BTGmoderatorLU wrote: ↑

Sat May 23, 2020 5:48 am

Source: Magoosh

An airport conveyor belt is 120 meters long, and moves at a constant rate of 2 m/s. A man carrying two suitcases walks at a constant rate of 4 m/s. How long will it take him to cross from one end of the conveyor belt to the other, if the man only walks for half the time he is on the conveyor belt?

A. 25 seconds
B. 30 seconds
C. 42 seconds
D. 45 seconds
E. 54 seconds

The OA is B

Solution:

If he walks but does not use the conveyor belt, it will take him 120/4 = 30 seconds to walk the length of the conveyor belt.

Therefore, he walks for 15 seconds on the conveyor belt before he stops and lets the conveyor belt move at its own speed.

For the 15 seconds he walks on the conveyor belt, he’s moving 4 + 2 = 6 m/s.

Therefore, he moves 6 x 15 = 90 meters on the conveyor belt while he is walking on it.

For the remaining 30 meters that he is not walking, it takes 30/2 = 15 seconds.

Therefore, it takes a total of 15 + 15 = 30 seconds to move from one end of the conveyor belt to the other.

Alternate Solution:

Let’s keep in mind that the rate of the man is 2 + 4 = 6 m/s when he is walking on the conveyor belt and 2 m/s when he is standing still. In the light of this information, let’s test each answer choice.

A) 25 seconds

If the entire trip took 25 seconds, he walks 12.5 seconds on the conveyor belt and stands still for another 12.5 seconds. Therefore, he covers a distance of (12.5)*6 + (12.5)*2 = 75 + 25 = 100 meters. Answer choice A is not correct.

B) 30 seconds

If the entire trip took 30 seconds, he walks 15 seconds on the conveyor belt and stands still for another 15 seconds. Therefore, he covers a distance of (15)*6 + (15)*2 = 90 + 30 = 120 meters. Answer choice B is correct.

Answer: B