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 Post subject: Can anyone help?Posted: Wed Nov 08, 2017 8:54 pm

Joined: Sat Jan 09, 2016 6:45 pm
Posts: 303
Mr Sahota is cycling around a track at a constant speed. He completes 5 laps every 4 minutes. How long will it take him to complete 8 laps? Give your answer in minutes and seconds?

Mr Aldham is also cycling around the track at a constant speed. He completes 5 laps every 6 minutes. How long is it between the first time that Mr Sahota overtakes Mr Aldham and the second time that Mr Sahota overtakes Mr Aldham?

Thank you!

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 Post subject: Re: Can anyone help?Posted: Wed Nov 08, 2017 9:00 pm

Joined: Mon Feb 12, 2007 2:21 pm
Posts: 16254
MCLC wrote:
Mr Sahota is cycling around a track at a constant speed. He completes 5 laps every 4 minutes. How long will it take him to complete 8 laps? Give your answer in minutes and seconds?
Mr Aldham is also cycling around the track at a constant speed. He completes 5 laps every 6 minutes. How long is it between the first time that Mr Sahota overtakes Mr Aldham and the second time that Mr Sahota overtakes Mr Aldham?

I'll start you off:

5 laps in 4 mins [240 secs]
1 lap in 4/5 min [48 secs]
8 laps in 32/5 mins [384 secs = 6 mins 24 secs]

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 Post subject: Re: Can anyone help?Posted: Wed Nov 08, 2017 9:03 pm

Joined: Sat Jan 09, 2016 6:45 pm
Posts: 303
Thanks Guest! DD managed to get that answer. I couldn't see her working and as I mentioned before I'm all but useless .

Is the second part 4 minutes?

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 Post subject: Re: Can anyone help?Posted: Wed Nov 08, 2017 9:15 pm

Joined: Mon Feb 12, 2007 2:21 pm
Posts: 16254
Sorry I'm a bit busy but I don't think it is .. will try and look at it properly later.

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 Post subject: Re: Can anyone help?Posted: Wed Nov 08, 2017 9:21 pm

Joined: Sat Jan 09, 2016 6:45 pm
Posts: 303
Thanks!

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 Post subject: Re: Can anyone help?Posted: Wed Nov 08, 2017 11:49 pm

Joined: Mon Feb 12, 2007 2:21 pm
Posts: 16254
I'm not sure the best way to explain this:
Mr S times to go each lap 48, 96, 144, 192, 240, 288
Mr A times to go each lap 72,144, 216, 288,

So 144 seconds - 2 mins 24 secs

I also thought about giving the 'lap' a distance - I chose 144m because it divides by 48 and 72

Mr S speed = 144/48 = 3 m/s
Mr A speed = 144/72 = 2 m/s

So difference in speed is 1 m/s and as the 'lap' is 144m it will be 144 secs before they meet again.

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 Post subject: Re: Can anyone help?Posted: Thu Nov 09, 2017 7:28 am

Joined: Sat Jan 09, 2016 6:45 pm
Posts: 303
Thanks as always guest!

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 Post subject: Re: Can anyone help?Posted: Fri Nov 10, 2017 12:51 pm

Joined: Fri Apr 15, 2016 4:35 pm
Posts: 70
First time when they both meet will be LCM of 48 and 72.
Second time when they meet will be double of LCM.
But you need to know how to calculate LCM for this.

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