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Maths Problem Involving Lateral Thinking


Date: 5/24/96 at 23:58:39
From: Anonymous
Subject: Maths problem involving lateral thinking

Binh and Duong are running on a circular track. They start at the same 
place, but run in opposite directions. Binh runs at 15 km/h
and Duong runs at 12 km/h. After they have each made several laps
they arrive simultaneously at their starting place, at which time they 
stop.

How many times will they have met on the track during their run, not 
including the start and the finish?

I solved this problem using a number line with 5 units in the positive 
direction and 5 units in the negative direction. I imagined that the 
circular track was laid out in a line. If A goes forward to the right 
5 units, then B goes forward to the left 4 units. If you keep using 
this model they will meet 4 times.

Is this solution correct? Is there another way to solve it?


Date: 5/30/96 at 14:49:35
From: Doctor Anthony
Subject: Re: Maths problem involving lateral thinking

Relative motion problems are usually best solved by bringing one of 
the moving objects to rest and making the other object move with the 
combination of its own velocity and the reversed velocity of the other 
object.  So if we bring D to rest by imposing a velocity of 12 km/hr 
in the opposite direction, then B will move with a total velocity of 
15+12 = 27 km/hr. 

To find the number of laps we must however consider that both are 
moving. If s is the circumference of the circle, then the two would 
meet back at the start after time ns/15 = ms/12 where n is number of 
laps that B does and m is the number of laps that D does.  

So n/15 = m/12

This gives n = (15/12)m and both m and n must be integers.
The least value that m can have is 4 for n to be an integer -

So n = 60/12 = 5

Now we must find how many laps B would do in the same time if
travelling at 27 (= 15+12) km/hr.  Time = distance/speed = 5s/15 =
s/3.  This is the time B actually traveled.  Thus, using
distance=speed*time, the distance B would travel at 27 km/hr is
speed*time = 27 * s/3 = 9s.

B would do 9 laps while D stayed at the starting post, so 
B and D would pass each other 8 times before meeting again 
at the starting post.

-Doctor Anthony,  The Math Forum
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