Maths - Please Help
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Maths - Please Help
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
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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- Posts: 12817
- Joined: Fri Sep 15, 2006 9:51 am
- Location: The Seaside
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
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
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 old children
Patricia
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 old children
Patricia
-
- Posts: 12817
- Joined: Fri Sep 15, 2006 9:51 am
- Location: The Seaside
55-40=15sec
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.
the time is 55x8=440sec. which/60=answer:
7mins and 20sec.
by the way, I think if you say :440sec. that's wrong.
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.
the time is 55x8=440sec. which/60=answer:
7mins and 20sec.
by the way, I think if you say :440sec. that's wrong.
:-)
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.
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.
Re: Maths - Please Help
Trick of the trade:
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
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