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A bicyclist started on his trip from city A to city B. In 1 hour and 36 minutes, a biker also left A and headed towards B, and he arrived there at the same time as the bicyclist. Find the speed of the bicyclist if it is less than the speed of the biker by 32 km/hour, and the distance between the two cities is 52 km.

I need help on what this means. thanks!

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closed as off-topic by user228113, hardmath, user91500, Lee Mosher, user296602 Jan 29 '16 at 5:22

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, hardmath, user91500, Lee Mosher, Community
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  • $\begingroup$ The title seems to be about "t[w]o bicyclists", but understanding the word problem may hinge on distinguishing the bicyclist and the biker (presumably someone riding a motor bike, and hence travelling at a faster speed). $\endgroup$ – hardmath Jan 29 '16 at 1:06
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You are expected to write a pair of simultaneous equations. Let $a$ be the speed of A in km/hour and $b$ be the speed of B. How long does it take each to ride $52$ km? You are given the difference in speeds and the difference in times. Each one becomes an equation, giving two equations in two unknowns.

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  • $\begingroup$ Excellent answer. So many people in this situation would just give the solution, which doesn't really help the OP learn particularly. +1 $\endgroup$ – Sam T Jan 23 '16 at 21:45
  • $\begingroup$ @SmileySam: Thanks. I do this sort of thing often. In this case it is what OP explicitly asked for. $\endgroup$ – Ross Millikan Jan 23 '16 at 21:47
  • $\begingroup$ Exactly. When I give answers, in general I say how to do it, and then leave the details to the OP. (Sometimes it's different if it's just one piece of rearranging or something, but in general.) Similarly, when I ask a question, I always ask for answers that give hints, not just full solutions. Anyway. :) $\endgroup$ – Sam T Jan 23 '16 at 22:17

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