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Since speed of cycling must be faster than speed of walking, Claire and Charles cannot be started at the same point, Charles must be somewhere behind Claire.

2012-07-08 03:38:51 補充：
Let Charles is d km behind Claire. Let distance from Claire to swimming pool = s. Distance from swimming pool to library = 3. And let speed of cycling of Charles = v. Since Charles overtakes Claire at the library, so time taken for both of them to the library is the same.

2012-07-08 03:41:52 補充：
That is (s + 3)/6 = (d + s + 3)/v .....(1). Time for Charles to swimming pool = (d + s)/v, time for Claire to swimming pool = s/6. But Claire starts 20 min (1/3 hr.) late, so (d + s)/v = s/6 + 1/3 ...........(2)

2012-07-08 03:45:51 補充：
From (1), (s + 3)/6 = (d + s)/v + 3/v, or (d + s)/v = (s + 3)/6 - 3/v. Sub. into (2), (s + 3)/6 - 3/v = s/6 + 1/3, s/6 + 3/6 - 3/v = s/6 + 1/3. That is 1/2 - 3/v = 1/3, 1/2 - 1/3 = 3/v, 1/6 = 3/v, so v = speed of Charles = 3 x 6 = 18 km/hr.(D).