(Can you please explain clearly how you found the solution to this question as I'm only a Year 7)

In a $5000$ m race, the winner finishes $200$ metres ahead of second and $600$ metres ahead of third. Assuming the runners run at a constant speed, to the nearest integer how many metres ahead of third will second finish?

I tried using $$\text{time}=\frac{\text{distance}}{\text{speed}}$$

formula and I replaced it with the distances between them and pro-numerals for the speed, but then I got stuck. I found out the answer in the internet to be $417$ m but I didn't understand how they got the answer because it was in Year 9 format. Can anyone help me?

P.S. This is the answer I didn't understand enter image description here

  • $\begingroup$ Do we know the finishing time of the winner? $\endgroup$ – N. F. Taussig Mar 5 '17 at 2:19
  • $\begingroup$ How can the answer be 417 meters when the question is how many seconds? That's like asking, how old are you? and answering, 13 kilograms. $\endgroup$ – Gerry Myerson Mar 5 '17 at 3:01
  • 1
    $\begingroup$ Oh sorry I'll edit $\endgroup$ – bio Mar 5 '17 at 4:54
  • $\begingroup$ In the length of time it takes #2 to run 4800 meters, #3 runs 4400 meters. So, in the time it takes #2 to run 5000 meters, how far will #3 run? $\endgroup$ – Gerry Myerson Mar 5 '17 at 9:37
  • $\begingroup$ @GerryMyerson By your logic, the answer should be 400m but it's 417. Why? $\endgroup$ – bio Mar 5 '17 at 9:55

Let $t_1$ be the finish time for the winner and $t_2$ be the finish time of the second.

\begin{align*} v_1 t_1 &= 5000 \\ v_2 t_1 &= 4800 \\ v_3 t_1 &= 4400 \\ v_2 t_2 &= 5000 \\ v_3 t_2 &= \frac{4400}{t_1} \times \frac{5000}{v_2} \\ &= \frac{4400 \times 5000}{4800} \\ v_2 t_2-v_3t_2 &= 5000 -\frac{4400 \times 5000}{4800} \\ &\approx 417 \; \text{m} \end{align*}

  • $\begingroup$ what are the v's for? $\endgroup$ – bio Mar 6 '17 at 0:33
  • $\begingroup$ $v_1$ velocity (or speed) of the first, $v_2$ velocity of the second, etc. $\endgroup$ – Ng Chung Tak Mar 6 '17 at 0:34
  • $\begingroup$ Well, it works, but it's much too hard. There's no need to introduce five variables – or any variables – into an arithmetic problem. $\endgroup$ – Gerry Myerson Mar 6 '17 at 5:25
  • $\begingroup$ @GerryMyerson Actually, the model answer (updated in the post) does involve $5$ variables as mine. $\endgroup$ – Ng Chung Tak Mar 6 '17 at 12:41
  • $\begingroup$ Maybe that's why bio couldn't understand it. $\endgroup$ – Gerry Myerson Mar 6 '17 at 21:21

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