Saturday, October 8, 2011

aptitude tricks 1

hi friends, i want to share some of the tricks on how to solve aptitude questions quickly.there are few points to remember:
1)There are no tricks which are unanimous.
2)You need to practice as many questions as you can and than have to make tricks which are suitable to you accordingly as mind level and level of preparation along with interests does vary from person to person so unanimous tricks won't do ...

3)The tricks also varies from questions to questions , so it won't be possible to provide you tricks for each and every type but you can post some type of questions over here and i will try to elaborate tricks to you as much as possible as i have also given such tests ...
4)You need to practice as much tests as possible before your exam and that will be suffice 

let us today discuss about train problems:
first let us solve a basic question.
1.  A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
A. 120 metresB. 180 metres
C. 324 metresD. 150 metres
this is the simplest question you can get in trains problem..read the question and you now know that speed and time is given and you need to find out length i.e. distance.we know the formula for distance which is speed *time. the only problem is that speed is given in km/hr and time in second.this is one of the common mistake we usually do.we should solve only when units are in same system.now we can either convert speed into m/sec or time into hour.but converting speed is more easy coz we need to multiply only by 5/18.
if you don't know how this 5/18 came then this is the explanation:
1 km=1000metre
1hr=60*60 seconds=3600 seconds
so, km/hr=1000/3600m/sec=5/18m/sec
 when we convert from km/hr to m/sec we multiply by 5/18 and when we convert from m/sec to km/hr we multiply by 18/5
now the solution
Speed= 60 x 5 m/sec = 50 m/sec.
18 3
Length of the train = (Speed x Time) = 50 x 9 m = 150 m.
3
NOW, if instead of stationary object like a pole,if there is some moving object like man,car etc.
2.  A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
A. 45 km/hrB. 50 km/hr
C. 54 km/hrD. 55 km/hr










in this question train's length and time to cross the man is given.this problem would have been same like the




above problem if the man would have been stationary. before we solve the question let us understand the concept of same direction and opposite direction.let us take an example, the distance between A and B is 100 km.
 CASE 1: a car from A starts moving at speed of 50 km/hr towards B and the car at B starts moving at speed 30 km/hr in same direction as of car from A was moving i.e. opposite to A.now the time required for the two trains to meet will be 100/(50-30)=5hr coz in the same direction the relative speed is the difference of two speed.
 CASE 2:if the car from B was moving toward A then the time required would be 100/(50+30)=1.125hr coz in opposite direction the relative speed is the sum of two speed.
NOW the solution:

Speed of the train relative to man = 125 m/sec
10

   = 25 m/sec.
2


   = 25 x 18 km/hr
2 5

   = 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45         x = 50 km/hr.
NOW LET ME GIVE YOU SOME IMPORTANT FORMULAS:
  1. Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.
  2. Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.
  3. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u - v) m/s.
  4. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
  5. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:
          The time taken by the trains to cross each other =(a+b)/(u+v)sec
      6.If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:
           The time taken by the faster train to cross the slower train =(a+b)/(u-v)sec
       7.If two trains (or bodies) start at the same time from points A and B towards each other and after     
          crossing they take a and b sec in reaching B and A respectively, then
          (A's speed) : (B's speed) =square root of a: square root of b


     
        now will discuss further topics in another blog:)























                               


1 comment: