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What method or formulae can I use to solve the maths below ?

Two trains are running, on separate tracks, round a model railway layout. One completes a circuit every 40 seconds and the other every 55 seconds. The trains start together at the station. How long, in minutes and seconds, will it be before they are at the station together again ?

The answer given is 7 mins 20 secs

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 Post subject: Posted: Sun Oct 29, 2006 9:03 pm

Joined: Fri Sep 15, 2006 8:51 am
Posts: 8119
Hi

I ended up doing a spreadsheet - not what should be done I am sure !!

I looked at the cumulated number of extra seconds the slow train took to go round the circuit for each time the fast train went round ie 15 seconds deficit per circuit
ie

15
30
45
60
75
90
105
120
135
150
165
180
195

Then found the first one that was wholey divisible by 55 - it was circuit 11 of the fast train and circuit 8 of the slow train - ie 165 seconds behind. This was reached by the both trains after 440 seconds
..... must be a simpler way

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 Post subject: Posted: Sun Oct 29, 2006 9:18 pm

Joined: Mon Jan 30, 2006 4:07 pm
Posts: 2660
Dear guest2910

Child should quickly jot some notes down....

multiples of 40....

40 80 120 160 200 240 280 320 360 400 440 480 520....

multiples of 55....

55 110 165 220 275 330 385 440...go no further....found first number the same in both lines = 440

Should take a child approx 30 seconds to find the answer....440 seconds...then quickly convert into mins and seconds.

Sometimes I think we, as adults.....my self included.... look for methods which are just too complicated!.....these are 10 year olds studying KS2 maths.....

Patricia

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 Post subject: Posted: Sun Oct 29, 2006 9:31 pm
Dear hermanmunster and Patricia,

Thanks for you quick response.

Patricia, you are right; the solution looks simpler than I thought !

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 Post subject: Posted: Sun Oct 29, 2006 9:39 pm

Joined: Fri Sep 15, 2006 8:51 am
Posts: 8119
blimey patricia - that's what I was trying to do just got RATHER complicated !! I guessed there had to be a simple way if ks2 kids were doing it!!

always knew A levels maths was going to be a handicap.

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 Post subject: Posted: Sun Oct 29, 2006 11:07 pm
Find the prime factors of 40

2x2x2x5

Find the prime factors of 55

5x11

There is a common factor of 5

Work out the sum

2x2x2x5x11 = 440

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 Post subject: 55-40=15secPosted: Wed Nov 15, 2006 12:07 am

Joined: Tue Nov 14, 2006 11:31 pm
Posts: 1
Location: LONDON
the second train slower than the first 15sec.

than think about how many 15sec. can be / by 40 (time of first train one turn around)

15x8(all even no. and 8 is the smallest one) =120/40=3

so the second train need to take 8 around that will meet the orther train at

the station.

7mins and 20sec.

by the way, I think if you say :440sec. that's wrong.

_________________

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 Post subject: Posted: Wed Nov 15, 2006 8:22 am
Guest has done the problem the correct way; just failed to convert to mins and secs. It is a Lowest Common Multiple question, and this is the most elegant way of answering the question.

Lowest Common Multiple/Highest Common Factor topic was 1st year Seniors when I was at school (many, many years ago). I remember it was the very first thing we did.

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Joined: Thu Nov 02, 2006 3:07 pm
Posts: 1149
Location: Finchley - Barnet

Divide both 40 and 55 by the largest possible common exact divisor number, i.e. 5 in this case

This gives 8 and 11 correspondingly. Then multiply 8 with 11 = 88 and then multiply

88 with the 5=440

In the case of prime numbers the first step does not apply, just multiply the two numbers to gether and get the answer.

All the best,

INEX

guest2910 wrote:
What method or formulae can I use to solve the maths below ?

Two trains are running, on separate tracks, round a model railway layout. One completes a circuit every 40 seconds and the other every 55 seconds. The trains start together at the station. How long, in minutes and seconds, will it be before they are at the station together again ?

The answer given is 7 mins 20 secs

_________________
sj355

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