Kevin\textquotesingle s speed $=\dfrac{70}{1\dfrac{1}{2}}=\dfrac{140}{3} \text{km/h}$  Let John\textquotesingle s speed be $x \text{km/h}$.  $$\left (x+ \frac{140}{3}\right ) \times 8=800 \Rightarrow x=100- \frac{140}{3}= \frac{160}{3}=53 \frac{1}{3}\text{km/h}$$. 
Two motor-cyclists John and Kevin were $800 \text{km}$ apart and travelling towards each other at a constant speed. They started at the same time, meeting after $8$ hours. If Kevin started $1\dfrac{1}{2}$ hours later than John, they would be $70 \text{km}$ apart $8$ hours after John started. At what speed was John travelling in $\text{km/h}$? 

[1] $$50$$ 
[2] $$51 \frac{2}{3}$$ 
[3] $$52 \frac{1}{2}$$ 
[4] $$53 \frac{1}{3}$$ 
[5] None of the above 