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1)An escalator is descending at constant speed.

A walks down and takes 50 steps to reach the bottom.

B runs down and takes 90 steps in the same time as A takes 10

steps.

How many steps are visible when the escalator is not operating?

2)Every day a cyclist meets a train at a particular crossing.

The road is straight before the crossing and both are travelling in

the same direction.The cyclist travels with a speed of 10 Kmph.

One day the cyclist comes late by 25 min. and meets the train 5km

before the crossing. What is the speed of the train?

A walks down and takes 50 steps to reach the bottom.

B runs down and takes 90 steps in the same time as A takes 10

steps.

How many steps are visible when the escalator is not operating?

2)Every day a cyclist meets a train at a particular crossing.

The road is straight before the crossing and both are travelling in

the same direction.The cyclist travels with a speed of 10 Kmph.

One day the cyclist comes late by 25 min. and meets the train 5km

before the crossing. What is the speed of the train?

It would appear for example that B should have been running UP the escalator and hence take (90*(50/10)) or 450 steps to reach the TOP since he is walking against the flow.

If we assume that both A & B reach the ends of the escalator at the same time .. then the number of extra steps A saved would be the number of extra steps B took (both due to escalator movement).

Then if we stop the escalator they will both have an equal number of steps to reach the top or bottom. A would have a few more (50+200) and B would have a few less (450-200) so they would both walk 250 steps (the actual number of steps on the escalator from either direction).

I am also preparing for Infy :)

Distance = Speed * Time

Cyclist:

Speed = 10 km/h

Distance = 5 km

Time = 5/10 h = 1/2 h = 30 min

The cyclist takes 30 min to travel 5 km.

The cyclist is 25 minutes late, so it takes the train 5 minutes to travel 5 km.

Hence, the speed of the train is 60 km/hr.

Number of visible steps of the non-operating escalator: n

Speed of the escalator: v

Number of steps taken by A to reach the bottom: 50

Number of steps taken by B to reach the bottom: 90 + (50 - 10) = 130

Walking down:

Time: (n - 50) / v

Running down:

Time: (n - 130) / v

[(n - 50) / v] : [(n - 130) / v] = 5 : 1

(n - 50) : (n - 130) = 5 : 1

(n - 50) * 1 = (n - 130) * 5

n - 50 = 5n - 650

n - 50 - n + 650 = 5n - 650 - n + 650

600 = 4n

600 / 4 = 4n / 4

150 = n

150 steps are visible.

But I am not convinced by the answer

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Mathmatically if we assume daily both meet at the 0km point at 0:00.

We know the distance to the crossing is 5km and it will take the cyclist 30min to reach 0km (5km@10kmph). Since they normally meet at the crossing (not this point) and are travelling in parallel. We know the cyclist is (overall) 25min late. We must assume the train is NOT late and hence will be at the crossing at 0:00 (5km and 5min hence). The train must be moving at 1km/min or 60kmph.