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The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per

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The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per

by BTGmoderatorDC » Fri Mar 27, 2020 6:46 pm

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The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a rate of 3 feet per second. Once Bill reaches the group of people, he stops walking and stands with them until the walkway ends. What is Bill’s average rate of movement for his trip along the moving walkway?

A. 2 feet per second
B. 2.5 feet per second
C. 3 feet per second
D. 4 feet per second
E. 5 feet per second


OA E

Source: Manhattan Prep
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Re: The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per

by Scott@TargetTestPrep » Sat Apr 04, 2020 2:27 pm
BTGmoderatorDC wrote: ↑
Fri Mar 27, 2020 6:46 pm
 The ‘moving walkway’ is a 300-foot long walkway consisting of a conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a rate of 3 feet per second. Once Bill reaches the group of people, he stops walking and stands with them until the walkway ends. What is Bill’s average rate of movement for his trip along the moving walkway?

A. 2 feet per second
B. 2.5 feet per second
C. 3 feet per second
D. 4 feet per second
E. 5 feet per second


OA E

Source: Manhattan Prep
We are given that Bill catches up to a group of people and then remains stationary until the walkway ends. Let’s determine the time, t, needed to catch up to the group of people who are 120 feet ahead of him.

Since Bill is moving at a rate of 6 ft/sec (walkway speed plus Bill’s foot speed), his distance is 6t. Since the group of people is moving at 3 ft/sec (walkway speed only) and the group is 120 feet ahead, the group’s distance is 3t + 120. When Bill catches up to the group, the two distances are equal. Thus, we can create the following equation and determine t:

6t = 3t + 120

3t = 120

t = 40

Therefore, it takes him 40 seconds to catch up to the group. During this 40-second period, he moves 6 x 40 = 240 feet. Since the walkway is 300 ft long, he has 300 - 240 = 60 feet remaining. However, since he is stationary after he catches up to the group, he is moving only at the walkway speed, which is 3 ft/sec. Therefore, in the last 60 feet, it takes 60/3 = 20 seconds to complete the distance, and thus it takes him a total of 40 + 20 = 60 seconds to complete the walkway from one end to another.

Since average speed = total distance/total time, his average speed = 300/60 = 5 ft/sec.

Answer: E

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