CAT 2001 — QA Question 24
Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top of the escalator after having taken 25 steps. While Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?
Answer & solution
- A
40
50
- C
60
- D
80
Assume Shyama takes 3 steps and Vyom takes 2 steps in 6 seconds and let the escalator moves up by x steps per second.
∴Shyama takes 6/3 = 2 seconds for a step and Vyom takes 6/2 = 3 seconds for a step
∴ Shyama took 25 × 2 = 50 seconds to go up
∴ Height of the stairway = (25 + 50x) steps
Vyom took 20 × 3 = 60 seconds to go up
Similarly, in Vyom’s case, the height of the stairway = (20 + 60x) steps
∴ 20 + 60x = 25 + 50x
∴ x = 1/2
∴ If the escalator was turned off, they would have to take (20 + 60 × 1/2) = 50 steps
Hence, option (b).