A train travels at a certain average speed for distance of 63km and then travels st a distance of 72km at an average speed of6km/hr more than its original speed. If it takes 3hrs to complete total journey, what is the original average speed?
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CASE I:
DISTANCE = 63 KM.
SPEED = x.......(say)
TIME = DISTANCE/SPEED = 63/x..............(i)
CASE II :
DIATANCE =72 KM.
SPEED = x+6
TIME= 72/[x+6].............(ii)
TOTAL TIME = 3.....GIVEN.
(i)+(ii) = 3.
[63/x] + [72/(x+6)] = 3.
divide both sides by 321/x + [24/(x+6)] = 1.
x2-39x + 126 =0.
x2+42x -3x +126 =0.
x(x+42) - 3(x+42) = 0.
speed cannot be negative
thus speed= 3 km/hr.
hope this help you
DISTANCE = 63 KM.
SPEED = x.......(say)
TIME = DISTANCE/SPEED = 63/x..............(i)
CASE II :
DIATANCE =72 KM.
SPEED = x+6
TIME= 72/[x+6].............(ii)
TOTAL TIME = 3.....GIVEN.
(i)+(ii) = 3.
[63/x] + [72/(x+6)] = 3.
divide both sides by 321/x + [24/(x+6)] = 1.
x2-39x + 126 =0.
x2+42x -3x +126 =0.
x(x+42) - 3(x+42) = 0.
speed cannot be negative
thus speed= 3 km/hr.
hope this help you
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