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Quick Handout for Time, Speed and Distance problems

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Quick Handout for Time, Speed and Distance problems [#permalink]

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TIME, SPEED AND DISTANCE - Part 1

Time, Speed and Distance problems typically appear in GMAT as word problems. While the chapter is simple enough to understand (There is just one formula effectively – Distance = Speed * Time :-D) the different contexts (Elevators, escalators, boats, trains and what not) actually make the job quite challenging for a test taker. More importantly, common sense and linking the problems with real world understanding helps the most rather than trying to memorize formulae.

There have been some very good TSD materials posted by Narenn (http://gmatclub.com/forum/time-speed-and-distance-simplified-150163.html) and a few others, so I won’t repeat them here. However, I want to particularly highlight on a few additional sections that might be tested in the exam – particularly in difficult 700+ level problems. Hope you’ll find it useful.

Frame of Reference (External vs. Internal)

A frame of reference is nothing but the point of observation. An external frame of reference typically means an observer is standing outside the moving bodies (E.g. An observer on a platform), an internal frame of reference means the observer is inside the moving bodies (E.g. a passenger inside a train). There is a ton of material available on this forum on external frame of reference (Train crossing a man/pole/platform/another train etc. So, I’d limit this discussion to internal reference frame alone.

Situation 1 – Stationary Object inside a moving body

Sample problem: A train traveling at a speed of 36 kmph and having a length of 100 metres crosses a platform 300 metres long. How long would Andy, a passenger on-board the train would take to cross the platform?

The key for these questions is to understand that we’re now looking at the problem from Andy’s perspective, who is a passenger on the train. So, we are no longer interested in the train’s time taken to cross the platform. Thus, to solve this problem, a simple way would be to consider as though Andy, a point object (And hence Length = 0) is traveling at a speed of 36 kmph and crossing the platform. Hence time taken would be 300/36*(5/18) = 30 seconds. Note that we ignored the length of the train because we are only interested in the time Andy took to cross the platform and not the train.

Situation 2 – Moving Object inside a moving body

Sample problem: A train traveling at a speed of 36 kmph and having a length of 100 metres crosses a platform 300 metres long. Andy, a passenger on board the train is standing in the last compartment and starts moving towards the engine at a speed of 18 kmph. How long would he take to cross the platform?
Many of you might think that two bodies moving in same direction, so relative speed is V1 – V2. You’d be wrong to think so in this question, because here we are referring to an internal frame of reference. Applying common sense would tell you that by the time Andy crosses the platform, he would have advanced some distance towards the engine and hence would take less time to cross the platform. So, to obtain a lesser time, we must add the speeds – V1+ V2. Hence the answer in this case would be 300/(36+18)*(5/18) = 20 seconds.

It would be similar if Andy was moving from the Engine towards the rear end of the train, with the only change -> the relative velocity would be V1 – V2.

Escalator Fundamentals

Escalators are another classic example of problems involving internal frame of reference. Since we covered the fundamentals above, test your understanding by trying the two problems below.

Sample Problem: R and S are walking on an escalator. R walked down the descending escalator and took 40 steps to reach the escalator base. S started from the escalator base at a speed twice that of R and started walking towards the top and reached the top at the same time as R reached the base. How many steps more than R did S cover when they crossed each other? Assume that the elevator moves 1 step/second.

(A) 10
(B) 20
(C) 30
(D) 40
(E) 50

Answer: (B)

2. A woman walked down a descending escalator in a mall at a certain speed and took 10 steps to reach the base. As soon as she reached the base she remembered that she left her purse in a shop upstairs and started rushing to the floor above at a speed 5 times her speed earlier. This time she covered 25 steps to reach the top. How many steps does the elevator have?

(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Answer: (C)

Average Speed

Again, average speed has been discussed aplenty in different forums, so I’m just going to summarize the result here. Whenever the distance travelled can be classified as m parts of the distance travelled at a velocity p and n parts of the distance travelled at q then the average speed of the journey can be classified as (m+n)pq/(mq+np).
Here’s an interesting problem to test your understanding of average speed.

IR Two Parts Analysis Question on TSD

A dog, standing at a hilltop starts descending down the slope and meets his owner who starts his upward journey from the hill-base at the same time the dog starts. The dog meets the owner, goes back to the top, comes back again to meet the owner and repeats this till the owner reaches the hill top. The owner’s up-hill speed is 6 m/sec, dog’s downhill speed is 15 m/sec, uphill speed is 9 m/sec. The hill slope is 720 metre.

Find the following:
i) The total distance travelled by the dog.
ii) The time taken by the dog for downward journey.

Answer choices
i)
A. 900 m
B. 1050 m
C. 1200 m
D. 1350 m
E. 1500 m

Answer: (D)

ii)
A. 30 sec
B. 45 sec
C. 60 sec
D. 75 sec
E. 90 sec

Answer: (B)

Relative Speed

There is a particular type of problem in TSD where many of us often gets stumped. At the first look the questions seems to provide inadequate information, but in fact the details are cleverly disguised. There is no special formula except D = S * T which makes the job even more frustrating. Here is a couple of examples of such problems. Typically, we are asked to find distance in most such questions, but they could be tweaked differently too. Te key for such problems is to assume some variables D and T and construct 2 equations. We then need to make use of the following facts depending on the question to construct the equations.

1. When Distance is constant - Time is inversely proportional to speed
2. When Time is constant - Distance is directly proportional to Speed
3. When Speed is constant - Distance is directly proportional to Time

Here are two frustrating problems :wink: Give it a try and I shall discuss the solutions in my subsequent post.

1. Two buses A and B leave cities P and Q at equal speeds from cities P and Q and travel at equal speeds. They meet at a point R after 2 hours and continue further to reach cities Q and P. Next day, Bus A starts 36 minutes early, and bus B starts 24 minutes late and they meet at a point S such that RS = 24 km. Find the distance between the cities P and Q.

A. 96 km
B. 144 km
C. 160 km
D. 184 km
E. 192 km

Answer: (E)

2. Two buses P and Q start from cities A and B towards each other. They meet at a distance of 22 km from City B, cross each other to reach city B and A and turn back. During the second time they meet at a distance of 10 km from City A. Find the distance between cities A and B.

A. 24 km
B. 56 km
C. 60 km
D. 72 km
E. 84 km

Answer: (B)

I plan to add some more content on circular tracks, relay races and clocks. Do let me know if you find the post useful. That would help me stay motivated to add further contents :)

P.S. – I shall post the answers to the questions after a few days. In the meantime, it would be good to see some attempts and analyses. I shall try pitching in from time to time :-D
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Re: Quick Handout for Time, Speed and Distance problems [#permalink]

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TIME, SPEED AND DISTANCE - PART 2

Linear Races

Races are another variation of the concept of Relative Speed as discussed in Part 1. Few concepts are important to understand here. The concept of head start, beating by a certain distance and beating by a certain time.

Head Start - Let there be 2 runners A and B. If B gives a head start to A, it simply means, A starts ahead of B. Head start can be measured in terms of distance or time. Say, in a 100m race B gives A a head start of 10m. For solving a problem, it is equivalent to considering A running 90m and B runs 100m. Similarly, if A is given a head start say, 10 sec, then it means B starts 10 seconds after A starts.

Beating by a distance X - If A beats B by a distance of 10m in a 100m race, it means when A finishes the race, B has completed 90m.

Beating by time T - If A beats B by 10 seconds, it means B completes the 100m race 10 seconds after A.

With this understanding, let's see a few problems.

Sample Problem: In a 1200m race, P beats Q by 300m and Q beats R by 400m. Find the distance by which P beats R.

What is being tested in this question? It is testing the test takers ability to establish a ratio relationship between P, Q and R using the distance traveled by P & Q and Q & R. Note that time is constant. When P beats Q, P and Q have run for the same time - say T1. Similarly when Q beats R, Q and R have traveled for the same time say T2.
Now, Distance = Speed * Time, so if time is constant, distances must be in proportion. In other word ratio of distance covered by P and Q should be equal to the ratio of their respective speeds. Now, P beats Q by 300m, so by the time P runs 1200m, Q runs 900m. Hence ratio of their speeds is 12:9 = 4:3.
Similar arguments can be applied for Q and R too and their respective speeds can be established as 1200:800 = 3:2.
Combine the two ratios we get P:Q:R = 4:3:2 (Since Q correspnds to 3 in both the ratios they can be combined as is)
Thus P:R = 4:2 or 2:1. Hence by the time P finishes the race, R would finish 600m. Hence P beats R by 600m.

Now, try the following two easy problems to solidify the concept.

Practice Problem 1: In a 200m race A beats B by 10m or 2 seconds. Find the speeds of A and B in m/sec.

(A) 10 and 8
(B) 10/9 and 8/9
(C) 100/19 and 5
(D) 15 and 10
(E) 9/5 and 7/5

Practice Problem 2: P is 50% faster than Q. In a race, P gave Q a head start of 200m. Both finished the race simultaneously. Find the length of the race.

(A) 300m
(B) 400m
(C) 500m
(D) 600m
(E) 700m

Circular Races

Circular races are a variation of linear races with the exception that the start and the end points are the same. So, we can think of the track as a wire ring, and if we cut the ring at the starting/ending point and straighten the wire, we get the above case of linear races. This visualization helps solve some of the problems very easily. Now for circular races, it is important to understand 2 cases which can be generalized further.

Case 1: Two people running around a Circular Track of length L

Same Direction

When 2 people a and b run around a circular track with respective speeds a and b where a>b, then

(i) Time taken to meet for the first time ever is L/(a-b)
(ii) Time taken to meet for the first time at the starting point is LCM[L/a, L/b]

Opposite Direction

When 2 people a and b run around a circular track with respective speeds a and b where a>b, then

(i) Time taken to meet for the first time ever is L/(a+b)
(ii) Time taken to meet for the first time at the starting point is LCM[L/a, L/b]

Case 2: Three people running around a Circular Track of length L

Same Direction

When 3 people a, b and c run around a circular track with respective speeds a, b and c where a>b>c, then

(i) Time taken to cross/meet for the first time ever is LCM [L/(a-b), L/(b-c)]
(ii) Time taken to cross/meet for the first time at the starting point is LCM[L/a, L/b, L/c]

Same Direction

For opposite direction, there is no generalized case, and if asked, it should be done considering 2 of the 3 people at a time. The time to meet at the starting point would however remain the same as above.

Logic for the above cases

As mentioned, earlier, for the first case, time to meet at the starting point, consider cutting open the wire frame and treating the race as a linear one. The only exception being every time a person reaches the end of the race track, we'll have to imagine that he starts from the beginning of the track for the next lap. With that assumption, it simply becomes a relative speed problem where the faster person has to cross the length of the race track once to cross the slower person. Hence distance traveled = L and relative speed is a-b, so time taken is L/(a-b). For the three people scenario we take 2 people at a time and consider the LCM because a we are looking for a certain specific multiple of the time when both a crosses B and B crosses C. Note that this is the condition for all 3 people crossing for the very first time. Similar argument holds tru for the opposite direction as well.

With this understanding, give the next two problems a try.

Practice Problem 1: Peter and Paul are running in opposite directions along a circular track of 600m with initial speeds of 3m/sec and 6 m/sec. They start at the same time from the same point. Whenever they meet, they exchange their speeds and carry on in their respective directions. Find the distance between them in metres when Peter completes 17/4 rounds.

(A) 150
(B) 200
(C) 300
(D) 120
(E) 175

Practice Problem 2: P and Q are running along a circular track having started at the same time from the same point, in the same direction. How much more distance would P have traveled compared to Q by the time they meet for the 11th time? The radius of the track us 7m and the speeds of P and Q are 22 and 11 m/sec respectively.

(A) 440m
(B) 384m
(C) 524m
(D) 484m
(E) 600m

Clocks

Clocks involve the same concept as Circular Tracks. Only two things need to be remembered for simplifying Clock problems.

1. Minute hand covers 360 degrees in 60 minutes => minute hand covers 6 degrees/minute
2. Hour hand covers 30 degrees in 60 minutes => hour hand covers 0.5 degrees per minute

These also imply that the relative speed is 5.5 degrees per minute.

Sample Problem: Find the smaller angle between the hands of a clock when the time is 3:40 PM.

Note that, at 3 PM, the hour hand has a head start of 3 * 30 degrees = 90 degrees.
Now in 40 minutes, minutes hand covers 40*6 = 240 degrees and the hour hand covers 90+40*.5=110 degrees. Hence the smaller angle between them is 240-110 = 130 degrees.

Practice Problem - IR Two parts Analysis Question

After how many minutes post 5 PM and before 6 PM will the hands of the clock be at 90 degrees to each other?

(i) First Time

(A) 80/11 minutes
(B) 100/11 minutes
(C) 110/11 minutes
(D) 120/11 minutes
(E) 140/11 minutes

(i) Second Time

(A) 300/11 minutes
(B) 450/11 minutes
(C) 480/11 minutes
(D) 500/11 minutes
(E) 600/11 minutes

Do let me know, if you found the post useful. If you're having any question on any of the materials covered, please post here as a reply. I shall try my best to clarify them :-D

P.S. - I have added the answers the the questions I included in the first post :). Please refer the above post.
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Re: Quick Handout for Time, Speed and Distance problems [#permalink]

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New post 03 Nov 2016, 18:36
“300/36*(5/18) = 30 seconds.” In situation 1, where did the 5/18 come from?

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Quick Handout for Time, Speed and Distance problems [#permalink]

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New post 03 Nov 2016, 19:26
I've tried without much avail to solve the following question

Sample Problem: R and S are walking on an escalator. R walked down the descending escalator and took 40 steps to reach the escalator base. S started from the escalator base at a speed twice that of R and started walking towards the top and reached the top at the same time as R reached the base. How many steps more than R did S cover when they crossed each other? Assume that the elevator moves 1 step/second.

(A) 10
(B) 20
(C) 30
(D) 40
(E) 50

Answer: (B)


I have formulas 2x-1 as speed of S and x+1 for R where X is the speed of R however I do not know how to arive at the answer of 20.

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Re: Quick Handout for Time, Speed and Distance problems [#permalink]

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New post 03 Nov 2016, 19:28
rachnfrosty wrote:
“300/36*(5/18) = 30 seconds.” In situation 1, where did the 5/18 come from?


F/18 comes from the fact that in order to convert the kph to m/s you have to multiply by 1000 metres and divide by 3600 seconds which equals 5/18.

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Re: Quick Handout for Time, Speed and Distance problems [#permalink]

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New post 03 Nov 2016, 21:20
rachnfrosty wrote:
“300/36*(5/18) = 30 seconds.” In situation 1, where did the 5/18 come from?


withfaith wrote:
F/18 comes from the fact that in order to convert the kph to m/s you have to multiply by 1000 metres and divide by 3600 seconds which equals 5/18.


Hey, the explanation above is right. Whenever, you have distance in metres and speed in km/hr, you are actually comparing units in the form of apples and oranges. So, you need to either convert km/hr to m/sec or convert the distance from metres to km.

km/hr -> m/sec: Multiply by 1000/3600 or 5/18
m/sec -> km/hr: Multiply by 3600/1000 or 18/5

Hope this helps :)

withfaith wrote:
I've tried without much avail to solve the following question

Sample Problem: R and S are walking on an escalator. R walked down the descending escalator and took 40 steps to reach the escalator base. S started from the escalator base at a speed twice that of R and started walking towards the top and reached the top at the same time as R reached the base. How many steps more than R did S cover when they crossed each other? Assume that the elevator moves 1 step/second.

(A) 10
(B) 20
(C) 30
(D) 40
(E) 50

Answer: (B)


I have formulas 2x-1 as speed of S and x+1 for R where X is the speed of R however I do not know how to arive at the answer of 20.


Hey, I shall help you with some hints here :)
In escalator problems, you must understand that whether it is a stationary escalator or a moving escalator, the length (or in this case the number of steps) at any point of time is always the same.

withfaith wrote:
I have formulas 2x-1 as speed of S and x+1 for R where X is the speed of R however I do not know how to arive at the answer of 20.

2x-1 and x+1 are not the speeds of R and S, but they are the relative speeds. You can say, speed of R is x and S is 2x.

You can easily establish the speeds of R and S quite easily here.
Assume that the escalator has S steps. Then the time taken by R and S to cover the distance is the same. Is Distance and Time both are equal, then the relative speeds of R and S must also be equal. Thus 2x-1=x+1 => x=2 steps/sec.
Thus R travels at 2 steps/sec and S at 4 steps per sec. You already know that the escalator travels at 1 step/sec and also the fact that both R and S start and finish at the same time. Think logically about when they should cross each other and you'll have the answer :)

Let me know if this helps. If not, I shall solve it in the next post.

Cheers :)
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Re: Quick Handout for Time, Speed and Distance problems [#permalink]

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New post 20 Oct 2017, 04:42
You didn't post any of the detailed solutions. A slope has no units whereas in the problem about a dog and its owner it says that the hill slope is 720 m . Please post a detailed solution.

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Re: Quick Handout for Time, Speed and Distance problems   [#permalink] 20 Oct 2017, 04:42
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