Mark drove 150 mi to visit Sandra. Returning by a shorten route, he found that the trip was individual 130 mi, but traffic slowed his speed by 6 miles. If the two trips took exactly the same time, what was his rate on the return trip?
I enjoy no idea how to begin this problem. Any backing would be appreciated!
Answers: Return speed is 39 mph
METHOD
If x is the outward speed in mph, then (x-6) is the inward speed surrounded by mph
The outward journey takes [150 divided by x] hours.
The inward tour takes [130 divided by (x-6)] hours
but these are equal, so
[150 divided by x ] equals [130 divided by (x-6)]
which is equivalent to
[150 multiplied by (x-6)] equals [130 multiplied by x]
which is equivalent to
150x - 900 = 130x
so, 20x = 900
thus x = 45
Since the ourward speed is 6mph faster than the return, the return speed is (45-6) mph
or 39 mph
I enjoy no idea how to begin this problem. Any backing would be appreciated!
Is this the answer to congestion and...
Answers: Return speed is 39 mph
METHOD
If x is the outward speed in mph, then (x-6) is the inward speed surrounded by mph
The outward journey takes [150 divided by x] hours.
The inward tour takes [130 divided by (x-6)] hours
but these are equal, so
[150 divided by x ] equals [130 divided by (x-6)]
which is equivalent to
[150 multiplied by (x-6)] equals [130 multiplied by x]
which is equivalent to
150x - 900 = 130x
so, 20x = 900
thus x = 45
Since the ourward speed is 6mph faster than the return, the return speed is (45-6) mph
or 39 mph
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