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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 ?

2 Answers

Jun 20, 2018

I got $10$ $10$ km.

Explanation:

If the slower runner ( $16$ $16$ km/hr) has run for $t_{1}$ $t_1$ of an hour,
and the faster runner ( $20$ $20$ km/hr) has run for $t_{2}$ $t_2$ of an hour.

We are told
[1] $XXX t_{2} + t_{1} = 1$ $color(white)("XXX")t_2+t_1=1$

The slower runner will have run $16$ $16$ km/hr $\times t_{1}$ $xx t_1$ hr $= 16 t_{1}$ $=16t_1$ km
and
the faster runner will have run $20$ $20$ km/hr $\times t_{2}$ $xx t_2$ hr $= 20 t_{2}$ $=20t_2$ km.

We are also told that
[2] $XXX 20 t_{2} - 16 t_{1} = 2$ $color(white)("XXX")20t_2-16t_1=2$ (with everything in km)

Multiplying [1] by $16$ $16$
[3] $XXX 16 t_{2} + 16 t_{1} = 16$ $color(white)("XXX")16t_2+16t_1=16$

Adding [2] and [3]
[4] $XXX 36 t_{2} = 18$ $color(white)("XXX")36t_2=18$

$\Rightarrow$ $rArr$ [5] $XXX t_{2} = 0.5$ $color(white)("XXX")t_2=0.5$

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

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