DOWNLOAD QUANTITATIVE APTITUDE TRICKS OF TIME-SPEED AND DISTANCE

posted by Unknown on 05:59 No Comments


Today, I am going to share with you a very simple but helpful technique which can be easily applied in questions of time-speed-distance provided the distance remains constant and speed or time follow a definite progression.Examiners make use of this technique to frame questions

They can be either in Arithmetic progression(A.P) or Harmonic progression(H.P).In simple terms ,it can be stated as

Given that the distance between two points say A and B remains constant then:

1. If the velocities are in A.P ,then the time taken will be in H.P respectively.
2. If the time taken are in H.P ,then the velocities will be in H.P respectively.

Now let us take few examples to clarify this:

ALSO READ :(Recommended Post)



The following questions are taken from real exam papers:

If a man cycles at 10 km/hr ,the he arrives at a certain place at 1 p.m.If cycles at 15km/hr,he will arrive at the same place at 11 a.m.At what speed must he cycle to  get there at noon?

 Sol: 
-We can see that the time given : 11 a.m, 12:00 (noon) and 1 p.m are in A.P. with a difference of 1 hour.

-Therefore according to statement 1:. If the velocities are in A.P ,then the time taken will be in H.P respectively.

-Velocities will be in H.P

velocity at 12 :00 will be


V=2*10*15/(10+15)

V=12km/hr.


Let take one more example ,this time stream problem:

ALSO READ :

Question:Priyanka can row  a certain distance upstream in 84 minutes.She can row the same distance downstream in 9 minutes less then she can row in still water.How long will priyanka take to row down with the stream?

Sol:The distance is constant.

We will take our usual anotations :speed of stream : u
 speed of priyanka(boat): v
Upstream speed :v-u
Still water speed:v
Downstream speed: v+u

-So the velocities are in A.P with a common difference of 'u'.
Therefore according to statement 1: If the velocities are in A.P ,then the time taken will be in H.P respectively.

 -Let the time taken in still water be t min.
Therefore in downstream it will be t-9 min.

Time in still water t=2*84*(t-9)/(84+t-9) min
t= 72 min.

Downstream time = t-9=72-9
                              =63 min.
Similarly there questions on Train and Escalators.

For more such tricks and shorcut download the below pdf: 

  DOWNLOAD PDF :   


IF LINK BUSY: CLICK HERE

0 comments:

Post a Comment