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PostPosted: Tue Dec 29, 2015 1:10 pm 
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Can anyone help with this question?

Paddy is cycling steadily round a track. He completes 5 laps in 6 minutes.

Julian is also cycling steadily round the track in the same direction. He complete 6 laps in 5 minutes.

At 10.00am exactly, Paddy is just ahead of Julian on the track. By 10.10 how many times will Julian have overtaken Paddy?

Thank you! :)


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PostPosted: Tue Dec 29, 2015 11:06 pm 
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Kitsilano wrote:
At 10.00am exactly, Paddy is just ahead of Julian on the track.


This sounds ambiguous to me as we cannot exactly quantify "just a head". Could you double check whether the question is correctly typed?


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PostPosted: Wed Dec 30, 2015 8:01 am 
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It is typed word for word exactly as it is written on the exam paper!

I can work out that they have done 12 laps and 8 1/3 laps. But I struggle to turn that into exact number of overtaking.


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PostPosted: Wed Dec 30, 2015 10:32 am 
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straight forward - to complete an overtake the difference needs to be a full integer difference - as the difference is only 3.67 there are only 3 complete overtakes


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PostPosted: Wed Dec 30, 2015 11:05 am 
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At a second glance, I realized "just ahead" is a simple way to trick us. It would be easier if we think if it as "how many times they meet" including the starting point. I'll explain below.

Paddy cycles 5/6 laps per minute.
Julian cycles 6/5 laps per minute.

So, the second meeting point would be 1 + (6/5)X = (5/6)X
Therefore, X = 30/11 minutes, which is slightly over 2 minutes.

Then we divide 10 minutes by 30/11 which will give us 11/3 (or 3 and 2/3).
That means Julian will overtake Paddy 3 times during the course of 10 minutes.

But Since Paddy was "just ahead" of Julian on the track, Julian overtook Paddy within the first few seconds and we need to add this as well.

So, it would be 4 overtakes.


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PostPosted: Wed Dec 30, 2015 12:38 pm 
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It's a poor question - where did it come from?


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PostPosted: Wed Dec 30, 2015 1:59 pm 
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It's a Latymer Upper School sample paper.

Thanks for the help.

Although I'm not sure I'm clear yet. The answer that he's only done 3.67 laps more makes sense to me - therefore overtaken only 3 times.

I get lost with this:

"So, the second meeting point would be 1 + (6/5)X = (5/6)X
Therefore, X = 30/11 minutes, which is slightly over 2 minutes."

What is X representing here?


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PostPosted: Wed Dec 30, 2015 2:19 pm 
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Kitsilano wrote:
I get lost with this:

"So, the second meeting point would be 1 + (6/5)X = (5/6)X
Therefore, X = 30/11 minutes, which is slightly over 2 minutes."

What is X representing here?


X would be time required for Julian to overtake Paddy.

3.67 is 11/3 (or 3 and 2/3). But since Paddy was just ahead of Julian at the beginning, you need to add 1 more.


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PostPosted: Wed Dec 30, 2015 4:50 pm 
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Well I'm surprised it was felt to be suitable for any exam - it wouldn't get into a text book with a vague phrase such as 'just ahead'.


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PostPosted: Wed Dec 30, 2015 10:31 pm 
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Guest55 wrote:
Well I'm surprised it was felt to be suitable for any exam - it wouldn't get into a text book with a vague phrase such as 'just ahead'.


Exactly, my take on that was that the compiler was trying to imply that they were effectively side by side so there was no initial pass. However poorly worded


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