SBI Clerk Time and Distance Questions

SBI Clerk Time and Distance Questions

The topic of Time and Distance is one of the topics asked in the Quantitative Section of the Prelims and Mains Exam of SBI. The SBI Clerk Recruitment 2020 is for a total post of Junior Assistant with 8000 vacancies. Tips to solve SBI Clerk Time and Distance Questions for Prelim Exam 2020 can be checked down the post.

SBI Clerk Time and Distance Questions

As the Examination is near, candidates are eager to know more about the topics expected to come in the upcoming SBI Clerk Prelim Exam. Time and distance Questions are asked very Frequently in the SBI Clerk Prelim Examination. Questions from Time and Distance can be tricky at times unless the whole concept is clear in your head.

To make the concept pearl clear, we bring for you the detailed concept of Average Speed, Relative Speed, and Train and Platforms. All the probable questions are covered in the Example of various concepts for SBI Clerk Time and distance Questions by our experts.

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Concept 1 – Average  Speed, Time and distance Question SBI Clerk

Average Speed Formula:

                         Average speed =   Total Distance covered/ Total Time Taken 

Case  1:

 Average speed = 2xy/x+y      (When the distance is constant)

Where x and y are the two speeds at which the same distance has been covered.

For n values    If the distance covered is constant (d1 = d2 = d3 = dn) in each part of the journey, then the average speed is the Harmonic Mean of the values.

               Speed_Avg   =  n  / (1/s1 + 1/s2 + 1/s3 … + 1/sn)

Example   A person goes from A to B at the speed of 40 kmph and comes back at the speed of 60  kmph. What is his average speed for the whole journey?

Solution:  Since the distance traveled on both sides is the same, we can use the formula of the harmonic mean of speeds

Average Speed: 2xy/(x+y) where x is the speed while going from A to B and y is the speed while coming back.

Average  Speed =  2*60*40 / (60+40) = 2*6*4 = 48

So using this formula, we get the answer as  48 kmph.

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Case  2:

 Average speed = (x + y)/2  (When time taken is constant )

Where, x and y are the two speeds at which we traveled for the same time.

For  n cases   If the time taken is constant (t1 = t2 = t3 = tn) in each part of the journey then the average speed is Arithmetic Mean of the values.

                 Speed_Avg = (s1 + s2 + s3 … +sn ) /  n

Example –    Myra drove at an average speed of 30 miles per hour for T hours and then at an average speed of 60 miles/hr for the next T hours. If she made no stops during the trip and reached her destination in 2T hours, what was her average speed in miles per hour for the entire trip?

Solution: Here, the time for which Myra traveled at the two speeds is the same.

               Average Speed = (a + b)/2 = (30 + 60)/2 = 45 miles per hour

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Concept  2 – Relative  Speed, Time and distance Question SBI Clerk

Relative speed is defined as the speed of a moving object with respect to another. When two objects are moving in the same direction, the relative speed is calculated as their difference. When the two objects are moving in opposite directions, the relative speed is computed by adding the two speeds.

Case 1: Relative Speed of 2 bodies = Sum of their individual Speeds if they are moving in   the opposite direction

Example – Jethalal starts from Pune to Mumbai and at the same time Roshansingh Sodhi starts from Mumbai to Pune. Their speed is 25 kmph and 35 kmph respectively.   If the distance between Pune and Mumbai is 120 km.

  1. When will these two gentlemen meet?
  2. The place where they meet has a hotel. How far is this hotel from Pune?

Solution – Since they’re moving in opposite directions,
   
(Relative Speed)Jetha and Sodhi
=    Jetha’s speed + Sodhi’s Speed
=    25+35
=    60kmph

Since they start together, the speed is 60kmph from the beginning.

Speed x time = distance
60 x time =120
Time = 120/60 =2 hrs.

Ans1:     They’ll meet after two hours.

From the answer1:  We know that Jetha and Sodhi will meet after two hours.
    Since Jetha is riding from Pune side, in two hours he will cover

And  D = S*T

       D =  25*2 = 50 km

Ans2:  50 km

Case 2:

Relative Speed of 2 bodies = Difference of their individual Speeds if they are moving in the same direction.

Example –   Two athletes are running from the same place at the speed of 6  km/hr and 4 km/hr. Find the distance between them after 10 minutes if they move in the same direction.

Solution:            

When they move in same direction,

Their relative speed = (6 – 4) km/hr = 2 km/hr

Time taken = 10 minutes

Distance covered = speed × time

                          = (2 × 10/60) km

                          = 1/3 km

                          = 1/3 × 1000 m

                          = 333.3 m

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Concept  3 – Trains and Platforms, SBI Clerk Time and Distance Questions

Important Points

  • When two trains are going in the same direction, then their relative speed is the difference between the two speeds.
  • When two trains are moving in the opposite direction, then their relative speed is the sum of the two speeds.
  • When a train crosses a stationary man/ pole/ lamp post/ signpost- in all these cases, the object in which the train crosses are stationary and the distance traveled is the length of the train.
  • When it crosses a platform/ bridge- in these cases, the object in which the train crosses is stationary and the distance traveled is a sum of the length of the train and the length of the object.
  • When two trains cross each other the total distance is the sum of the length of both the trains.
  • When a train crosses a car/ bicycle/ a mobile man- in these cases, the relative speed between the train and the object is taken depending upon the direction of the movement of the other object relative to the train- and the distance traveled is the length of the train.

Examples

  1.  A train traveling at 60 kmph crosses a man in 6 seconds. What is the length of the train?

Solution:   Speed in m/sec = 60 (5/18) = 50/3     m/sec
                      Time is taken to cross the man = 6 secs
                     Therefore, Distance = (50/3)
6 = 100 metres (i.e. the length of the train)

2)     A train traveling at 60 kmph crosses another train traveling in the same direction at 50 kmph in  30 seconds. What is the combined length of both the trains?

Solution :         Speed of train A = 60 kmph = 60 (5/18) = 50/3          m/sec
                              Speed of train B = = 50 kmph = 50
(5/18) = 125/9     m/sec


The relative speed =(50/3)-(125/9)= 25/9    m/s (we have subtracted the two values because both            the trains are going in the same direction)


Time is taken by train A to cross-train B = 30 secs
Distance = Speed
Time


Distance =25/9
30 = 250/3 metres   (i.e. the combined length of both trains)

3)    How long will a 150 m long train running at a speed of 60 kmph take to cross a bridge of 300 m?

Solution:      Total Distance = 300 + 150 = 450 m

Speed = 60 kmph = 60 (5/18)=(50/3)    m/sec

Distance = Speed Time

450 =(50/3) Time

Time = 27 seconds

Concept  4)   Boats and Streams

Important Terms:

Still Water: If the speed of the water is zero, i.e. water is stationary, then it is called still water.

Stream: The moving water in the river is known as a stream.

Upstream: If a boat or a swimmer moves in the opposite direction of the stream, then it is called upstream.

Downstream: If a boat or a swimmer moves in the same direction of the stream, then it is called downstream.

When the speed of a boat or a swimmer is given, it usually means speed in still water.

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Formulae:

Suppose the speed of the boat in still water is u km/hr and the speed of the stream is v km/hr.

Then:

Upstream Speed:  It is the speed of the boat against the stream = (u – v) km/hr

Downstream Speed:  It is the speed of the boat with the stream = (u + v) km/hr

If upstream is denoted as US and downstream as DS, then:

Speed of boat or swimmer in still water = 1/2 (Ds + Us)
Speed of stream = 1/2 (D
s – Us)

Examples

1. A man rows downstream 32 km and 14km upstream. If he takes 6 hours to cover each  distance, then the velocity (in kmph) of the current is:

Solution:   Rate downstream=(32/6)kmph; Rate upstream=(14/6)kmph


                  Velocity of current=1/2(32/6-14/6)kmph=3/2kmph=1.5kmph

2. In one hour, a boat goes 11km along the stream and 5km against the stream. What is the speed of the boat in still water (in km/hr)?

  Solution:        Speed in still water=1/2(11+5) kmph  =  8kmph

3. A man can row 50 km upstream and 72 km downstream in 9 hours. He can also row 70 km upstream and 90          km downstream in 12 hours. Find the rate of current.

  Solution:   Let x and y be the upstream and downstream speed respectively.

Hence, 50/x + 72/y = 9 and 70/x + 90/y = 12
Solving for x and y we get x = 10 km/hr and y = 18 km/hr

We know that Speed of the stream = 1/2 * (downstream speed – upstream speed)
= 1/2 (18 – 10) = 4 km/hr.


4)     A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

Solution: Let the man’s rate upstream be x kmph and that downstream be y kmph.

Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.

                    X *  8(4/5)  =  4* y

                     44 x / 5  = 4 y

                     y = 11/5 x

Ratio =  (y + x ) / 2  :  ( y- x) / 2

           =   16 x / 5   :  6 x / 5

          =   8:3   Ans

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