In the entire TSD --- the only theory is D = S*T, the important thing is how youare going to use this formulae in different scenario.In cat 70% of the questions -- which had appeared can be solved using ----"Proportionality b/w Time, Speed and Distance" ---This is probably the most important topic in this chapter. And also the most thought-intensive. It provides the foundation to solve many tough problemsorally!---- so today we will learn how to use this concept to solve Questions "ORALLY" --- in a minimum time--- so while solving the questions you all musttry to solve orally ( means avoiding writing complicated equation) --- try solving by eq also --- you must mastered both techniques.we say that in questions there are two scenarios, it requires some efforts on your part to identify the two scenarios.The following are some common instances of how to tune your thought process towards identifying the two scenarios (Complete questions are not given, just the crucial info is given, so dont be tooargumentative about the conclusions)1.Today I travelled 20 % faster from home to office The two scenarios are everyday and today. And while it is directly given that speedsare not constant, you need to realize the statedassumption that distance covered, home to office, will not change between everyday and today and thus, distance is constant2. Two friends started simultaneously from their homes towards each other to meet The two scenarios are that of the two individuals. Since they start simultaneously and they meet, both of them are travelling for the same time.Thus, the time is constant when we consider the case of the two individuals separately.3. A train takes 10 secs to cross a pole and 15 seconds to cross a platform Whilewe will see this scenario in details later on,the two scenarios are obvious one is train crossing pole and other is train crossing platform. And in the two scenarios,the speed of the train is going to be constant. It will be on immense help to think on the lines of two scenarios. And then using proportionality.Most questions in the following text will be done by the traditional approach and then solved by the approach of using proportionalityTime is inversely proportional to speed, when distance is constant.This is to say that, over a same distance, if the ratio of speeds is a : b, theratio of the time taken will be b : a.And this should be obvious, because over a same distance, if I double my speed (ratio of speeds 1 : 2), the time taken will be half (ratio of time 2 : 1).If I travel at 1/3rd the usual speed (ratio of speed 3 : 1), I would take thricethe time taken earlier (ratio of time 1 : 3)If I reduce my speed to 3/5thof the usual speed (ratio of speed 5 : 3), the timetaken will be 5/3 times the usual time (ratio of time 3 : 5).1. cat 2009 my set ----A boy walks at 1/3rd his original speed and reaches school 20 minutes late.Find the time taken by boy usually and the time taken at the reduced speed.( oral approach-- no pen pencil)In such problems, late by 20 minutes implies that 20 moreminutes will be taken to travel the same distance, or in otherwords, the difference in the time taken at the usual speed andthe reduced speed will be 20 minutes.Had my usual speed been s, the reduced speed would be 1/3s .Thus the ratio of usual speed to reduced speed would be. (Fromnext problem onwards, this step will be done directly).Since time is inversely proportional to speed, the ratio of thetime is 3 : 1. We also know that the difference in the time takenwill be 20 minutes. Thus we are looking for two numbers that are in the ratio 3: 1and the difference between them is 20 minutes. 31 ( there is a gap of 2but we need a gap of 20 --> so multiply by 10) ---> 3010.2. Travelling at 3/7th of his usual speed, a person is 24 minutes late in reaching his office.Find the usual time taken by him to cover this distance. ( solve orally)2. speed (37)====> time (73) gap 0f 4 but we need a gap of 24 somultiply ratio by 6 ===> 42-----183. If a man walks at the rate of 30 kmph, he misses a train by 10minutes. However, if he walks at the rate of 40 kmph, he reachesthe station 5 minutes before the departure of the train. Find thedistance to the station.The ratio of speeds is 3 : 4 and since distance is constant, the ratio of the time taken will be in the ratio 4 .Missing the train by 10 minutes and reaching early by 5 minutes implies that thetime taken at speed of 40 kmph is 15 minutes less than thetime taken at speed of 30 kmph.Thus, we need to find two numbers in ratio 4 with a difference of 15. Thus, a difference of 1 on the ratio scale is a difference of15 in actual values. Hence the multiplying factor is 15.Thus time taken at 30 kmph is4 15 = 60 minutes and at 40 kmph is 3 15 = 45 minutes. now you can use d = s*t = 30 *60/60 = 30 km4. A train meets with an accident and travels at 4/7th of its regularspeed hereafter and hence it reaches its destination 36 minutes late. Had the accident occurred 30 kms further, the train would havebeen late by only 21 minutes. Find the regular speed of the train.4. Lets say the point where the accident occurred was A when thetrain was late by 36 minutes and was point B when the trainwas late by 21 minutes. Let the destination be D.Comparison over distance AD: The two scenarios are at regularspeed and at reduced speed. Ratio of speeds 7 : 4. Ratio oftime 4 : 7. Difference in time is 36 minutes. Thus, time taken atregular speed for the distance AD is 4 12 = 48 minutes.Comparison over distance BD: The two scenarios are at regularspeed and at reduced speed. Ratio of speeds 7 : 4. Ratio of time4 : 7. Difference in time, over this distance is 21 minutes. Thus,time taken at regular speed for the distance BD is 4 7 = 28minutes.At its regular speed, the train takes 48 minutes to travel A to Dand takes 28 minutes to travel B to D. Hence it must be taking20 minutes to travel from A to B, a distance of 30kms. Thus, its regular speed = 30/(20/60) = 90 km/hr4. Shortcut:While in the above solution we had the comparisons (at regular speed & at reduced speed) made twice, once over distance AD and once over distance BD.We could make do with just onne comparison as well.Consider the stretch AB. When accident occurs at A, this stretch is travelled atreduced speed. And when accident occurs at B,this stretch is travelled at regular speed. And the difference in the time takenjust over this stretch is 36 21 = 15 minutes. (At destination,why is the train less late, 21 mins, when accident occurs at B as compared to more late, 36 mins, when accident occurs at A?The train would have been lesser late because, over AB, it would have not have got late) Comparing the two scenarios,ratio of speeds over AB is 4 : 7. Hence ratio of time will be 7 : 4 and difference in time taken over AB is 15 minutes. The multiplying factor to gofrom ratio scale to actual values will be 5. Thus, at regular speed it would take 4 5 = 20 minutes to cover AB, a distance of 30 km.extra tym to travel the 30 km stretch = 3t/4 = 15,so t = 20.. 30km in 20 mins or90k/hr5. To late by 15 min--- distance travelled is 30: so to late by 1 min -- distance travelled is 30/15 = 2 ; so to late by 36 min( since begining)--- distance travelled is 36*2 = 72 km6. If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. Ifhe cycles at 15 km/hr, he will arrive at the same placeat 11 a.m. At what speed must he cycle to get there at noon? ( cat 2004)6. ratio of speed--- 10:15= 2:3 ===> ratio of time ---3:2 ( there is a gap of 1but we need a gap of 2hrs as 11am &1pm) so timeis 6hrs & 4hrs. now let spped at 12noon is " s" ----> 10/s =5/6 ===> s= 12km/hr7.Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively.Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? (cat 2006)let A,B,K represents speed & a,b,k Represents time of Arun, Barun & Kiranmal respectively. Here distance is constant (1) A/B = b/a ===> b/a = 3/4( gap of 1 but we need gap of 2--- as barun started 2 hrs late) so b= 6, a =8 ;(2) A/K =k/a ===> 30/60 = k/8 ==> k = 4hr ===> kiran mala started 4hrs late.8. The Ghaziabad-Hapur-Meerut EMU and the Meerut-Hapur-Ghaziabad EMU start at the same time from Ghaziabad and Meerut and proceed towards each otherat 16 km/hr and 21 km/hr, respectively. When they meet, it is found that one train has travelled 60 km more than the other.The distance between two stations is:[IIFT 2007]1) 445 km2) 444 km3) 440 km4) 450 km8. distance is constant , s1/s2 = d1/d2 ===> d1/d2 = 16/21 ( there is a gap of 5-- but we need gap of 60 so multiply by 12)d1= 16*12 &d2=21*12 ..total = 4449. Two trains start simultaneously, one from Bombay to Kolkataand other from Kolkata to Bombay. They meet each other atNagpur which is at a distance of 700 kms from Bombay. If thedistance between Bombay and Kolkata is 1600 km, find theratio of their speeds.9. Since the trains started simultaneously, the time they have beentraveling till they meet is equal. And hence the distance theycover will be in ratio of their speed. Since the train from Bombayhas covered 700 km and the train from Kolkata has covered1600 700 = 900 kms, the ratio of their speeds will be 700 :900 i.e. 7 : 9.10. A police-man starts chasing a thief. The ratio of the speeds ofthe thief and the policeman is 9 : 11 and when the policemancatches the thief it is found that the policeman has covered 60meters more than the thief. How much distance did the policehave to run to nab the thief?10. Since the chase starts with both of them running simultaneously, from this point onwards to the time the police has caught the thief,they are running for same duration. Thus the distance covered will be proportional to their speeds. So we are searching for two distancesin the ratio 9 : 11 and the difference being 60 meters i.e. 2 of the ratio scalecorresponds to 60 mts, implying that the multiplying factor is 30. Hencedistance run by police will be 11 30 = 330.11. In the movie Ghulam, Aamir is able to spot the approaching train when it is2 km away. He has to run towards the train and reach the redkerchief hung on a pole 400 meters away from him before the train reaches the pole. How fast must Aamir run if the speed of the trainis 36 kmph so that he just manages to reach the kerchief at the same time as thetrain reaches it?11. From the point when the distance between them is 2000 m, Aamir has to covera distance of 400 m and the train will cover therest of the 1600 m i.e. the distance will be in the ratio 1 : 4. Since they arerunning for the same duration,the speed will be proportional to the distance covered. Since the speed of the train is 36 kmph, the speed of Aamir should be 9 kmph.12. A car overtakes an auto at point A at 9 am. The car reaches point B at 11 amand immediately turns back. It again meetsthe auto at point C at 11:30 am. At what time will the auto reach B?Since the car takes 2 hours to travel AB and 0.5 hours to travel BC, The ratio of the distances AB : BC will be 2 : 0.5 i.e. 4 : 1.(Distance is proportional to time) Since C lies in between A and B, the ratio ofthe distance AC to CB will be 3 : 1.The auto has traveled AC in 2.5 hours (from 9 am to 11:30 am). To travel CB (one-third distance of AC)he will take 2.5 hrs /3 =150 min/ 3 = 50 more minutes. Thus the auto will reachB at 12:20 pm.13. There is a tunnel connecting city A and B. There is a CAT which is standingat 3/8 the length of the tunnel from A.It listens a whistle of the train and starts running towards the entrance where,the train and the CAT meet.In another case, the CAT started running towards the exit and the train again met the CAT at the exit. What is the ratio of their speeds?[CAT 2002]here time is constant----> so use s1/s2 = d1/d2 ; let speed of train be x ... when train runs x then cat runs 3 and when cat runs5 the tranruns x+8 equae it x=12 ratio of speeds =12:3 i.e. 4:1.14. Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves stationA for station B, a south passenger train S reaches station A from station B. Thespeed of the superfast train is twice that of a normal expresstrain E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behindthe normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what shouldbe the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduledtime at the station A on that day? [CAT 2002]1) 1 : 32) 1 : 43) 1 : 54) 1 : 6Let the speed of superfast & passenger train is N & S respectively. Since thereis one single track so combined time taken by superfast & passengertrain is !hr. Ratio of speed N/S = 4:1 ===> ratio of time = 1:4===> time taken by superfast train = 1/5*60 = 12 min & time taken by passenger train= 48 min. Now one day train started 20 min late --since speed of superfast is doubled time taken is --6 min .so time taken by passenger train to reac on time = 60 - (20+6) = 34min : so newratio is 6:34 = 1:6 (approx)15. Two boats start from opposite banks of river perpendicular to the shore. Oneis faster then the other. They meet at 720 yardsfrom one of the ends. After reaching opposite ends they rest for 10mins each. After that they start back.This time on the return journey they meet at 400yards from the other end of theriver. Calculate the width of the river.Also find ratio of speed of boats.15. when they meet for the first time , together they will cover 1 round in which 1st boat will contribute 720; when they meet for the 2nd timetogether they will contribute 3 round -- so contibution by 1st boat is 720*3 =2160 ,but since they are meeting from other end .so length is 2160-400 = 176015. Alternative method: he distances traveled by the boats are proportional to their speeds___ since the speeds are constant, the ratio of the distances is constantlet w="width of river" ___ river is >720 yd widelet x and y be the boat distancesat first meeting ___ x=720, y=w-720at second meeting ___ x=w+400, y=2w-400ratio of distances is constant, so 720/(w-720)=(w+400)/(2w-400) on solving thisW = 1760$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$--APPLICATION OF AM/HM IN TSD -- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$Using AM & HM in proportionality relationsQuite often in questions we find that the given speeds (or time taken) are in anArithmetic Progression. And if distance covered at the speeds is constant, thentime taken (or speeds) will be inversely proportional i.e. they will be in Harmonic Progression.For those who have forgotten, the Arithmetic Mean of a and b is(a + b)/2 and the Harmonic Mean of a and b is 2ab/ a + b .Though it requires a little trained eyes to identify the above, it will be useful if you keep a watch for it. See the following data to realise that either timetaken or speeds are in an Arithmetic Progression.E.g.: A, B, C leave point P, one after the other in the given order, with equaltime intervals between their departure. If all three imultaneously meet at Q, given that speed of A and C is 40 kmph and 60 kmph, find speed of B. The time taken by A, B, C over constant distance PQ will be of the type t, t x and t 2x i.e.in an AP. Thus, their speeds will be in a Harmonic Progression. The required speed will bethe HM of 30 and 60 i.e. 2 *30 *60/90 = 40 kmphwith this techniques lots of difficult question can be solved easily. If u had understand the concept than i can start taking application of this...1. If I travel at 15 kmph, I reach office at 10 am, if I travel at 10 kmph, I reach office at 10:30 am. At what speed should I travel so that I reach office at10:15. Assume I leave home at same time and take the same route. ( use am/hm -no other method)1. Leaving at same time and reaching at 10 am, 10:15 am and 10:30 am suggests that the time travelled are in AP. Thus, speeds are in HP and required speed is the HM of 10 & 15i.e.2 *10 *15/25 = 12 kmph2. If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. Ifhe cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speedmust he cycle to get there at noon?[CAT 2004]1) 11 km/hr2) 12 km/hr3) 13 km/hr4) 14 km/hr3. Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively. Barun startstwo hours after Arun. If Barun and Kiranmala overtake Arun at the same instant,how many hours after Arun did Kiranmala start? [CAT 2006]1) 32) 3.53) 44) 4.55) 53. speed ---30---40---60 (HP) wheras time t---(t-2) -----???, since speed is inHP , so time Must be in AP ,so ans is t-4 ===> 4hrs4. Anuva takes 20min less than the usual time to reach her office if her speedincreases by 5km/hr , and takes 30 min more than the usual time if her speed decreases by 5 km/hr. What is her usual speed.4. Anuva takes 20min less than the usual time to reach her office if her speed increases by 5km/hr , and takes 30 min more than the usual time if her speed decreases by 5 km/hr. What is her usual speed.Speed times-5 T + 30s Ts+5 T 20Since speed is in AP, so time must be in HPSo T = 2(T+30)(T-20) / (T + 30) + (T 20)Solving this u can easily find T & then S also & it;s not quadratic, so there isno calculation5. A man can walk up a moving up escalator in 30 second. The same can walk down this moving up escalator in 90 seconds. Assume his walking speed is same Upwards & downwards. How much time he will take to walk up the escalator when escalator isnot moving ? (cat 1994)speed in ap ...tym shud be in hp...speed=x+y,x,x-y...(ap) so for tym 2*30*90/120=456. A man can row UP stream in 84 minute. He can row the same distance in 9 min less than he could row it in still water. How long will he take to row down withthe stream?upstream = b-r, still water = b, downstream = b+r ; so time must be in hp ; 84,t, t-9 are in hp ===> solving this u will get t = 72; so ans is 72-9 = 637. a boy is walking along the direction of 2 parallel railway tracks. on one ofthese tracks, trains are going on 1 direction at equal intervals. on the other track, trains are going in the opp direction at the same equal intervals. the speed of every train is same . In one direction, a train crosses the boy every 20 mins. and in the opp direction the train passes the boy every 30 mins. if the boystands still beside the tracks, at intervals of how many will two consecutive trains going in the same direction cross him?when boy & train are travelling in same direction - relative speed = t-b; for opposite direction =t+b; when boy is stand stil, train will move at = t ; so heret-b,t,t+b are in AP ===> time must be in HP ==> t = 2*20*30/50 = 24 min ( this is the beauty of AM/HM concept)8. Everyday I cover the distance from my home and office at a usual speed and take a certain time at the usual speed. When I increase my usual speed by 5 kmph,I take 10 minutes less than usual. If I reduce my usual speed by 5 kmph, I take15 minutes more than usual. Find the distance from home to office.Can you realise that if usual speed is s, then the three speeds are (s 5), s and (s + 5) i.e. in AP? Thus, the time taken are in HP. The time taken as per thedata in question is (t + 15), tand (t 10) and hence we have t= 60 min; Thus timetaken are 75 mins, 60 mins and 50 mins. Speeds will be in ratio of 1/75 : 1/60: 1/50 i.e. 4 : 5 : 6. And we know the difference in speeds are 5 kmph. Thus speeds are 20 kmph,25 kmph and 30 kmph. Now distance can be found using any combination of speed and time.Nitin Sir s link to old sessions :https://www.facebook.com/notes/alphanumeric-mba-prep-study-group/link-for-quantlrdi-session-archive/314878828571422$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$--- ESCALATORS --- $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$Escalator is similar to Boats & StreamsDistance = no. of steps in escalator (when escalator is not moving)Here speed is usually given as no. of steps per secondLet speed of escalator is e steps/secLet speed of man/woman = m steps per secondCase 1: when escalator & man are moving in same direction, effective speed = (m+ e) steps/secCase 2: when escalator & man are moving in opposite direction, effective speed =(m -e) steps/secAnother important concept:Case 1: when escalator & man are moving in same direction, --- No. of steps covered by man is always less than actual no. of steps in escalator.Case 2: when escalator & man are moving in opposite direction, --- No. of stepscovered by man is always more than actual no. of steps in escalator1. You walk upwards on an escalator, with a speed of 1 step per second. After 50steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator.The Question: How many steps do you need if the escalator stands still?Case 1: when escalator & man are moving in same direction, --- No. of steps covered by man is always less than actual no. of steps in escalator.Case 2: when escalator & man are moving in opposite direction, --- No. of stepscovered by man is always more than actual no. of steps in escalator.say escalator speed x steps/sec.so total steps = 50+50x.(from upward condition, in 50 sec escalator will cover 50x).Total time to reach up is 50 sec.total time to reach down = 25 sec.(125 steps, 5 steps/sec)total steps = 125- 25x (in 25 sec escalator will cover 25 x)50+50x = 125 - 25x75 x = 75=>x =1,so total steps = 50 + 50*1= 1002. A walks down an up-escalator and counts 150 steps. B walks up the same escalator and counts 75 steps. A takes three times as many steps in a given time as B.How many steps are visible on the escalator?Approach 1:Let N = no. of steps in escalator (when escalator is not moving)Speed of a / speed of b = 3:2 , let speed of a = 3x & speed of b = 2xCase 1: A walk down on up- escalator,N /(3x-e) =1 50/3x-------------(1)Case 2: B walk up on up- escalator,N/(2x+e) = 75/2x --------(2)Dividing eq 1 by eq 2 , you will get e= xPutting e = x you will get N= 120 steps.RememberCase 1: when escalator & man are moving in same direction, --- No. of steps covered by man is always less than actual no. of steps in escalator.Case 2: when escalator & man are moving in opposite direction, --- No. of stepscovered by man is always more than actual no. of steps in escalator.Let T be time B takes to make 25 steps. Then B takes 3T to make 75, and A takes2T to make 150. Suppose the escalator has N steps visible and moves n steps in time T. Then A covers N + 2n = 150, N - 3n = 75. Hence N = 120, n = 15.. If he runs up 8 steps, then he needs 37.5 seconds to reach the top.If he runs up 14 steps, then he needs 28.5 seconds to reach the top.The 6 additional steps take 9.0 seconds.Therefore, each step takes 1.5 seconds.Total steps in escalator = 8 + 37.5 / 1.5 = 33 or Total steps in escalator = 14+ 28.5 / 1.5 = 33.If Colin did not run up any steps at all,he would reach the top of the escalator in 49.5 seconds (i.e., 33 steps 1.5 seconds/step).Alternative Solution through Equations:Let the total number of steps in the escalator be x.The escalator moves at a constant speed given bySpeed of escalator = (x 8)/37.5 = (x 14)/28.5The above equation may be solved as follows.28.5 (x 8) = 37.5 (x 14); orx = (14 37.5 8 28.5) / (37.5 28.5) = 33.Now, Speed of escalator = (33 8)/37.5 = (33 14)/28.5 = 1/1.5 steps/second.Time to reach top = Total Steps / Speed = 49.5 seconds.Approach 3 :Let escalator speed = e steps/secNo. of steps = 8+37.5x = 14+28.5xX = 2/3No. of steps = 8 + 37.5*2/3 = 33 stepsNow, Speed of escalator = (33 8)/37.5 = (33 14)/28.5 = 1/1.5 steps/second.Time to reach top = Total Steps / Speed = 49.5 seconds.4. Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyoms steps. Shyamagets 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 wouldthey have to take to walk up? (cat 2001)a. 40 b. 50 c. 60 d. 804. Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyoms steps. Shyamagets 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 wouldthey have to take to walk up? (cat 2001)a. 40 b. 50 c. 60 d. 805. 2 kids, John and Jim, are running on an escalator (a moving stairway). John is running three times as fast as Jim, and by the time they are off the escalator, John has stepped on 75 stairs while Jim has stepped on 50 stairs. How many stairs does the escalator have? How is its speed related to the speed of the boys?Were they running with or against the escalator?5. The answers are: the length is 100 stairs, the boys were running along the escalator which was moving with the same speed as the slow boy. Solution: in the time the fast boy stepped on 75 stairs, the slow one could step on only 25, so, since he stepped on 50, he spent twice as much time on the escalator as the fastone. Therefore his speed relative to the ground was half that of the fast boy, therefore the escalator s speed was the same as the speed of the slow boy, and hecounted exactly half the stairs. Another way is to use algebra (omitted).Approach 2:Assume A takes 1 step per unit time. Then B will take. Also, assume the the escalator is moving at E stepsLet T be the total number of steps.Let ta be time taken by A on the escalator, tb = timer.Since A takes 50 steps therefore we have:50 = T/(1+E) units of time.similarly,75 /3= T/(3+E) units of time.Solving for T, E we get E = 1 step per unit time; T =3 steps per same unit timeper unit time.taken by B on the escalato100 steps6. A man can walk up a moving up escalator in 30 second. The same can walk down this moving up escalator in 90 seconds. Assume his walking speed is same Upwards & downwards. How much time he will take to walk up the escalator when escalator isnot moving ? (cat 1994)7. An escalator is descending at constant speed. A walks down and takes 50 stepsto reach the bottom. B runs down faster and takes 90 steps to reach the bottom.If B takes 90 steps in the same time as A takes 10 steps then how many steps arevisible when the escalator is not operating?B = 9A = 150 + 50x = 90 + 10x=> x = 1total steps = 100Case : When A & B walk on unmoving surface10 Steps of A = 90 Steps of B .................... [Eq.1]=> Speed of A : Speed of B :: 1 : 9=> Time of A : Time of B :: 9 : 1Case : When A & B walk on esclatorSteps taken by B : 90TIme taken by B : TSteps taken by A : 50Time taken by A : 5T [From [Eq.1]]Time of A : Time of B :: 5 : 1let e be the number of steps moved by the esclator when A takes 1 stepSo(1+ e) / (9 + e) = 1/5=> e = 1 step for every step of AA takes 50 steps=> Esclator has moved by 50 stepsTotal number of steps = 50 + 50 = 1008. On an upward moving escalator Amit, Sanjeev and Vicky take 10 steps, 8 stepsand 5 steps respectively to reach the top. On the same upward moving escalator Amit takes 30 steps to come down from the top.Find the ratio of the time taken bySanjeev and Vicky to reach the top.10 + 5x = 30 15x=> x = 1and total number of steps = 158 + S = 15=> S = 75 + V = 15=> V = 10so , ratio = S : V = 7 : 10no. of steps = 10+x = 30 3x===> 4x = 20 ===> x = 5; so total no. of steps in escalator is 15;now sanjeev takes 8steps(remaining 7 are taken by escalator on his behalf); amit5 steps ( (remaining 10 are taken by escalator on his behalf); so required ratio is 7:10