How do I find the time of an scheduled event if it is given between a range?

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I'm going in circles with this riddle. It doesn't appear too complicated but I got tangled with the right interpretation of the time mentioned.

The problem is as follows:

At a TV station a program director decides to set a new schedule for the morning news show. He decides that the show is to start after 5 am but before 8 am. If we know that the elapsed time between 5 am until 25 minutes before the show begins is equal to two thirds of the time which will be to 8 am, but in 25 minutes. What time does the show starts?.

The existing alternatives in my book are as follows:

$\begin{array}{ll} 1.& \textrm{7:17 AM}\\ 2.& \textrm{5:37 AM}\\ 3.& \textrm{6:27 AM}\\ 4.& \textrm{5:47 AM}\\ 5.& \textrm{6:17 AM}\\ \end{array}$

What I tried to do was the following. It's a bit tricky but I thought that the unknown time to be $x$ and built the equation from there given these interpretations:

Time elapsed between $25$ minutes before the start of the show and $\textrm{5 AM}$:

$\left(x-\frac{1}{4}\right)-5$

Time which is two thirds which will be to $8$ am but in $25$ minutes:

$8-\left(\frac{2}{3}+\frac{25}{60}\right)$

For this part and as well for the previous I'm working using hours and not minutes so by the end I can get a straight answer.

Since it mentions that both are equal then it is just plugging in together:

$\left(x-\frac{1}{4}\right)-5=8-\left(\frac{2}{3}+\frac{25}{60}\right)$

$x=13+\frac{1}{4}-\frac{2}{3}-\frac{5}{12}$

$x=13+\frac{3-8-5}{12}$

$x=13-\frac{16}{12}=13-\frac{4}{3}=12-1-\frac{1}{3}=11-\frac{1}{3}$

Then I interpreted the last line as:

$11 - \frac{1}{3}\times 60 = 11 - 20m$

Hence the time would be:

$\textrm{10 AM 40 mins}$

But this is clearly outside the boundary which is established on the problem.

Then I thought that the right interpretation would be:

Time which is two thirds which will be to $8$ am but in $25$ minutes:

$\frac{2}{3}\left(8-\frac{25}{60}\right)$

Hence:

$\left(x-\frac{1}{4}\right)-5=\frac{2}{3}\left(8-\frac{5}{12}\right)$

Multiplying by $36$ all:

$36x-9-180=192-10$

$36x=189+182$

$36x=371$

Simplifying the last result yields: $\textrm{1h 18m 20s}$. This result is not even close. Can somebody help me which is the part where I got lost or what I did not interpreted correctly?. I reviewed the equations over and over and still I cannot figure out what I did wrong. Can somebody help me with this?

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Apparently this is a matter of getting the right interpretation, having that ah-ha moment and also a bit of playing around like Where's Wally?

In the passage the key as mentioned in the comments is what to do with but in $25$ minutes. After the edit, it is more likely that it meant the time after the show started.

To better illustrate this situation below are the variables:

$\textrm{Time when the show started = x}$

$\textrm{But in 25 minutes =} x + \frac{25}{60}$

Time elapsed between 5 AM and 25 minutes before the show started:

$\left(x-\frac{25}{60}\right)-5$

Two thirds of the time which will be to 8 AM, but in 25 minutes.

$\frac{2}{3}\left(8-\left(x+\frac{25}{60}\right)\right)$

Finally all that is left to do is to equate both expressions:

$\left(x-\frac{25}{60}\right)-5=\frac{2}{3}\left(8-\left(x+\frac{25}{60}\right)\right)$

Simplifying:

$\left(x-\frac{5}{12}\right)-5=\frac{2}{3}\left(8-\left(x+\frac{5}{12}\right)\right)$

$\frac{12x-5-60}{12}=\frac{2}{3}\left(\frac{96-12x-5}{12}\right)$

$12x-65=\frac{2}{3}\left(91-12x\right)$

$36x-195=182-24x$

$60x=377$

$x= 6 \frac{17}{60}$

Hence the time when the show began was $\textrm{6:17 AM}$

This checks with the alternatives. Needless to say that probably this was the intended meaning.