the distance between two stations a and b is 230 km. two motorcyclist ...
Speed of first cyclist = X km/hr
speed of other cyclist = (X - 10) km/hr
distance covered by first cyclist
after 3 hours = 3X km
distance covered by other cyclist
after 3 hours = 3(X - 10) km
= (3X - 30) km
according to condition,
3X + 3X - 30 + 20 = 230
6X -10 = 230
6X = 230 + 10
6X = 240
X = 240/6
X = 40
therefore,
speed of first cyclist = X
= 40 km
speed of other cyclist = X - 10
= 40 - 10
= 30 km
the distance between two stations a and b is 230 km. two motorcyclist ...
Problem:
The distance between two stations A and B is 230 km. Two motorcyclists start simultaneously from A and B in the opposite directions. The distance between them after 3 hours is 20 km. If the speed of one motorcyclist is less than that of the other by 10 km/h, find the speed of each motorcyclist.
Solution:
Let the speed of the faster motorcyclist be x km/h. Then, the speed of the slower motorcyclist would be (x - 10) km/h.
After 3 hours, the distance covered by the faster motorcyclist would be 3x km and the distance covered by the slower motorcyclist would be 3(x - 10) km. The total distance covered by both of them would be 3x + 3(x - 10) = 6x - 30 km.
Given that the distance between them after 3 hours is 20 km. So, we have the equation:
6x - 30 = 230 - 20
6x - 30 = 210
6x = 240
x = 40
Therefore, the speed of the faster motorcyclist is 40 km/h and the speed of the slower motorcyclist is (40 - 10) = 30 km/h.
Answer:
The speed of the faster motorcyclist is 40 km/h and the speed of the slower motorcyclist is 30 km/h.
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