Time rate problem involving trains don't know if my answer is correct Suppose trains leave from NY for AT every 20 mins starting at 1:00 am, and trains leave AT for NY every 30 mins starting at 1:10 am. If it takes each train 10 hours to travel from one city to the other, how many trains traveling from AT to NY are passed by the train that leaves NY at 1 pm and arrives in AT at 11 pm?
So every 30 mins:

*

*1:10 am

*1:40 am

*2:10 am

*...

*12:40 pm

*1:10 pm

*...

*10:40pm

In 10 hours from 12:40 pm > 10:40 pm since 10hrs = 600 mins 600/30 = 20 trains passes when that single train travels from NY to AT from 1pm to 11pm.
 A: I think the easiest way to see it is that the train that leaves NY at $1$ PM will pass all the trains that leave AT before $11$ PM.  That is all the trains that leave from $1:10$ PM through $10:40$ PM.  There are $20$ of them.
A: Writing $0110$ for 1:10 am and $1310$ for 1:10 pm,
the trains leave AT at $0110,$ $0140,$ $0210$, $0240, \ldots ,$
and we want to know how many of these trains pass (and are passed by) the train that leaves NY at $1300.$
The train that leaves AT at $0240$ arrives in NY at $1240$ and therefore does not pass the train that leaves NY at $1300.$
But the train that leaves AT at $0310$ arrives in NY at $1310$ and therefore passes the train that leaves NY at $1300.$
Since the train that leaves NY at $1300$ arrives in AT at $2300,$ it passes the train that leaves AT at $2240$ but not the train that leaves AT at $2310.$
So now we know the first and last trains from AT that pass the $1300$ from NY.
Count those two trains and all trains in between.
A: Using an equation. Given the problem parameters every train AT→NY that leaves from the time the NY→AT train leaves (and until it arrives) qualifies.
The decimal time notation for $30$ mins is $0.5$ (hour), and 1:10 is $1$ hour + $\dfrac{10}{60}$ hour, or $\dfrac{7}{6}$ hour. We can write the "$t^{th}$ train number leaving" equation, that gives the departure time for all trains AT→NY
$$\dfrac{7}{6}+0.5\cdot t$$
Then we want to know how many trains leave between $13$h ($1$pm) and $23$h ($11$pm), that is all the $t$ that meet
$$13 \le \dfrac{7}{6}+0.5\cdot t\lt 23$$
or $$\dfrac{71}{3}\le t\lt \dfrac{131}{3}$$thus $$t=\dfrac{131}{3}-\dfrac{71}{3}=20$$
