Answer :
The speed of the slower bus is 49 mi/h.
The speed of the faster bus is 62 mi/h.
In the question, we are given that two buses leave a station at the same time and travel in opposite directions. One bus travels 13 mi/h slower than the other.
We are asked to find the rate of each bus if the two buses are 222 miles apart after 2 hours.
We assume the speed of the slower bus to be x mi/h, making the speed of the faster bus be (x + 13) mi/h.
We know that, speed = distance/time, or, distance = speed*time.
Thus, for the:
Slower bus:
Speed = x mi/h.
Time = 2 hours.
Thus, distance = 2x mi.
Faster bus:
Speed = (x + 13) mi/h.
Time = 2 hours.
Thus, the distance = 2(x + 13) mi.
The distance between them can be thus shown as:
2x + 2(x + 13) miles, which is given to be 222 miles, leading to the linear equation in one variable:
2x + 2(x + 13) = 222,
or, 2x + 2x + 26 = 222,
or, 4x = 222 - 26 = 196,
or, x = 196/4 = 49.
Thus, the speed of the slower bus = x mi/h = 49 mi/h.
The speed of the faster bus = (x + 13) mi/h = (49 + 13) mi/h = 62 mi/h.
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