Question

Two trains, A and B, leave a train station at at the same time but in different directions, one towards east the other towards north. Trains B travels 20 km/h faster than train A. What is the average speed of each?

After 6 minutes, the two trains are 10 km apart.

Answer 1

Speed of train A `=60\ \text{km/hr`

Speed of train B `=80\ \text{km/hr`

Explanation:

`x\ \text{km/hr` be the average speed of train `A` then the speed of train `B` will be `(x+20)\ \text{km/hr` then

The distance traveled by train `A` moving with average speed `x\ \text{km/hr` in `6\ \text{min}=6/60=0.1\ \text{hr`

`=x\times 0.1`

`=0.1 x\ \ text{km`

Similarly, the distance traveled by train `B` moving with average speed `(x+20)\ \text{km/hr` in `6\ \text{min}=6/60=0.1\ \text{hr`

`=(x+20)\times 0.1`

`=0.1 x+2\ \ \text{km`

Now, the distance between two trains `A` & `B` after `6\ min` is given by the hypotenuse of right triangle with legs `0.1x` & `0.1x+2`

`10=\sqrt{(0.1x)^2+(0.1x+2)^2}`

`100=0.01x^2+0.01x^2+4+0.4x`

`x^2+20x-4800=0`

`x^2+80x-60x-4800=0`

`x(x+80)-60(x+80)=0`

`(x+80)(x-60)=0`

`x=-80\ \ or \\ x=60`

But, `x>0` hence we take `x=60`

The average speed of train `A`

`=60\ \text{km/hr`

The average speed of train `B`

`=60+20`

`=80\ \text{km/hr`