Finding time duration when clocks hand interchange position 
Archana began her work somewhere between 3 and 4 pm. When she
  completed her work, somewhere between 5 and 6 pm she noted that hour
  and minute hand positions at beginning of her work were exactly the
  same for minute and hand when she finished. How long did she take?

How to solve this question?
Following are the outcome of a google search

careerride.com gives the choices as  a) $7/13$ min, b) $6/13$
  min, c) $10/13$ min, d) $3/13$ min
careerbless.com gives answer as $24/13$ hour

I am not able to solve this question and also decide the right answer due to the difference in the answers I mentioned.  
I think the problem can be solved using the below concept
minute hand moves 6 degrees in every minute.
hour hand moves 0.5 degree in every minute.
But not able to proceed.
Please help and guide me how to approach this problem.
 A: Leaving between $3$ and $4.$ So, taking the initial position of the hands as
position of hour hand = $3+M/12$
position of minute hand $=M$
Returning between $5$ and $6.$ So, taking the final position of the hands as
position of hour hand $=5+m/12$
position of minute hand $=m$
Since hands interchange
$3+M/12 = m \ ...(1)$
$5+m/12 = M \ ...(2)$
Solving,
$m = 492/143$
$M = 756/143$
initial position: $3 : 756/143$
means, initial time is $3$ hr and ($756/143 \times 5$) minutes
i.e., initial time is $3$ hr and $(3780/143)$ minutes
(we have to multiply with $5$ as M was merely the position. for example, M$=1$ means $5$ minutes)
Final position: $5 : 492/143$
means, final time is $5$ hr and ($492/143 \times 5$) minutes
i.e., final time time is $5$ hr and $(2460/143)$ minutes
So, the difference is $1$ hr and $7260/143$ minutes
i.e., difference in time (in hour)
$=1+(7260/143)/60$ hour
$=1+(121/143)$ hour
$=1+(11/13)$ hour
$=(24/13)$ hour
So, in my calculation, I got the same answer as provided by careerbless.com
(http://www.careerbless.com/qna/discuss.php?questionid=3510)
@trueblueanil, Please let me know if my calculations is wrong or not. I was trying to use the same calculations what you have mentioned. (What I understand is, we had to multiply the M and m with $5$ to get value of minutes because M and m was merely pointing to the digit in the clock)
A: 
Suppose that the hour and the minute hands are initially separated by an angle $\alpha$.
Let us measure the angles as a fraction of $360^\circ$.
The minutes hands moves $12$ times faster than the hours one, so in the time that 
the H hand moves through $\alpha$, the M hand will have moved by $12 \alpha$, and we shall have
(see the sketch) that
$$
12\,\alpha  \equiv  - \,\alpha \quad \left( {\bmod \;1} \right)\quad  \Rightarrow \quad 13\,\alpha  \equiv 0\quad \left( {\bmod \;1} \right)\quad  \Rightarrow \quad \alpha  = n/13
$$
With the same approach, in the time that the H hand spans $\beta$, the M hand will have spanned $12$ times that angle, and this shall equal $\beta + \alpha$
$$
12\beta  \equiv \beta  + \alpha \quad \left( {\bmod \;1} \right)\quad  \Rightarrow \quad 11\beta  \equiv \alpha \quad \left( {\bmod \;1} \right)\quad  \Rightarrow \quad \beta  = n/143 + m/11
$$
where clearly $n$ and $m$ are  integers.
We want to keep $0<\alpha, \beta <1$, therefore $0< n < 13$ and $0< m < 11$.
To express the angles (in fraction of a turn) into hours unit is just a matter to multiply by the number of hours / turn, i.e. by $12$, so:

$$
\left\{ \begin{array}{l}
 t_{\,0}  = 12\,\beta  = 12/143\;n + 12/11\;m\quad (h) \\ 
 \Delta \,t = 12\,\alpha  = 12/13\;n\quad (h) \\ 
 t = 12\,\left( {\beta  + \alpha } \right) = 144/143\;n + 12/11\;m\quad (h) \\ 
 \end{array} \right.
$$  

where the meaning of the symbols should result clear.
Below are reported some values obtained with the formulas above

Thereby it results clear that:


*

*the $\Delta t$ can even be something less than $1$ hour;

*with $n=0$ the various $m$ will give the positions of overlap ;

*starting between $3$ and $4$ o'clock and ending betwee $5$ and $6$ gives the solution in bold.

A: Simple new solution
I take it that the hour and minute hands intechange, as given in the header (in the body, it is vague)
Obviously, the duration has to be more than one hour, so if only one of the choices is more than one hour, you can opt for it straight away.
But even if there is more than one solution $>1\;hr$, there is a much easier way to find it

Let the initial gap between the hour hand and minute hand, at $\approx 3:27,say. =z$ (in hours)
In the time the hour hand covers $z$ hrs, the minute hand has covered two hours less $z$, i.e. $2-z$
Now the speed of the minute hand is $12$ times that of the hour hand, thus
$\dfrac{2-z}{z} = 12$, which yields $z = 2/13$ hrs
and time taken for the hands to interchange $= (2 - z) = 24/13$ hrs   
