# 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 ?

Jun 20, 2018

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]$\textcolor{w h i t e}{\text{XXX}} {t}_{2} + {t}_{1} = 1$

The slower runner will have run $16$ km/hr $\times {t}_{1}$ hr $= 16 {t}_{1}$ km
and
the faster runner will have run $20$ km/hr $\times {t}_{2}$ hr $= 20 {t}_{2}$ km.

We are also told that
[2]$\textcolor{w h i t e}{\text{XXX}} 20 {t}_{2} - 16 {t}_{1} = 2$ (with everything in km)

Multiplying [1] by $16$
[3]$\textcolor{w h i t e}{\text{XXX}} 16 {t}_{2} + 16 {t}_{1} = 16$

Adding [2] and [3]
[4]$\textcolor{w h i t e}{\text{XXX}} 36 {t}_{2} = 18$

$\Rightarrow$ [5]$\textcolor{w h i t e}{\text{XXX}} {t}_{2} = 0.5$

and the fastest runner will have run $0.5$ hr. $\times 20$ km/hr $= 10$ km

Jun 20, 2018

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