Determining value for 'k' for minute and hour hand given a condition in swapping problem

Question

In problems where minute and hour hand interchange their places, I read a couple of posts here.

The following post gives a general formula to get the values of all such possible cases in a day:

After swapping the positions of the hour and the minute hand, when will a clock still give a valid time?

And the following 2 involved application of the same concept but were resolved using different approaches:

1. Clock hands moving to take each others places.
2. Minute hand and hour hand interchange

When I tried the above 2 questions using the general formula, I could not, without the help of excel. I need to be able to do it using pen and paper only and that too in say 2-3 mins.

Here's the general formula that was obtained from the post:

In degrees:- Hour Hand = $\frac{360k}{143}$ and Min Hand = $\frac{360k}{143}*12$

With respect to the 2nd question where the hour hand was between 4 and 5 while the minute hand between 5 and 6 on a clock before the swap, I made the following futile attempt:

$120<\frac{360k}{143}<150$ and $150<\frac{360k}{143}*12<180$

From first we'll get 12 values of k but 2nd I don't know how to work with as it should also be the remainder of 360. Please help so that for any given case like min hand between this and hour hand between that, I can find the value of k.

Answer 1

Between $4$ and $5$, let the gap by which the minute hand is ahead be $z$ hours. Then in the time the hour hand moves to $z$, the minute hand moves by $1-z$

so $\frac{1-z}{z} =12, z = \frac1{13},\; and \;(1-z) = \frac{12}{13}$ hrs.

This is simple enough to compute without calculator !

PS:

OP apparently also wanted the starting and finishing times, and has posted an answer after finding out which particular value of k applies here but it isn't really needed. Here's a simple way.

---

Starting between $4$ and $5$ and ending between $5$ and $6$,counting on the dial, we have the equations

$4 +M/12 = m ..... [I]$

$5+m/12 = M ... [II]$

which yields $M = \frac{768}{143},\;\; m = \frac{636}{143}$

To convert the dial positions for minutes to minutes, we need to multiply by $5$, yielding times as $\approx 4:26.85\;\; and\;\; 5:22.24$

Hope this adds a simple tool to your armoury !

Answer 2

One of the ways to get/know the value of k is by pattern method.

In degrees:- Hour Hand = $\frac{360k}{143}$ and Min Hand = $\frac{360k}{143}*12$

The following can be observed: (keeping the above formula in mind will be helpful)

1. There are a total of 143 (unique) values that k can take, that are from 0 to 142.
2. From k = 143 we'll again start getting the repetition of the above results.
3. For every hour, from 0 to 10, we have 12 different values for swapped time.
4. For 11th hour, we have only 11 values (as 12th value is same as and counted in for the 0th hour).
5. For every hour (except 11), the swapped time's hour hand values are equally distributed from 0 to 11 (thus 12 in total confirming pt 3).
6. Even for the 11th hour, the swapped time's hour values are equally distributed but from 0 to 10 only.

So, we can generalize these results to come up with a formula that can help us get the value of k as follows:

For hour hand between $H_1$ and $(H_1+1)$ hr digits on the clock and the minute's hand between $M_1$ and $(M_1+1)$ hr digits on the clock, $$k=12\times H_1+M_1$$ For example, for your 2nd question, $H_1=4$ and $M_1=5$, so $k=12\times4+5=53$.

Note: $H_1$ & $M_1$ will be $0$ instead of $12$ when between $12$ and $1$ on the clock.

Since it helps seeing the pattern ourselves:

From left to right, the image contains the value of k, corresponding time before and after swap respectively.