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Time, Speed & Distance # 2

by Gaurav Jain

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AVERAGE SPEED


Average Speed = \frac{Total Distance Traveled}{Total Time Taken}


Means,If an object travels D1, D2, D3, …… Dn distances, with speeds S1, S2, S3, …… Sn in T1,T2,T3 …… Tn times respectively, then Average Speed

= \frac{D1 + D2 + D3 +....Dn}{T1 + T2 + T3 +....Tn}





= \frac{S1T1 + S2T2 + S3T3 + .... SnTn}{T1 + T2 + T3 + ....Tn}


= \frac{D1 + D2 + D3 +....Dn}{\frac{D1}{S1} + \frac {D2}{S2} + \frac{D3}{S3} + ..... \frac{Dn}{Sn}}

SOME IMPORTANT FORMULAE - from Test's point of view



1)
If an object travels a distance D1 and D2 at a speed of S1 and S2 Miles per Hour, respectively, in time T1 and T2 then the total time taken 'T' is given by:



T = T1 + T2 = \frac{D1}{S1} + \frac{D2}{S2}


The total distance covered 'D' is given by:


D = D1 + D2 = S1T1 + S2T2


2)
If a certain distance D, from point X to point Y is traveled at 'a' km/hr and the same distance is covered i.e. from Y to X at 'b' km/hr, then the average speed during the whole journey can be calculated as:



Average Speed = [\frac{2ab}{a + b}]\frac{km}{hr}


Also, if T1 and T2 are the times taken to travel from X to Y and from Y to X respectively, the distance D from X to Y can be calculated as:


D = (T1 + T2)[\frac{ab}{a + b}]km
D = (T1 - T2)[\frac{ab}{a - b}]km
D = (a - b)[\frac{T1*T2}{T1 - T2}]km

3)
While a certain distance 'D' is covered, if A man changes his speed in the ratio X : Y, then the ratio of time taken becomes Y : X.


4)
If an object travels a distance D from X to Y with Speed 'a' in time T1 and travels back from Y to X with \frac{m}{n} of the usual speed 'a', then the change in time taken to travel the same distance can be calculated as:


Change In Time = (\frac{n}{m} - 1)*T1 ; for n > m

Change In Time = (1 - \frac{n}{m})*T1 ; for m > n

*5)
IF two objects, say A and B, start at the same time in opposite directions from two different points (P and Q) and arrive at the opposite points in 'a' and 'b' hours respectively after having met.


Then the ratio of their speed is:


\frac{As Speed}{B = \frac{\sqrt{b}}{\sqrt{a}}

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10 Comments
    suyash2804
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    suyash2804 – Wed, 08 Feb 2012 08:04:32 -0000

    having problems in solving advanced questions of time,spped and distance……………..can you please e-mail me some advanced problems with solution…………

    Regards,
    Suyash Mishra

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    suyash2804
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    suyash2804 – Wed, 08 Feb 2012 08:02:47 -0000

    having problems in solving advanced questions of time,spped and distance……………..can you please e-mail me some advanced problems with solution…………

    Regards,
    Suyash Mishra

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    PareshSinha
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    PareshSinha – Thu, 01 Dec 2011 07:22:03 -0000

    Nice article…Thanks Gaurav…few of my doubts got cleared.

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    Dips88
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    Dips88 – Mon, 29 Mar 2010 06:48:19 -0000

    good article

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    chandan0rules1
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    Chandan Mallik – Sun, 27 Sep 2009 19:05:24 -0000

    sir,
    i'm new to this online community……

    Please help me prepare an ideal schedule for CAT 2010..

    Thanks..

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    deepa mittal
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    Deepa Mittal – Thu, 29 Oct 2009 10:48:52 -0000

    CAT community contains lot of lessons and tests to help you with the prepartion. Make sure that you are going through this community regularly. Please go through the book s:

    How to prepare for CAT Quant by Arun Sharma.

    How to prepare for CAT DI

    These books are really good and will be helpful to you.

    Besides CAT, you must also consider other exams like XAT, IIFT, FMS and SNAP to get into the next level top B-Schools.

    In case you have any query, you can get in touch with Learnhub's Expert Team

    Go through the following links:

    CAT Online 2007 Exam

    CAT Online 2008 Exam

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    Murali_Santhi
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    Murali – Mon, 22 Dec 2008 07:02:43 -0000

    Nice Job Gaurav!!!! Thanks a lot…..

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    shampy88
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    shampy88 – Fri, 28 Nov 2008 07:39:04 -0000

    please tell in more detailed abt usual speed concept

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    Sureshbala
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    Suresh – Fri, 30 Jan 2009 12:31:29 -0000

    Dear shampy,

    Please Click here to go through a lesson that discusses the basic concepts of Speed and Distance

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    gauravjain26
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    Gaurav Jain – Sun, 28 Sep 2008 19:02:49 -0000

    Nice shortcuts


    I don't think discussing shortcuts is a right thing when basics are discussed. for that separate lesson is there.

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    Sureshbala
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    Suresh – Mon, 29 Sep 2008 05:44:47 -0000
    I guess that these are not shortcuts, instead we are trying to use the concepts explained by you in the lesson 1. (Relationship between SPEED, TIME and DISTANCE). What I personally think is that this is how our teachers should have taught us Time and Distance. Unfortunately, this is not the case with most of the schools and we ended up framing equations, applying formulae and finally landed in a situation where applying a concept is somehow treated as a shortcut. So I think, instead of teaching by using some formulae (which most of the people does) we should straight away start teaching by applying the concepts (so that a student can imagine the same problem in real life situation) and this is how we can make a huge difference in the process of learning with the aid of LEARNHUB.

    Dear Gaurav, I am sure you are a person of great calibre and definitely a person like you can make this huge difference in the art of teaching. I hope you will understand me.

    Regards

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    Sureshbala
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    Suresh – Sun, 28 Sep 2008 06:12:49 -0000

    Let's look at question 3….

    Mr. X traveled 120 km by vehicle A, 450 km by vehicle B and 60 km by vehicle c. The whole journey took 13 hours and 30 minutes. Speed of vehicle B is 3 times the speed of vehicle C and 1.5 times the speed of vehicle A. Find the speed of vehicle B.

    Given that Speeds ratio of A B and C is 2:3:1

    Given that Distance ratio of A B and C is 4:15:2

    Hence the Time ratio of A B and C will be \frac {4}{2} : \frac {15}{3} : \frac {2}{1} = 2:5:2

    Since it is given that total time taken is 13.5 hrs i.e 27/2 hrs, time taken by B is 5/9×27/2 = 15/2 hrs.

    Hence B covered a distance of 450 km in 15/2 hours.

    So Speed of B is 60km/hr

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    Sureshbala
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    Suresh – Sun, 28 Sep 2008 06:01:43 -0000

    I just want to discuss some of these questions……
    The basic point that a CAT aspirant has to keep in mind is that "No formula Approach"
    Speed = \frac {Distance}{Time} is the only formula that we are going to use and of course the relationship betwenn Speed, Distance and Time.

    Coming to the second question discussed above…….

    Walking at \frac {3}{4}th of the speed , Mr.X is 2 hrs late. Find his usual time.

    Since distance is constant here and we know that SPEED and TIME are inversely proportional let's make use of that statement.

    Since speed has become 3/4, the time becomes 4/3 i.e he will be late by 1/3 of this normal time which is given as 2 hours.

    So 1/3 of his normal time is 2 hrs .
    Hence his normal time is 6 hrs.

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gauravjain26
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About: This is GJ. I can describe myself as a cool, friendly and fun loving guy with a predetermined goal to fulfill my objectives. I believe hard work and Sincerity, backed by good character, discipline and self-confidence, is must to solve a problem in its best solution. I consider myself a voracious learner, no matter from whom I am learning..
Last Updated At Dec 13, 2012
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