The 'moving walkway'

sanju09 wrote:The 'moving walkway' is 300-foot long 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 combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary until the walkway ends. What is Bill's average rate of movement for his trip along the moving walkway?
(A) 2 ft/sec
(B) 2.5 ft/sec
(C) 3 ft/sec
(D) 4 ft/sec
(E) 5 ft/sec

Bill's rate alone = (combined rate for Bill and the walkway) - (rate of the walkway alone) = 6-3 = 3 feet per second.
Time for Bill to walk to the people 120 feet in front of him = d/r = 120/3 = 40 seconds.

In the 40 seconds it takes Bill to reach the group of people, the distance traveled by the walkway = r*t = 3*40 = 120 feet.

Since Bill has traveled 120 feet, and the walkway has traveled 120 feet, the remaining distance = 300-120-120 = 60 feet.
Time for the walkway alone to carry Bill the remaining distance = d/r = 60/3 = 20 seconds.

Average speed for the entire trip = (total distance)/(total time) = 300/(40+20) = 300/60 = 5 feet per second.

The correct answer is E.

GMATGuruNY wrote:

sanju09 wrote:The 'moving walkway' is 300-foot long 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 combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary until the walkway ends. What is Bill's average rate of movement for his trip along the moving walkway?
(A) 2 ft/sec
(B) 2.5 ft/sec
(C) 3 ft/sec
(D) 4 ft/sec
(E) 5 ft/sec

Bill's rate alone = (combined rate for Bill and the walkway) - (rate of the walkway alone) = 6-3 = 3 feet per second.
Time for Bill to walk to the people 120 feet in front of him = d/r = 120/3 = 40 seconds.

In the 40 seconds it takes Bill to reach the group of people, the distance traveled by the walkway = r*t = 3*40 = 120 feet.

Since Bill has traveled 120 feet, and the walkway has traveled 120 feet, the remaining distance = 300-120-120 = 60 feet.
Time for the walkway alone to carry Bill the remaining distance = d/r = 60/3 = 20 seconds.

Average speed for the entire trip = (total distance)/(total time) = 300/(40+20) = 300/60 = 5 feet per second.

The correct answer is E.

I only wish there be a "LIKE" button also.

sanju09 wrote:The 'moving walkway' is 300-foot long 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 combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary until the walkway ends. What is Bill's average rate of movement for his trip along the moving walkway?
(A) 2 ft/sec
(B) 2.5 ft/sec
(C) 3 ft/sec
(D) 4 ft/sec
(E) 5 ft/sec

The Princeton Review

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Time taken for Bill to meet the group = 120/(6-3) = 40s
Actual distance covered by Bill in this time = 6*40 = 240 ft
Remaining distance = 300 - 240 = 60 ft
Time taken for Bill to cover this remaining distance = 60/3 = 20s

So total time taken for Bill to climb the escalator of 300 ft = 60s

Average speed = 300/60 = 5 ft/sec

Choose E

Cheers