Kalsdale To Lamsville (Posted on 2006-06-19)

	Submitted by K Sengupta
Rating: 3.0000 (1 votes)
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The required distance between Kalsdale and Lamsville is 792 Kilometres, the initial velocity of the train was 98 Kilometres per hour and the train arrived at Cadsville precisely (49 + 47/49) minutes behind the scheduled time. EXPLANATION: Let the the engine snag occurred at a distance of x Kilometres from Lamsville and the initial velocity of the train be y Kilometres per hour. Let us suppose that the train arrived at Lamsville precisely C hours behind schedule. Accordingly, by conditions of the problem: (I) x/(5y/7) = x/y +C; (II) (x-98)/( 5y/7) = (x-98)/y + C - 2/5 Subtracting (II) from (I), we obtain: (98* 2)/5y = 2/5, giving y = 98. Hence, the initial velocity of the train was 98 Kilometres per hour.(Kmph) Also y=98 in (I) gives; 2x/ (5*98) = C, so that: x = 245C Hence, at the outset, the train traveled a distance of 6*98 = 588 Kilometres from Kalsdale prior to the occurrence of the engine snag at an initial velocity of 98 Kmph. Accordingly, it follows that the distance between Kalsdale and Lamsville (D) = 588 + 245C. Since the train reached Lamsville less than one hour behind the scheduled time, it follows that: 588 < D < (588+245); giving: 588 < D < 833-------------------------(#) We readily observe that the only multiple of 264 satisfying relationship (#) occurs at D = 264*3 = 792. Consequently, C = (792 - 588)/ 245 = 204/245 hour, and 144/204 hour = (204/245)*60 = (49 + 47/49) minutes Hence, the required distance between Kalsdale and Lamsville is 792 Kilometres, the initial velocity of the train was 98 Kilometres per hour and the train arrived at Cadsville precisely (49 + 47/49) minutes behind the scheduled time.

	Subject	Author	Date
	Maybe	Bender	2006-06-20 11:01:31
	solution	Charlie	2006-06-19 15:01:34