If Bill gives Peter an hour's head start, he overtakes him when they havew both walked 12km. If peter wlaks at twice his normal speed, he can give bill an hours' head start and overtake him when they have walked 24km. Find the speed at which both Bill and Peter walked. (Assume they each walk at a constant speed).
Let Bill and Peter be walking at x and y kph respectively. Then Bill walks 12 km in 12/x hour and Peter walks 12 km in 12/y km. Since Petr had an hour’s headstart over Bill , hence 12/x = (12/y)-1 or, 12/y = (12/x)+1…(1).Further, If Peter walks at 2y kph, then he can walk 24 km in 24/2y or, 12/y hour. Also, Bill walks 24 km in 24/x hour. Now, since, this time, Bill had an hours' head start over Peter, hence 12/y = (24/x)-1…(2). From these 2 equations, we get (12/x)+1 = (24/x)-1 or, (12+x)/x = (24-x)/x or, 12+x = 24-x or, 2x = 24-12 = 12 so that x = 12/2 = 6. On substituting x = 6 in the 2nd equation, we get 12/y = (24/6)-1 = 3 or, 3y = 12 so that y = 4.
Thus Bill and Peter are walking at 6 and 4 kph respectively.
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