In the previous post, we saw escalator questions pertaining to movement of an individual along or opposite an escalator. In this concluding post, I will try to explain movement of two individuals along an escalator.

The basics remain the same as was discussed in the previous post. Distance is always in terms of steps, and distance covered individually by the person and the escalator would add up to the number of visible steps on the escalator.

Let’s start with an example covering one of the question types:

Prasad takes 50 steps and Sriram takes 75 steps to move along an escalator while the escalator is moving down. Given that the time taken by Prasad to take 1 step is equal to time taken by Sriram to take 3 steps, how many steps are visible on the escalator when it is stationary?

Here, we know the distance covered individually by Prasad and Sriram and the ratio of their speeds. What we need to find is the speed of the escalator and the time for which it is moving so that we can find the distance covered by the escalator in either of the cases.

As the distance covered individually is already known, you just need to process the information pertaining to their speeds and figure out if something is possible. When Prasad takes 1 step, Sriram takes 3 steps. Let the speed of Prasad be S steps/sec. Then the speed of Sriram becomes 3S steps/sec. We already know the individual distance covered. As we saw in the last part of the previous article, we can figure out the time taken by Prasad and Sriram to complete the entire journey. Also, let the speed of the escalator be x steps/sec. We can form the equation easily as the effective distance covered by both of them would be the same.

(S+x)(50/S)=(3S+x)(75/3S)

50S+50x=75S+25x

25S=25x

x=S

We can reverse substitute and find the answer to be 100.

Just keep in mind that there could be at most two variables in these questions and so, you can freely assume the speed of the escalator as a variable post which you are almost done. If the speed of the individuals are known, you can simply insert those values. If they are not known, there will be some relationship given between them which can be used to put the second variable. Let’s solve another example:

Deepti and Pavitra walk up an escalator. Deepti takes 9 steps in the same time that Pavitra takes 16 steps. Deepti gets to the top of the escalator after having taken 30 steps while Pavitra, takes 40 steps to reach the top. If the escalator was turned off, how many steps would they have to take to walk up?

Again, people who have understood the topic would pounce on this. Distances covered by Deepti and Pavitra individually are known and the ratios of their speeds is known. So again, we can take 2 variables and solve it as we did in the previous question. Let speed of Deepti be 9S steps/sec. Speed of Pavitra would be 16S steps/sec. Let speed of escalator be x steps/sec.

(9S+x)(30/9S)=(16S+x)(40/16S)

On solving, we get:

x=12S and so, number of steps = 70.

Difference in the number of steps:

The final bit that you need to know in this topic is how the movement occurs between the two individuals in question. It is extremely similar to an upstream downstream example and we end exactly where we started. Consider the following question:

Prasad is moving along an escalator at the speed of 2 steps/sec starting from the bottom. Sriram is traveling against the same escalator at the speed of 4 steps/sec starting from the top. Prasad takes 20 steps to reach the top and both of them start and end at the same time. What is the difference in the number of steps that they have taken when they meet?

This might sound extremely cryptic when we look at it first but if you think carefully, you can see that this is nothing but a linear motion question wherein, at the first meeting point, we have to find the difference in the distances covered by both the individuals. Once you figure this out, you are sorted.

We know the individual speeds, and we know the distance covered by Prasad. What we do not know is the number of steps on the escalator (ie. the total distance) and the number of steps taken by Sriram. But we know an additional bit of information that the time taken is same.

Now, the distance covered is the same and the time taken is the same. So, the effective speeds must be the same too. So, we get a simple equation, assuming the speed of the escalator to be x steps/sec.

2+x=4-x

x=1 steps/sec

So, we get the total steps easily to be:

Distance covered by Prasad + Distance covered by the escalator

20 + 10*1 = 30 steps in total.

To find the difference in the number of steps taken, it would be easier if you can visualize this as a different type of a problem wherein, Prasad and Sriram start from the opposite ends of a path. There is a wind blowing at the speed of 1 unit per second assisting Prasad. Speed of Prasad is 2 units/sec and speed of Sriram is 4 units/sec. Find their meeting point if they cover the entire distance, equal to 30 units in the same time.

Again, we can see that this is nothing but a time-speed-distance question of two people moving in opposite directions. Effective speed of Prasad is 3 units/sec and effective speed of Sriram is also 3 units/sec. So, they will meet at the midpoint of the 30 units stretch.

So it means that Prasad has covered 15 units and Sriram has covered 15 units with and against the assistance of wind respectively. As the ratio of the units covered by Prasad to the total units covered is 2:3, Prasad has covered 10 units ie. 10 steps whereas, Sriram has covered 15 steps plus an additional 5 steps to negate the escalator and so, 20 steps. So, the difference is of 10 steps.

The above type is something that is highly unexpected in CAT but is a simple application of a logical principle and so, can be dealt with easily. You can also look at it from another angle wherein, you can find the total steps taken individually by Prasad and Sriram and then halve the difference as they are meeting at the midpoint. Prasad has taken 20 steps whereas Sriram would have taken 30+10=40 steps in total. As the difference has been constantly reducing in the same proportion, by the time they reach the midpoint, the difference would be of 10 steps.

So, with a few simple techniques, you can easily crack any escalator question within a minute or so and gain those 3 marks. This is one of the types where you have an edge over a non-serious aspirant and is a big bonus when it comes to the final tally.

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