If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what is the average speed for the entire trip?
1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
2) The distance from Townsend to Smallville is 165 miles.
The OA is A
Source: Manhattan Prep
If a car traveled from Townsend to Smallville at an average
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Average speed is a WEIGHTED average: (total distance)/(total time).
Since we're given that this is a round trip, the distance of each leg of the trip will be the same. Since we know the average speed for one leg of the trip, all we need to know to calculate the overall average speed is the average speed of the return trip.
1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
We can use this to calculate the speed of the return trip:
Rate x Time = Distance
1st leg: 40t = d --> t = d/40
If that's 50% longer than the time for the other trip, let x = time for return trip:
d/40 = (1.5)x --> d/40 = (3/2)x
x = d/60
Thus, the rate for the return trip must be 60 mph.
... Or we could pick numbers. Let's say the distance = 120 miles.
40t = 120
t = 30
If this is 50% more than the time for the return trip, the time for the return trip = 30/(3/2) = 20.
r(20) = 120
r = 60 mph
If we know both rates, we can solve for the average speed for the entire trip:
Algebraically:
total distance = 2d
total time = (d/40) + (d/60) --> 5d/120
(total distance)/(total time) = (2d)/(5d/120) = 240/5 = 48.
Picking numbers:
total distance = 120 + 120 = 240
total time = 30 + 20 = 50
(total distance)/(total time) = 240/5 = 48.
Sufficient.
2) The distance from Townsend to Smallville is 165 miles.
Just knowing the distance allows us to solve for the time of the first leg of the trip, but gives us no information about the rate or the time of the 2nd leg of the trip. Thus, we can't infer anything about the average speed for the entire trip. Insufficient.
The answer is A.
Since we're given that this is a round trip, the distance of each leg of the trip will be the same. Since we know the average speed for one leg of the trip, all we need to know to calculate the overall average speed is the average speed of the return trip.
1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
We can use this to calculate the speed of the return trip:
Rate x Time = Distance
1st leg: 40t = d --> t = d/40
If that's 50% longer than the time for the other trip, let x = time for return trip:
d/40 = (1.5)x --> d/40 = (3/2)x
x = d/60
Thus, the rate for the return trip must be 60 mph.
... Or we could pick numbers. Let's say the distance = 120 miles.
40t = 120
t = 30
If this is 50% more than the time for the return trip, the time for the return trip = 30/(3/2) = 20.
r(20) = 120
r = 60 mph
If we know both rates, we can solve for the average speed for the entire trip:
Algebraically:
total distance = 2d
total time = (d/40) + (d/60) --> 5d/120
(total distance)/(total time) = (2d)/(5d/120) = 240/5 = 48.
Picking numbers:
total distance = 120 + 120 = 240
total time = 30 + 20 = 50
(total distance)/(total time) = 240/5 = 48.
Sufficient.
2) The distance from Townsend to Smallville is 165 miles.
Just knowing the distance allows us to solve for the time of the first leg of the trip, but gives us no information about the rate or the time of the 2nd leg of the trip. Thus, we can't infer anything about the average speed for the entire trip. Insufficient.
The answer is A.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education