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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Challenge: At noon, Joe and a hummingbird leave Point A and
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27 Oct 2018, 10:37
27 Oct 2018, 10:37
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70% (02:12) correct
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At noon, Joe and a hummingbird leave point A and travel towards point B, which is 12 miles away. At the same time Sue leaves point B and travels toward point A. Joe, Sue and the hummingbird travel at constant speeds of 3 miles per hour, 4 miles per hour, and 84 miles per hour respectively. When the hummingbird reaches Sue, it immediately changes direction and heads toward Joe. When it reaches Joe, it turns around and heads toward Sue. If this process continues until Sue and Joe meet, what is the total distance traveled by the hummingbird?
A) 49 miles
B) 84 miles
C) 121 miles
D) 144 miles
E) 168 miles
A) 49 miles
B) 84 miles
C) 121 miles
D) 144 miles
E) 168 miles
Re: Challenge: At noon, Joe and a hummingbird leave Point A and
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27 Oct 2018, 20:03
27 Oct 2018, 20:03
Expert Reply
GreenlightTestPrep wrote:
At noon, Joe and a hummingbird leave point A and travel towards point B, which is 12 miles away. At the same time Sue leaves point B and travels toward point A. Joe, Sue and the hummingbird travel at constant speeds of 3 miles per hour, 4 miles per hour, and 84 miles per hour respectively. When the hummingbird reaches Sue, it immediately changes direction and heads toward Joe. When it reaches Joe, it turns around and heads toward Sue. If this process continues until Sue and Joe meet, what is the total distance traveled by the hummingbird?
A) 49 miles
B) 84 miles
C) 121 miles
D) 144 miles
E) 168 miles
A) 49 miles
B) 84 miles
C) 121 miles
D) 144 miles
E) 168 miles
If one can understand that the TIME traveled by all 3 is the SAME, the question becomes very easy..
So in the given time J and S cover a combined distance of 12 miles..
The speed combined is 3+4=7 miles per hour, so J and S cover 7 miles in an hour
Thus they will take 12/7 hours to cover 12 miles.
So the hummingbird is also traveling for 12/7 hours
@ 84 Miles per hour, it covers \(84*\frac{12}{7}=12*12=144\) miles
D
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Challenge: At noon, Joe and a hummingbird leave Point A and
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28 Oct 2018, 14:24
28 Oct 2018, 14:24
GreenlightTestPrep wrote:
At noon, Joe and a hummingbird leave point A and travel towards point B, which is 12 miles away. At the same time Sue leaves point B and travels toward point A. Joe, Sue and the hummingbird travel at constant speeds of 3 miles per hour, 4 miles per hour, and 84 miles per hour respectively. When the hummingbird reaches Sue, it immediately changes direction and heads toward Joe. When it reaches Joe, it turns around and heads toward Sue. If this process continues until Sue and Joe meet, what is the total distance traveled by the hummingbird?
A) 49 miles
B) 84 miles
C) 121 miles
D) 144 miles
E) 168 miles
A) 49 miles
B) 84 miles
C) 121 miles
D) 144 miles
E) 168 miles
The main idea here is that all 3 travelers travel for the SAME amount of time.
So, let's first determine the time it takes for Joe and Sue to meet.
This is a shrinking gap question.
To determine the rate at which the gap shrinks, we'll ADD the travel speeds of Joe and Sue
That is, shrink rate = (3 miles per hour) + (4 miles per hour) = 7 miles per hour
At noon, the gap between Joe and Sue is 12 miles.
time = distance/speed
So, the time it takes to reduce the gap to zero = 12/7
That is, it takes 12/7 hours until Joe and Sue meet.
During those 12/7 hours, the hummingbird is flying at a constant speed of 84 mph
distance = (speed)(time)
So, the distance traveled by the hummingbird = (84)(12/7) = 144 miles
Answer: D
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