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Comment by misterprime | Hacker News Reader

The struggle is real. Someone help me out here please:

>5. Two old ladies left from A to B and from B to A at dawn heading to-wards one another (along the same road). They met at noon, but did not stop, and each of them carried on walking with the same speed. The first lady came (to B) at 4pm, and the second (to A) at 9pm. What time was the dawn that day?

I think dawn was at 6am. This assumes that Old Lady A travels the full distance in 10 hours and Old Lady B takes 15 hours to travel the same distance.

I don't know how I arrived at that answer other than some sort of trial and error. I don't like it.

Any advice?

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Say the middle point is M. We know that for each person the ratio of time to go between M->A should be the same as to go between M->B since the distance is fixed only the rates are different.

First lady got to B at 4 pm so her M->B is 4. Second lady got to A at 9 pm so her M->B is 9. Since they both left in the morning, if we call the first lady's M->A time to be x, we know that the second lady's M->B has to be x as well.

Using the fact that the ratio of M->A : M->B should be the same for both of them, we can set up the equation

4/x = x/9 ==> x^2 = 36 ==> x = 6

Since getting to M was at noon and 12-6=6, dawn was at 6 am.

(For what it's worth, I did trial and error too cause I was lazy and didn't feel like thinking... then noticed the nice pattern and back justified it. :P)

The text is confusing to me.

>and each of them carried on walking with the same speed

I thought the "same speed" meant lady1's speed = lady2's which made no sense...

That's exactly how I read it. It was the first obstacle to overcome. "OK, they're walking different speeds, and the statement means that neither of them changed their own unique speed over the course of their walk."

It took a bit of convincing myself, but eventually I became confident that it did not mean that they were each walking at the same rate as each other after their lunch meeting.

But even then, I struggled with figuring out how I arrived at 6AM.

I like the method of switching to the 24 hour clock (as opposed to the AM/PM clock), and setting their arrival times as 16:00 and 21:00, with their meeting at 1200, then solving it.

just summing it up, v{i} are ladies speeds:

    MA = v1*x = v2*9
    MB = v1*4 = v2*x
    → v1 = 9*v2/x
    → 9*4*v2/x = v2*x
    → x = 6

After their noon rendezvous: One lady traveled some distance in 9 hours, and the other lady traveled some other distance in 4 hours. Before their noon rendezvous: you switch both the distances and the ladies. If x is the number of hours before noon that they traveled, you thus get that the ratio of 9 to x (what the first lady covered in 9 hours, the second covered in x) is that of x to 4 (what the first lady covered in x hours, the second covered in 4). Thus x is the geometric mean of 9 and 4, or in other words, x=6 and sunrise was 6 hours before noon at 6am.

Simply translate the words into equations:

  (16 - t)v = d
  (21 - t)w = d
  (12 - t)(v + w) = d

Then, if you substitute v and w in the third line, it turns out that d cancels, and you are left with a quadratic formula of t, one of the two solutions being 6 (and the other one making less sense).

Let C be the point where they meet in the middle and s be dawn.

It takes Lady 1 `12 - s` time to walk AC and 4 time to walk BC. It takes Lady 2 `12 - s` time to walk BC and 9 time to walk AC.

Let's find the ratio of time it takes Lady1 to walk a distance to time it takes Lady2 to walk a distance. The ratio should be the same for all distances since their speed doesn't change.

Thus, (12-s)/9 = 4/(12-s) <=> s^2 -24s + 108 = 0. The quadratic formula gives us s = 6, 18. Since 0 <= s < 12, s = 6.

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