SOLUTION: Two motorcyclists left point A for point B at the same time. The speed of one motorcyclist was 1.5 times greater than the speed of the other one. The motorcyclist, who reached poin

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Question 1014461: Two motorcyclists left point A for point B at the same time. The speed of one motorcyclist was 1.5 times greater than the speed of the other one. The motorcyclist, who reached point B first, immediately went back. He met the other motorcyclist in 2 hours and 24 minutes after the departure from point A. The distance between point A and point B is 120 mi. What is the speed of the cyclists and the distance from the point they met and point B?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
More than one way to analyze and handle the description. The slow cyclist at rate r went distance
d. The fast cyclist in the same amount of time, instead of returning to A covering the full
round trip, went d miles less than the full found trip; so that he met the slow cyclist on the
way back.
                      rate          time          distance

slow cyclist          r             2&2/5          d

fast cyclist          (3/2)r        2&2/5          2*120-d

total(one way)                                      120

24 minutes is 2%2F5 hour;
"one and five tenths" is the same as 3%2F2, about the factor relating their two speeds or rates.

When the tabulated data makes sense, two equations will be ready to formulate.

system%28r%2A%282%262%2F5%29=d%2C%283%2F2%29r%282%262%2F5%29=2%2A120-d%29
Simplify this system and solve for r and d.

system%28%2812%2F5%29r=d%2C%283%2F2%29%2812%2F5%29r=240-d%29

Substitute for d into the second equation.
%2836%2F10%29r=240-%2812%2F5%29r
%2836%2F10%2B12%2F5%29r=240
%2836%2F10%2B24%2F10%29r=240
%2860%2F10%29r=240
6r=240
r=240%2F6
highlight%28r=40%29------the rate or speed of the slower mototcyclist. You can find the
distance which each motorcyclist had traveled from this.