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
Best Answer
Let them meet after $x$ hours. We know that $25x+20=30x$, hence $x=4$. So one train travelled $25\cdot4$ km, the other $30\cdot4$ km, in total $220$ km.