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27 May 2022, 05:15
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:36) correct
27%
(01:31)
wrong
based on 116
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Alex drives to and from work each day along the same route. If he drives at a speed of 80 miles per hour on the way to work and he drives at a speed of 100 miles per hour on the way from work, which of the following most closely approximates his average speed in miles per hour for the round trip?
A. 80.0
B. 88.9
C. 90.0
D. 91.1
E. 100.0
A. 80.0
B. 88.9
C. 90.0
D. 91.1
E. 100.0
27 May 2022, 06:28
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Bunuel
APPROACH #1: Assign a variable the distance from home to work
Average speed = (total distance traveled)/(total travel time)
Let d = the distance from home to work.
So, total distance travelled = d + d = 2d
Time = distance/speed
Travel time from home to work = d/80
The travel time from work to home = d/100
So, total travel time = d/80 + d/100 = 5d/400 + 4d/400 = 9d/400
So, average speed = 2d/(9d/400) = 2d x (400/9d) = 800d/9d = 800/9
No need to actually calculate 800/9.
Instead, recognize that 810/9 = 90, which means 801/9 = 89, which means 800/9 is a little bit less than 89
Answer: B
APPROACH #2: Assign a "nice number" to the distance from home to work
We want a number that works well with 80 mph and 100 mph.
So let's say the distance from home to work = 400 miles
So, total distance travelled = 400 + 400 = 800
Time = distance/speed
Travel time from home to work = 400/80 = 5 hours
The travel time from work to home = 400/100 = 4 hours
So, total travel time = 5 hours + 4 hours = 9 hours
So, average speed = 800/9 = a little bit less than 89
Answer: B
27 May 2022, 11:14
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Since Alex drives to and from work along the same route, the distance he travels is same in both the onward and return journeys. As mentioned in the question, this is a round trip.
When the distance is same for the onward and return journeys (i.e. in round trips),
Average Speed = \(\frac{2*a*b }{ a + b}\),
where, a = speed for onward journey and b = speed for return journey
In Alex’s case, speed for onward journey = 80 miles per hour and speed for return journey = 100 miles per hour
Therefore, average speed of Alex for the round trip = \(\frac{2 * 80 * 100 }{ 180}\) = \(\frac{16000 }{ 180}\) = \(\frac{1600 }{ 18}\).
Note that 18 * 9 = 162, therefore, 18 * 90 = 1620. Since the numerator is 1600, the result should be slightly less than 90; there is one obvious choice in that case and that is 88.9
The correct answer option is B.
When the distance is same for the onward and return journeys (i.e. in round trips),
Average Speed = \(\frac{2*a*b }{ a + b}\),
where, a = speed for onward journey and b = speed for return journey
In Alex’s case, speed for onward journey = 80 miles per hour and speed for return journey = 100 miles per hour
Therefore, average speed of Alex for the round trip = \(\frac{2 * 80 * 100 }{ 180}\) = \(\frac{16000 }{ 180}\) = \(\frac{1600 }{ 18}\).
Note that 18 * 9 = 162, therefore, 18 * 90 = 1620. Since the numerator is 1600, the result should be slightly less than 90; there is one obvious choice in that case and that is 88.9
The correct answer option is B.
