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