algebra precalculus - To find the total distance between two points A and B with just speed of train and difference in distance between two trains?

Two trains approach each other at 25 km/hr and 30 km/hr respectively from two points A and B. Second Train travels 20 km more than first. What is the distance between A and B ?

My approach:

Since $Distance = Speed \cdot Time$
[I just added the time it took to cover the distance taken by the Second train($30$ km/hr) to cover $20$ km to the First train($25$ km/hr) to get distance as constant]

(First Train)

For $20$ km at $25$ km/hr, the time taken would be

time $= \frac{20}{25} = \frac45$

So First train taken would take $4/5$ time more to cover the distance $d$

$d = 25 \cdot (t + 4/5 )$ ---> (1)

$d = 30 \cdot t$ ---> (2)

Now since distance is constant and speed is inversely proportional to time,

Ratio of speeds $= \frac{25}{30} = \frac56$

Ratio of times $= \frac{t+4/5}{t} = \frac{\frac{5t+4}{5}}{t} = \frac{5t+4}{5t}$

So

$\frac56= \frac{5t+4}{5t}$

Since it is inversely proportional

$5(5t+4)=6(5t)$

$25t+20=30t$

$25t-30t=-20$

$-5t=-20$

$t= 4$

So applying $t=4$ in (2)
$d=30 \cdot 4 = 120$ km but its wrong

The correct answer is $220$ km

I don't understand! help

Answer

I think best way to describe all the data is in a table like this

$\begin{array}{cccc} & \text{V} & \text{T} & \text{D}\\ \text{Train 1} & 25\,\left[\frac{km}{h}\right]\\ \text{Train 2} & 30\,\left[\frac{km}{h}\right] \end{array}$

Denote the distance the first train traveled until the meeting
by $x$. Therefore the second train traveled $x+20$. Lets add this
to the table:

$\begin{array}{cccc} & \text{V} & \text{T} & \text{D}\\ \text{Train 1} & 25\,\left[\frac{km}{h}\right] & & x\,[km]\\ \text{Train 2} & 30\,\left[\frac{km}{h}\right] & & x+20\,[km] \end{array}$

Now we can complete our table using that $T=\frac{D}{V}$

$\begin{array}{cccc} & \text{V} & \text{T} & \text{D}\\ \text{Train 1} & 25\,\left[\frac{km}{h}\right] & \frac{25}{x}\,\left[h\right] & x\,[km]\\ \text{Train 2} & 30\,\left[\frac{km}{h}\right] & \frac{30}{x+20}\,\left[h\right] & x+20\,[km] \end{array}$

Assuming both trains left both points at the same time we get that
$T_{1}=T_{2}$ where $T_{1},T_{2}$ is the time travel for each train
until the meeting. So
$$\frac{25}{x}=\frac{30}{x+20}\qquad\Rightarrow\qquad x=100\,\left[km\right]$$
Therefore the distance between $A$ and $B$ is $D_{1}+D_{2}$ where $D_{1},D_{2}$
is the distance each train traveled untill the meeting. So the distance
is
$$\left(100\right)+\left(100+20\right)=220\,\left[km\right]$$