Question

Two trains starts from stations $A$ and $B$ and travel towards each other at speed of $50km/hr$ and $60km/hr$ respectively. At the time of their meeting, the second train has travelled $120km$ more than the first. The distance between $A$ and $B$ is

• A. $990km$
• B. $1200km$
• C. $1320km$
• D. $1440km$

Answer 1

The correct option is C $1320km$

Speed of train $A=50kmph$
Speed of train $B=60kmph$
Since, time is constant
$Speed\propto$ Distance covered
$\frac{S_A}{S_B}=\frac{D_A}{D_B}$
$\frac{50}{60}=\frac{D_A}{D_B}\Rightarrow \frac{5}{6}=\frac{D_A}{D_B}$
Given that train $B$ has travelled $120km$ extra.
$6x-5x=120$
$x=120$
The distance between $A$ and $B=6x+5x=11x$
$=11\times 120=1320$
Alternate Method:
Let train $A$ start form station $A$ and $B$ from station $B$.
Let the trains $A$ and $B$ meet after/ hours.
$\therefore$ Distance covered by train $A$ in $t$ hours $=50t$
Distance covered by train $B$ in $t$ hours $=60tkm$.
According to the question,
$60t-50t=120$
$\Rightarrow t=\frac{120}{10}=12hours$.
$\therefore$ Distance $AB=50\times 12+60\times 12$
$=600+720=1320km$.