# mathematics

The distance between two towns A and B is360km.A minibus left town A at 8.15a.m and travelled towards B at an average speed of 90km/hr. A matatu left town B at 10.35a.m on the same day and travelled towards A at an average speed of 110km/hr.
a)i)how far from A did they meet? (4mk)

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1. 10:35-8:15 = 2:20 = 2.33 h. delay.
90T + 110(T-2.33) = 360
200T-256 = 360
T = 3.08 h.
d = 90 * 3.08 = 277 km from Town A.

t

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2. I like the method

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3. The distance that the minibus covered before the matatu started the journey is 10:35-8:15 which gives 2hrs&20min. So the distance is 2.333333333 times 90km/hr which gives 210km. Therefore relative distance is 360-210 which gives 150km . while relative speed is 110 plus 90km/h, which gives 200km/hr . but relative time is equal to relative distance divide by relative speed which is 150km divide by 200 , which gives 45 min or 3/4hrs . this is the time they took to meet .but distance they covered to meet from A is 3/4 times90km/hr which gives 67.5.which we add to 210 which gives 277.5 which is the answer

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