Question

Two runners start from opposite ends of a course, one running at 16 km/h, the other at 20 km/h. When they meet, their two running times total 1 hour. If the slower runner has gone 2 fewer kilometers than the faster, how has the faster runner gone ?

Answer 1

I got `10` km.

Explanation:

If the slower runner (`16` km/hr) has run for `t_1` of an hour,
and the faster runner (`20` km/hr) has run for `t_2` of an hour.

We are told
[1]`color(white)("XXX")t_2+t_1=1`

The slower runner will have run `16` km/hr `xx t_1` hr `=16t_1` km
and
the faster runner will have run `20` km/hr `xx t_2` hr `=20t_2` km.

We are also told that
[2]`color(white)("XXX")20t_2-16t_1=2` (with everything in km)

Multiplying [1] by `16`
[3]`color(white)("XXX")16t_2+16t_1=16`

Adding [2] and [3]
[4]`color(white)("XXX")36t_2=18`

`rArr` [5]`color(white)("XXX")t_2=0.5`

and the fastest runner will have run `0.5` hr. `xx 20` km/hr `=10` km

Answer 2

Their two running times total 1 hour so they ran for 30 mins each

If you run at 16 km/h for 30 mins then you would travel 8 km

if you run at 20 km/h for 30 mins then you would travel 10 km

The faster one runs 10 km