Maths Problem Involving Lateral ThinkingDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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