

set up sliders and set n=3. flip between m=3 and m=4 to see the first 'short' crossing.

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