Two people walking towards each other and a passing Train Two people are walking towards each other alongside a railway track. A freight train overtakes one of them in $20$ seconds and exactly $10$ minutes later meets the other person coming from the opposite direction. The train passes this person in $18$ seconds. Assume that speeds are constant throughout. How long after the train has passed the second person will the two people meet?
Now this is a rather surprising question because:

*

*when I was taught this in class, teacher arrived at $89.7$ minutes,

*I arrived at $92$ minutes $22$ seconds,

*mostly answers on internet show $90$ minutes as answer,

*a google book mentions $92$ minutes $42$ seconds as the answer and

*another google book mentions the answer as "cannot be determined".

Let the speed of train, 1st person and 2nd person be $z$, $x$ and $y$ m/s.
Then everyone arrives at $z=10x+9y$.
That's where similarity ends. Now, while finding the required time taken ($T$) for the two people to meet is where all the problem befells:

*

*$T= \frac{10(10x+9y)-10x-0.3(x+y)}{x+y}$

*$T= \frac{600(10x+10y)+20(9x+9y)-638(x+y)}{x+y}$

*$T= \frac{600(z+y)-600(x+y)}{x+y}$

*(Assumes the speed of 2 people to be same)

*Only answer is mentioned as: Cannot be determined

Please help. (Hoping consensus between MSE members)
 A: I agree that there is insufficient information.
$10$ minutes + $20$ seconds = $620$ seconds from the time the front of the train is level with Person-1, the front of the train is level with Person-2.
Assume the speeds of Person-1, Person-2, and the Train are $a,b,$ and $t$, respectively.
Assume that from the moment the front of the train is level with Person-1, the distance between the front of the train and Person-2 is $D$.  Note that this is also the distance, at that moment, between Person-1 and Person-2.
Let $E$ denote the length of the train.
So, you have $5$ variables:

*

*The speeds, $a,b,t.$

*The distances $D,E$.

So, the challenge is to express both $a$ and $b$ in terms of $D$, so that you can compute $~\displaystyle \frac{D}{a+b}.$
Once you do that, you will have the time it took for Person-1 and Person-2 to meet, starting from the distance $D$.  From this, you will have to deduct the $20 + 600 + 18$ seconds that Person-1 and Person-2 were walking towards each other, until the train passed Person-2.
So, the desired computation will be
$$\frac{D}{a+b} - 638. \tag1 $$
From the premises:

*

*$\displaystyle E = 20(t-a).$

*$\displaystyle E = 18(t+b).$

*$\displaystyle D = 620(t+b).$
Variable $E$ and be eliminated in favor of variable $D$.
Also, both $a$ and $b$ can be eliminated in terms of the combination of variables, $D$ and $t$.  Unfortunately, with only $3$ equations in $5$ variables, you can't eliminate variable $t$, in terms of variable $D$.
Therefore, there is insufficient information.
Edit
The above conclusion is wrong.  See my Addendum.

Addendum
See the comment following my answer,
from InanimateBeing.  When I first posted my answer, I did consider that I needed $(a+b)$, rather than explicit values for $a$ and $b$.  However, I couldn't find a way to get there.
Their comment nails the problem, and indicates how the problem should be completed.  Hijacking their concepts:

*

*Since $20(t-a) = 18(t+b),$ you have that 
$2t = 20a + 18b \implies t = 10a + 9b.$


*Therefore, $(t + b) = 10a + 10b.$ 
This is the part that originally escaped me!


*Therefore, 
$D = 620(t + b) = 620(10a + 10b) = 6200(a + b).$


*Therefore, 
$\displaystyle \frac{D}{a+b} = 6200.$


*Therefore, 
The final computation is $6200 - 638 = 5562~$ seconds.
A: Let the speed of the train be $v$, and the speed of the first person be $v_1$ and the speed of the second person be $v_2$.  At $t = 0$, the front of the train crosses the first person, so the expressions for the positions of the three objects are
$ p(t) = v t $ ( front of train )
$ p_1(t) = v_1 t  $ ( 1st person)
$ p_2(t) = d - v_2 t $ ( 2nd person )
At t = 20, the rear of the train crosses the first person, so
$ v (20) - L = v_1 (20) $
from which
$ L = 20 (v - v_1) \hspace{20pt}(1) $
At t = 620, the front of the train crosses the second person, so
$ v (620) = d - v_2 (620) \hspace{20pt}(2)$
At t = 638 , the rear of the train crosses the second person, so
$ v(638) - L = d - v_2 (638) \hspace{20pt}(3)$
Combining $(2)$ and $(3)$, we deduce that
$ L = 18 (v + v_2 ) \hspace{20pt}(4) $
Now the time at which the two persons cross is $t_1$, that satisfies
$ v_1 t_1 = d - v_2 t_1 $
Thus $t_1 = \dfrac{d}{v_1 + v_2} $
Let's see how many unknowns we have.  We have
$ v , v_1, v_2, L , d $ unknown, and we have only $3$ equations relating them.
So, unless some variable gets cancelled out in the process we'll get a value for the answer otherwise it cannot be determined.
From $(1)$ and $(4)$, we have
$ L = 20( v- v_1) = 18(v + v_2) $
Therefore,
$ \dfrac{v + v_2}{v - v_1} = \dfrac{10}{9} $
From which
$ v = 10v_1 + 9v_2 $
Equation (2) gives us
$d = 620 (v + v_2) = 620 (10v_1 + 10v_2) = 6200 (v_1 + v_2) $
The two persons meet at $t_1 = \dfrac{d}{v_1 + v_2} = \dfrac{6200 (v_1 + v_2)}{v_1 + v_2} $
Hence
$ t_1 = 6200 \text{ sec} $
The time difference is $ 6200 - 638 =   5562 \text{ sec} $
Dividing by $60$ gives the required time in minutes, and this come to $92.7 \text{ minutes}$, which is equal to $92 \text{ minutes} $ and $ 42 \text{ seconds} $.
A: Let $v_1$ be the speed of the first people, meeting the tail of the train at point $P_1$ in the space-time diagram below, and $v_2$ be the speed of the second people ($v_2>0$), meeting the tail of the train at point $P_2$, $618$ seconds after $P_1$ ($10$ minutes and $18$ seconds). If $V$ is the speed of the train and $L$ its length, the given info on "passing by" times gives
$$
L=(V-v_1)\cdot 20\ \text{s} =(V+v_2)\cdot 18\ \text{s},
$$
that is:
$$
V-v_1=9(v_1+v_2).
\tag{1}
$$
Points $P_1$ and $P_2$ are separated by an unknown distance $\Delta x$, which is covered by the train in a time interval of $618\ $s. This means that:
$$
\Delta x=V\cdot 618\ \text{s}.
\tag2
$$
On the other hand, the same distance $\Delta x$ is covered together by the two people, the second one walking for the unknown time $\Delta t$ we must find, and the first one for the time $\Delta t+618\ \text{s}$, hence:
$$
\Delta x=\Delta x_1+\Delta x_2=
v_1\cdot(\Delta t+618\ \text{s})+v_2\cdot \Delta t.
\tag3
$$
Comparing $2$ and $3$ we thus find:
$$
(V-v_1)\cdot 618\ \text{s}=(v_1+v_2)\cdot \Delta t.
\tag4
$$
We can now happily substitute $(1)$ here and, as by magic, all speeds disappear and we are left with
$$
\Delta t= 618\ \text{s}\cdot 9 = 5562\ \text{s}.
$$

A: Let $V$ ("units"/sec) be the speed of the train, and $v_1$ and $v_2$ respectively be  the speed of the first and second person overtaken.
Since time taken is inversely proportional to speed, we can write
$\dfrac{V+v_1}{V-v_2} = \dfrac{20}{18} = \dfrac{10}{9}$
Now using the principle that when data is missing, we can put any values that satisfy the given relation(s),
It is easily seen that $\dfrac{10+1}{10-0.1} = \dfrac{11}{9.9} = \dfrac{10}9$,
and the rest is a walk in the park
In $10$ min + $18$ sec, the train has moved $10*618= 6180$ units, but person $1$ has reduced the distance by $0.1*618=61.8$, leaving $6118.2$ units to be covered @ $1.1$ unit/sec, giving the answer of $ 5562\;sec = 92$ min $42$ sec

NOTE:
In fact, I could have given an even simpler answer, with the relation
$\dfrac{1+k}{1-k} = \dfrac{10}{9}$, (both men walk at the same speed),
but I had the feeling that people  would jump on me and  say this is an unwarranted assumption. Not so. The question doesn't say that the two men are walking at different speeds, it is simply silent on it.
Of course, if you have to prove or show steps, it is a different matter, but otherwise we are not making assumptions, we are saying that in the absence of data, we can use any values that satisfy the relation(s)
