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Problems related to escalators baffle a lot of students preparing for CAT. Probably because not much time has been devoted to explaining this simple concept. Although we keep telling in our CAT classrooms continuously that escalator questions are similar to ‘upstream’ ‘downstream’ questions that students solve in time, speed and distance, the future MBAs still get confused because of the way that the questions are put. Today, we are going to see the theory behind escalator questions and solve them through both equations and ratios. In order to benefit the most, please solve these questions on your own before looking at the solutions.
An escalator is a moving stair, i.e. it moves continuously up or down. At any given point in time, the total number of stairs are fixed in an escalator. For example, for an escalator going up that is showing 50 steps, if 2 stairs disappear at the top, 2 stairs come out at the bottom. Most of the escalator questions involve people moving on an escalator. Remember that the speed of climbing stairs for a person does not vary whether the escalator is moving or still. For example, if Bantu can climb stairs at 2 stairs per second, his speed of climbing stairs on a moving escalator would stay the same.
Common Problem Scenarios in Escalators
1. A person moving on a descending escalator and/or an ascending escalator and questions related to total number of steps, speed of the escalator etc.
2. Two persons moving in the same directions on an ascending/a descending escalator and questions related to total number of steps, speed of the escalator, speeds of either person etc.
3. Two persons moving in opposite directions on an ascending/a descending escalator and questions related to total number of steps, speed of the escalator, speeds of either person etc.
Theory of Escalators in Time Speed and Distance
The basic thing to understand is that
1. if you’re moving ‘with’ the escalator, you’d have to climb less steps on your own because the escalator also will ‘push you forward’ some number of steps on its own. For example, if you climb 70 stairs on your own, and escalator pushes out 30 stairs in your favor in the same time, you’d have climbed 100 stairs in total.
2. If you’re moving ‘against’ the escalator, you’d have to climb more steps on your own because the escalator will try to ‘pull you back’ some number of steps on its own. For example, if you climb 70 stairs on your own, and escalator pushes out 30 stairs against you in the same time, you’d have climbed 40 stairs in total.
Remember!
With the escalator = Less steps to climb than the total
Against the escalator = More steps to climb than the total
Moving With the Escalator
Suppose a person can climb P stairs/second and the escalator is churning out E stairs/second. Let the total number of stairs in the escalator is L. Let the person climb D stairs on his own on the escalator. The time taken to do this would be $\frac{D}{P}$ seconds. In this time, the escalator will churn out $\frac{D}{P} \times E$ stairs.
Total number of steps = steps climbed oneself + steps produced by escalator OR
$L = D + \frac{D}{P} \times E$
Moving Against the Escalator
Given the same values as above, i.e. P, E, L, and D, the person is climbing more stairs than the total as he is working against the escalator. Therefore, the equation will be.
Total number of steps = steps climbed oneself – steps produced by escalator OR
$L = D – \frac{D}{P} \times E$
Now let’s look at some problems!
Problem 1: A man and his wife walk up a moving escalator. The man walks twice as fast as his wife. When he arrives at the top, he has taken 28 steps. When she arrives at the top, she has taken 21 steps. How many steps are visible in the escalator at any one time.
Solution: Since the man’s wife is slower, she would have taken more number of steps if they were moving ‘against’ the escalator. Therefore, they are moving ‘with the escalator, whereby the man’s wife takes less number of steps but more number of steps come out meanwhile as she is slow. Let the man’s speed be 2v and wife’s speed be v. Let the speed of the escalator be u. Then, total number of steps are equal in both cases.
$ 28 + \frac{28}{2v} \times u = 21 + \frac{21}{v} \times u$ or $\frac{u}{v}= 1$
Keeping the value on any side of the equation, the total length $= 28 + 14 = 42$
Alternative solution: The times taken by man and his wife are $\frac{28}{2v}$ and $\frac{21}{v}$ which are in the ratio $\frac{2}{3}$. Therefore, if x is the length of steps which escalator pushes out during the wife’s climbing, $\frac{2x}{3}$ is the length which comes out during the man’s climbing. Again, total number of steps are the same.
$28 + \frac{2x}{3} = 21 + x$ or x = 21. Therefore, total steps $= 28 + \frac{2}{3} \times 21 = 42$
Problem 2: A woman is walking down a downward-moving escalator and steps down 10 steps to reach the bottom. Just as she reaches the bottom of the escalator, a sale commences on the floor above. She runs back up the downward moving escalator at a speed five times that which she walked down. She covers 25 steps in reaching the top. How many steps are visible on the escalator when it is switched off?
Solution: Let the woman’s speed be v ‘with’ the escalator and 5v ‘against’ the escalator. Let the speed of the escalator be u. Then, total number of steps are equal in both cases.
$10 + \frac{10}{v} \times u = 25 – \frac{25}{5v} \times u$ or $\frac{u}{v}= 1$
Keeping the value on any side of the equation, the total length $= 10 + 10 = 20$
Alternative solution: The times taken by the woman ‘with’ the escalator and ‘against’ the escalator are $\frac{10}{v}$ and $\frac{25}{5v}$ which are in the ratio 2:1. Therefore, the steps thrown out of the escalator are in the ratio 2: 1. Hence,
10 + 2x = 25 – x or x = 5. Therefore, total number of stairs = 10 + 10 = 20
Problem 3: A famous mathematician who is always in a hurry walks up an up-going escalator at the rate of one step per second. Twenty steps bring him to the top. The next day he goes up at two steps per second, reaching the top in 32 steps. How many steps are there in the escalator?
Solution: The times taken on two days are 20s and 16s which are in the ratio 5:4. Therefore, the steps thrown out by the escalator are in the ratio 5:4. Therefore, 20 + 5x = 32 + 4x or x = 12. therefore, total steps are 20 + 60 = 80.
Problem 4: A man can walk up a moving ‘up’ escalator in 30s. The same man can walk down this moving ‘up’ escalator in 90s. Assume that his walking speed is same upwards and downwards. How much time will he take to walk up the escalator, when it’s not moving?
Solution: The times taken ‘with’ and ‘against’ the escalator are in the ratio 1:3. Therefore, the steps thrown out of the escalator are also in the ratio 1:3. Also, the number of stairs that he climbs on his own will also be in the ratio 1:3. Therefore, let the stairs he climbs are L and 3L, and steps produced by the escalator be x and 3x. Hence, L + x = 3L – 3x or L = 2x or x = L/2. Therefore total steps = L + L/2 = 3L/2. Now, time taken for L is 30s, therefore time taken for 3L/2 will be 45s.
Problem 5: Ram always walks down on a moving escalator to save time. He takes 50 steps while he goes down. One day due to power failure of 10 secs (when the escalator comes to a halt) he takes 9 secs more than usual time to get down. Find the visible steps of the escalator.
Solution: Let Ram cover N stairs in those 10 seconds, after which rest of the remaining distance he climbs down in a normal manner once the escalator starts. Therefore, climbing down those N stairs with escalator ON and without escalator ON brings a difference of 9s. Therefore, he normally covered the distance in 10 – 9 = 1s everyday.
OR $\frac{v + u}{v} = \frac{10}{1}$ or $\frac{u}{v} = 9$ Therefore the speed of the escalator is 9 times the speed of Ram. Everyday he climbs down 50 stairs. Therefore, in that time the escalator pushes out $50 \times 9 = 450$ stairs. Therefore, total number of stairs = 50 + 450 = 500.
Problem 6 Rohan walked down a descending escalator and took 40 steps to reach the bottom. Sohan started simultaneously from the bottom, taking 2 steps to every 1 step taken by Rohan. Time taken by Rohan to reach the bottom from the top is the same as time taken by Sohan to reach the top from the bottom.
1. How many steps more than Rohan did Sohan take before they crossed each other?
2. If Rohan were to walk at the speed of Sohan, what % of the initial time would he able to save?
Solution:Let Rohan’s speed be v and Sohan’s speed be 2v. Since the time taken for both is same the relative speed of both is same. Let the escalator’s speed be u. Therefore, v + u = 2v – u or v = 2u. Therefore, the speed of the escalator is half the speed of Rohan. Therefore, when Rohan covers 40 steps on his own, the escalator throws 20 steps. Therefore total steps on the escalator = 40 + 20 = 60.
1. Both Rohan and Sohan will meet halfway, i.e. 30 steps, as their relative speeds are same. Rohan’s speed is double that of escalator, therefore, to cover 30 steps, Rohan will cover 20 steps and 10 steps would be covered by escalator. Sohan’s speed is four times that of escalator and he is going against the escalator. Therefore, to cover 30 steps, he will have to climb 40 steps and the escalator will bring him down by 10 steps in the same time. Therefore difference in Sohan’s number of steps and Rohan’s number of steps = 40 – 20 = 20.
2. If Rohan walks at four times the speed of escalator, if escalator gives out x steps, Rohan climbs down 4x steps. Now 4x + x = 60 or x = 12. Therefore, previously the escalator was giving our 20 steps but now it is only giving out 12 steps. Therefore time save $= \frac{20 – 12}{20} \times 100 = 40\%$
Problem 7: Jordan wanted to climb down from the first floor to the ground floor of a shopping mall, whereas Karina wanted to climb up from the ground floor to the first floor. Both use the same escalator which was ascending from the ground floor to the first floor and both walked their respective destinations. Both of them started simultaneously from the top and the bottom of the escalator respectively and crossed each other after exactly 21 seconds. If instead, Karina had walked at $\frac{1}{3}rd$ of his speed while Jordan maintained his speed, they would have crossed each other after exactly 28 seconds from the start. Further if both Jordan and Karina had climbed up from the ground floor to the first floor using the same ascending escalator, the number of steps taken by Karina to reach the first floor would be 20% less than the number of steps taken by Jordan for the same.
1. If Jordan were to stand still on the same escalator, how long would it take for the escalator to take him from the ground floor to the first floor?
a. 42 sec b. 63 sec c. 84 sec d.105 sec e. None
2. Jordan walked down from the first floor to the ground floor using the same escalator. However, after some time the escalator stopped moving due to a power failure. Find the total time taken by Jordan to reach the ground floor, given that the time for which he walked on the moving escalator was the same as that for which he walked on the stationary escalator.
1. $25\frac{11}{13}sec$
2. $37\frac{1}{3}sec$
3. $48 sec$
4. $56 sec$
5. $70 sec$
Solution: 1. The student need to understand that when both persons are moving towards each other on an escalator, the relative speed is same as if they are moving towards each other on a stationary plane, i.e. $v_1 + v_2$ because the speed of one person is $v_1 – u$ and the other person is $v_2 + u$ and since they are moving in opposite directions, the relative is the sum of these two, i.e. $v_1 + v_2$
Let Jordan’s speed $= v_1$ and Karina’s speed $=v_2$ and escalator speed $= u$. Let the total number of stairs be D. Taking both the cases,
$\frac{D}{v_1 + v_2} = 21$ and $\frac{D}{v_1 + \frac{v_2}{3}} = 28$ Dividing one by another and simplifying we get $\frac{v_1}{v_2} = \frac{5}{3}$. So let Jordan’s speed be 5v stairs per second and Karina’s speed be 3v stairs per second.
If both of them go with the escalator, Karina climbs $20\%$ less stairs than Jordan (given). Let Jordan climb 5x stairs and Karina 4x. Time taken for Jordan would $\frac{5x}{5v} = \frac{x}{v}$ and for Karina it would be $\frac{4x}{3v}$. The times are in the ratio 3:4 so escalator would throw out stairs in the ratio 3:4 in these times. Let the escalator steps 3y and 4y for Jordan and Karina.
Therefore, $5x + 3y = 4x + 4y$ (total steps are same) OR x = y. Karina travels 4x and in the same time escalator gives out 4x, therefore their speeds are the same. Therefore, speed ratios of Jordan, Karina and Escalator are 5: 3: 3.
1. Now, when moving in opposite direction total speed of Jordan (against escalator) = 5x – 3x = 2x and total speed of Karina (with escalator) = 3x + 3x = 6x. Their total speeds are in the ratio 1: 3. Therefore they will travel the distances in the same ratio. Therefore Karina travels $\frac{3}{4}$ of the total stairs in 21 secs (first case). therefore, to travel the total number of stairs she would need $21 \times \frac{4}{3} = 28 sec$ Therefore, in 28 sec Karina reaches from bottom to top in which half the distance is covered by her and half the distance is covered by the escalator. Since escalator covers half the distance in 28 sec, it would cover the whole distance in 56 sec.
2. Jordan’s with escalator still = 5v and his speed with escalator moving (against) = 5v – 3v = 2v. Therefore, to take equal time he has to cover distances in the ratio 5: 2. As seen above, Jordan covers $\frac{1}{4}th$ of the distance in 21 sec, therefore he would cover the whole distance in $21 \times 4 = 84sec$ and he would cover $\frac{2}{7}th$ of the distance in $81 \times \frac{2}{7} = 24sec$. Therefore, total time taken = 24 (with escalator) + 24 (still escalator) = 48sec.
Problem 8: Ann walked up a descending escalator and took 154 steps in 100 seconds to reach the top. Karen started simultaneously from the top, taking 3 steps for every 4 steps of Ann, and reached the bottom in 40 seconds.
1. How many steps from the bottom were they, when they crossed each other?
A. 22 B. 25 C. 30 D. 33 E. 36
2. Find the number of steps between Ann and Karen exactly $18\frac{2}{11}$ seconds after they started walking on the escalator.
A. 21 B. 49 C. 28 D. 54 E. 42
Solution: Speed of Ann $= \frac{154}{100}$ stairs/seconds Therefore, speed of Karen $= \frac{3}{4} \times \frac{154}{100}$ stairs/seconds.
1. Number of stairs covered by Ann in 40 seconds $= \frac{3}{4} \times \frac{154}{100} \times 40 = 46.2$ Let the number of stairs thrown by the escalator in 40 seconds be x. Therefore, the number of steps thrown by the escalator in 100 seconds = 2.5 x.
Therefore, 154 – 2.5x = 46.2 + x or x = 30.8. Therefore, total stairs = 77. The speed ratio of Karen and escalator is 46.2 : 30.8 = 3 : 2, Therefore, speed ratios of Ann, Karen and Escalator are 4: 3: 2. The speed of Ann and Karen are in the ratio (4 – 2) : (3 + 2) = 2: 5. Therefore, distance from the bottom when they meet $= \frac{2}{2 + 5} \times 77 = 22$ steps
2. Distance covered = Relative speed $\times$ time $= (1.54 + 1.155) \times \frac{200}{11} = 49$. Therefore, stairs left between = 77 – 49 = 28
Problem 9: On an upward moving escalator, Amar, Akbar and Anthony take 10 steps, 8 steps, and 5 steps, respectively, to reach the top. On the same upward moving escalator Amar takes 30 steps to come down from the top. Find the ratio of the time taken by Akbar and Anthony to reach the top.
Time taken by Amar to climb down 30 steps would be thrice the time taken to climb 10 steps. Therefore, if x is the number of steps thrown by the escalator when Amar climbs 10 steps, 3x would be the number of steps thrown by the escalator when he climbs 30 steps. Total number of stairs are the same. Therefore,
30 – 3x = 10 + x or x = 5. So total number of steps = 10 + 5 = 15. For Akbar and Anthony, the steps given out by the escalator are 15 – 8 = 7 and 15 – 5 = 10. Therefore, ratio of their time = 7: 10
Problems related to escalators baffle a lot of students preparing for CAT. Probably because not much time has been devoted to explaining this simple concept. Although we keep telling in our CAT classrooms continuously that escalator questions are similar to ‘upstream’ ‘downstream’ questions that students solve in time, speed and distance, the future MBAs still get confused because of the way that the questions are put. Today, we are going to see the theory behind escalator questions and solve them through both equations and ratios. In order to benefit the most, please solve these questions on your own before looking at the solutions.
The best site on escalator questions. Stunning methods. Really awesome. Covered all types of possible questions from escalator. Very useful for tough exams like CAT, XAT, GMAT, GRE, SSC etc. Thank you so much.
thanks for your feedback Jayesh .. which exam are you preparing for ?
excellent guys…..your concepts are to the point….keep it up…!!!