An algebra problem by Paul Fournier

Algebra Level 3

Two cyclists, k k kilometers apart, and starting at the same time, would be together in x x hours if they traveled in the same direction, but would pass each other in y y hours if they traveled in opposite directions. What is the ratio of the speed of the faster cyclist to that of the slower?

(x+y)/(x-y) x/y x/(x-y) (x+y)/x (x+k)/(y+k)

Paul Fournier by Paul Fournier , 70, Canada

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1 solution

Greg Grapsas
Jan 17, 2018

Since everything is a variable I simply set up a specific example of 100km distance and 60 and 40 km/hr speeds...x is therefore 5hr when going in same direction since 60-40 =20 miles closing distance per hr or 5 hr to meet. Alternatively 60+40=100 is closing speed when going in opposite direction so y= 1hr to pass each other.(5+1)/(5-1)=6/4 = 60/40. In general (x+y)/(x-y)

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Assuming cyclists have speeds with magnitudes v1 and v2 with v1>v2

(v1+v2) * t =x
(v1-v2) * t =y

v1 = (x+y)/2t
v2 = (x-y)/2t

v1/v2=(x+y)/(x-y)

Eric Scholz - 2 years, 2 months ago

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