a, b and c are three linear towns. c divides the line. It takes Ron 72 mins to travel from a to c. It takes Zak 50 mins to travel from b to c. They complete their journeys ab and ba @ the same time and travel @ constant speeds. What time do they finish if they start- a @ 12.00 and b @ 12.22 .They met @ c @ 1.12. Thanks.
I came up with speeds of 72 and 94, then tried plugging these into time= dist/speed. Didn' t work.The answer is 2.12. Understand the concept, b travels faster
Excuse me? b doesn't travel at all, b is a
town! Are you confusing b with Zak? Yes, both Ron and Zak travel the same distance, the distance from a to b, but Zak started later so took less time and is traveling faster than Ron.
, there are a lot of unknowns. The soln wasn't given originally.
I am not sure I understand your question. Ron is traveling from a to b, Zak from b to a. Is that correct? He starts at 12:00 and take 72 minutes to get to c, arriving there at 1:12. Zak starts from b at 12:22 and takes 50 min to get to c, arriving there at the same time, 1:12. And the question asks when Ron arrives at b and when Zak arrives at a.
Is that correct?
Let x be the distance, in m, from a to c, y the distance, in m, from c to b. Ron travels at x/72 m/min, Zak at y/50 m/min. It will take Ron a total of (x+ y)(72/x) min to go from a to b, it will take Zak a total of (x+y)(50/y) min to go from b to a.
Since each arrives at his destination at the same time, but Zak started 22 min later, so it takes Zak 22 minutes less than Ron: (x+ y)(72/x)= (x+ y)(50/y)+ 22. Multiply on both sides by x and y to get rid of the fractions: 72y(x+y)= 50x(x+y)+ 22xy. 72xy+ 72y^2= 50x^2+ 50xy+ 22xy. 72y^2- 50x^2= 0 (the "xy" terms cancel!). Dividing by 2, 36y^2- 25x^2= (6y- 5x)(6y+ 5x)= 0. So either 6y- 5x= 0 or 6y+ 5x= 0. The first gives 6y= 5x, the second 6y= -5x. Since x and y are both distances, and so positive, 6y= -5x is impossible and we must have 6y= 5x. That is, we can replace y with (5/6)x. Can you finish the problem now?
