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
$$
\left(100\right)+\left(100+20\right)=220\,\left[km\right]
$$
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