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 Post subject: Mia is cyclingPosted: Sat Jan 09, 2016 10:42 am

Joined: Sat Jan 09, 2016 10:32 am
Posts: 5
Can anyone help to solve these question?
Mia is cycling round the track. she does one lap every 72 seconds. Gareth is cycling round the track in opposite direction. If they meet each other every 40 seconds, how long does it take Gareth to do a lap of the track?

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 1:35 pm

Joined: Sun Oct 19, 2014 11:52 am
Posts: 60
90 sec?

cycling "round" I assume a circle.

So, Mia does

360 degrees in 72 sec and meets Gareth in 40 sec

In 40 sec, Mia covers (360*40)/72=200 degrees

Remaining distance i.e 360-200 = 160 degrees is covered by Gareth.

If he takes 40 secs to cover 160, then for full lap (360 degrees), he will take :

(40*360)/160 = 90 sec

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 4:34 pm

Joined: Sat Jan 09, 2016 10:32 am
Posts: 5
Thanks but it is difficult question for 11+. The other part of this question is also confusing.

Q-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?

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 7:00 pm

Joined: Sat Jul 24, 2010 9:45 pm
Posts: 1560
zia wrote:
Thanks but it is difficult question for 11+. The other part of this question is also confusing.

It does sound very confusing. Is this really an 11+ question? Does it come from a book of some sort or a past/sample paper?

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 7:16 pm

Joined: Sun Oct 19, 2014 11:52 am
Posts: 60
Here's my try...

5 laps in 6mins vs 6 laps in 5mins

One is slow and the other is fast. The "relative" speed is what we are interested.

(6 laps/ 5mins ) - (5 laps/ 6 mins) = 11 laps/ 30mins

If you think of an object cycling at this relative speed, when it completes one lap, that will actually mean an "overlap" for the faster and slower cyclists.

In 30 mins there are 11 such overlaps... but we are interested in 10mins i.e 1/3rd... so 1/3rd of 11 is 3 and a bit. So 3 overlaps.. the 4th has not been done yet.

But this is a horrible question to ask at 11+ and very ambiguous, as the I'm not sure whether or not to count the initial overlap when they start. So answer could be 4 as well.

Really bad question setting. Which paper is this from?

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 7:18 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 13992

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 7:46 pm

Joined: Sat Jan 09, 2016 10:32 am
Posts: 5
This question is in Latymer School sample paper. Does anyone know that the real paper is exactly the same as the sample paper?

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 Post subject: Re: Mia is cyclingPosted: Sat Jan 09, 2016 8:00 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 13992
I would hope that more care goes into a 'real paper' - the phrase 'just ahead' is a poor choice of words.

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 Post subject: Re: Mia is cyclingPosted: Sun Jan 10, 2016 1:14 am

Joined: Wed Jan 18, 2012 11:41 am
Posts: 6842
Location: Essex
zia wrote:
This question is in Latymer School sample paper. Does anyone know that the real paper is exactly the same as the sample paper?

Latymer has the same CEM exam as CCHS, the Redbridge grammar schools and ten or so others. These questions aren't in the Latymer CEM familiarisation booklet (dated 2015) I found just now. Are they from an old paper?

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 Post subject: Re: Mia is cyclingPosted: Tue Jan 12, 2016 8:02 pm

Joined: Tue Jan 12, 2016 7:43 pm
Posts: 1
It's a Latymer Upper paper rather than a Latymer (Edmonton) paper.

It's a tricky question - no harm in that - but I don't think the wording's ambiguous. Hard to decipher perhaps in part two, but I imagine the process being tested is the idea of proportional reasoning (which the Latymer Upper setter seems to like a lot...)

No need to assume the track is circular in part i, target2016. Mia completes 40/72 (=5/9) of a lap between every meeting, which means Gareth must complete 4/9 of a lap in the same time. So 4/9 of a lap takes him 40 seconds, so one lap will take him 90 seconds.

In part two, Julian completed 12 laps in the time given (easy enough to calculate, and I imagine some credit would be earned?) and Paddy completes 8 1/3 laps (harder to calculate perhaps). So Julian completes 3 2/3 more laps than Paddy, and since 'just ahead' could by no stretch of the imagination be interpreted as as much as 2/3 of a lap, he must overtake him 4 times.

I imagine complete success rate on this question would be low, but with over 1000 candidates they need to be able to differentiate at the top end...

I reckon it's a nice little question for clever girls and boys - certainly encourages thought rather than regurgitating rote-learned methods!

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