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mart [117]
4 years ago
5

At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path. The bicyclist head

ing north is riding 5 km/hour faster than the bicyclist heading south. At 10:45, they are 47.25 km apart. Find the two bicyclists’ rates.
Mathematics
1 answer:
g100num [7]4 years ago
6 0
 <span>Remember the distance formula, d = rt. After 1.75 hours, they are 47.25 km apart from each other. Let us use A for the distance that the bicyclist heading north has traveled since they met. The distance that the other bicyclist has traveled will, therefore, be (47.25 - A) 



A = r(1.75) 
A = 1.75r 
(47.25 - A) = (r + 5)(1.75) 
47.25 - A = 1.75r + 8.75 
47.25 - A - 8.75 = 1.75r 
38.5 - A = 1.75r 
38.5 - A = A 
38.5 = 2A 
19.25 = A 



So A has traveled 19.25 km in 1.75 hours; this gives him a speed of 11 km/h. Bicyclist B has, therefore, traveled 47.25 km - 19.25 km = 28 km in 1.75 hours, giving him a rate of 16 km/h, which fits the provided information.</span>
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