Two buses were coming from two different places situated just in the opposite direction. The average speed of one bus is 5 km/hr more than that of another one and they had started their journey in the same time. If the distance between the places is 500 km and they meet after 4 hours, find their speed.
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given : the distance between the two places from where the buses start = 500 km, The time after whuch they meet each other = 4hr. and the Avg. speed of one bus is 5 km/hr more Than the other bus.
Let the bus A start from a place P and the bus B start from the place Q and let they meet at a point R after the journey of 4hr. the distane PQ = 500 km.
Let the avg. speed of the bus A = x km/hr the the avg. speed of the second bus B= (x + 5)km/hr.
The distance travelled by the bus A in 4hr. = PR = 4x The distance travelled by the bus B in 4hr. = QR = 4(x + 5)
PR + QR = PQ
4x + 4(x + 5) = 500
Or, 4x + 4x + 20 = 500
Or, 8x = 500 - 20
Or, 8x = 480
Or, x = 480/8 = 60
Therefore x = 60 km/hr and x + 5 = 60 + 5 = 65
Hence The average speed ot the two buser are : 60km/hr and 65km/hr
To start, you have represent the rate of each bus.
Let x = rate of the first bus and x + 5 = rate of the 2nd bus.
Note that distance = rate*time and t = 4. The distance they traveled in 4 hours was equal to 500. So the working equation should be:
x*4 + (x + 5)*4 = 500
4x + 4x + 20 = 500
8x + 20 = 500 subtracting 20 from both sides
8x = 480 divide both sides by 8
x = 60 kph
x + 5 = 65 kph
The speed of the buses are 60 kph and 65 kph.
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