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A cyclist begins to bike from point A at $6$:$00$ am to point B. At $8$:$00$ am a car begins to drive from point A also to point B and drove at a velocity $60$ km/h faster than the bike. The car met the bike after they had driven/biked $\frac{5}{16}$ from the the distance between A and B. Immediately the car continued driving. He got to point B and waited $2$ hours there. Then, drove back to point B. After the car had driven $\frac14$ of the distance between A and B he met the bike that was still on his way to point B.

Find the velocity of the cyclist and the distance between A and B.

In this type of problems I like to do charts. I put $x$ as the velocity of the cyclist, $t$ the time of the cyclist to meet the car and $y$ the distance between A and B. In addition, $t_2$ is the time that takes the bike to get from A to B. So what I did is the following:

$$\begin{array}{|c|c|c|} \hline \text{Until they met}& \text{$v$ velocity} & \text{$t$ time} &\text{$d$ distance} \\ \hline \text{Cyclist} & x & t & \frac{5}{16}y\\ \hline \text{Car} & x + 60 & t-2 & \frac{5}{16}y\\ \hline \text{Cyclist} & x & t_2 & y \\\hline \end{array}$$

I'm stuck here. I don't know how to continue. I don't know how to use the other data like he waited $2$ hours, in the way back he met again the bike, and so far.

Can anyone help me please?

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  • $\begingroup$ Since when do bikers bike? $\endgroup$ – sqtrat Mar 26 '16 at 11:52
  • $\begingroup$ @sqtrat Hahaha right! I corrected it. Sorry for my bad translation. $\endgroup$ – Pichi Wuana Mar 26 '16 at 11:55
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Let the biker's velocity be $x$. Then $x+60$ is the speed of the car. Let the distance between $A$ and $B$ be $d$. Now, according to the first statement, to cover $\frac{5}{16}$ of the distance from $A$ to $B$, the biker took $2$ hours more than the car. So we write :$$\frac{5d}{16x} - \frac{5d}{16(x+60)} = 2$$.

For the second part, let's calculate the time between the two meetings using the bike. So the bike first met at $\frac{5d}{16}$, and next he meets him at $\frac{3d}{4}$, so the distance travelled by the biker was $\frac{7d}{16}$, and the time it took was $\frac{7d}{16x}$.

Now use the car. The car traveled the remaining $\frac{11d}{16}$ of the distance, waited for $2$ hours, and came back $\frac{d}{4}$ distance, therefore he traveled $\frac{15d}{16}$ distance and waited two hours before the next meeting, so the times taken was $\frac{15d}{16(x+60)} + 2$. Now the time waited by the car and the bike were equal hence we write down the second equation: $$\frac{15d}{16(x+60)} + 2=\frac{7d}{16x}$$.

I'll leave you to solve these for yourself.

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